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On partial derivatives of Bernstein–Stancu polynomials for multivariate functions A. Yu. Veretennikovab, N. M. Mazutskiyc a Institute for Information Transmission Problems of the Russian Academy of Sciences (Kharkevich Institute), Moscow, Russia b Nikol'skii Mathematical Institute, Peoples' Friendship University of Russia, Moscow, Russia c Faculty of Mechanics and Mathematics, Lomonosov Moscow State University, Moscow, Russia
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    § 1. IntroductionFor a new (at the time) proof of Weierstrass’s theorem on polynomial approximation by use of Bernoulli’s law of large numbers, Bernstein [1] proposed to use the polynomials
$$
\begin{equation*}
{B}_{n}(f,x)=\sum_{k=0}^{n} f\biggl(\frac{k}{n}\biggr) \binom{n}{k} x^{k} {(1-x)}^{n-k}, \qquad 0\leqslant x\leqslant 1,
\end{equation*}
\notag
$$
subsequently named after him. The artificial (at first sight) interpretation of the variable $x$ as a probability and an application of the law of large numbers produced the convergence theorem
$$
\begin{equation*}
\sup_{| x | \leqslant 1}| B_n(f,x)-f(x)| \to 0, \qquad n\to\infty,
\end{equation*}
\notag
$$
for any function $f\in C[0,1]$. This idea was further developed in different directions as follows:
There are several modifications of Bernstein polynomials (see, in particular, [2]–[9]). For various aspects of the theory of Bernstein polynomials, see the books by Lorentz [2] and Videnskii [5], and also Bustamante’s book [10] and [11]. In [12] it was shown that natural analogues of Bernstein polynomials can be used for simultaneous approximation of multivariate functions and their derivatives of certain order or index on cubes and simplexes under the assumption that the derivatives of this order or index are continuous (that is, not necessarily all derivatives of, say, the second order, but only the ones that exist). In our paper, based on a certain modification of Bernstein–Stancu polynomials (see [3]), we examine the problem of approximation of the second-order mixed derivative in the integral norm without making the assumption that this derivative is continuous. Results of this type can be useful in applications if the derivative in question is not continuous or if its continuity is not known a priori. Note that precisely second-order derivatives (along with first-order ones) and their approximations are important, in particular, in stochastic analysis. For the completeness of presentation we also formulate and prove an auxiliary result on the approximation of first-order partial derivatives in the same norm. § 2. On previous resultsBefore formulating the main result of this paper, we recall Bernstein’s fundamental theorem on the approximation of a continuous function on $[0, 1]$ and some results on the approximation of derivatives, apparently obtained independently by Chlodovsky and Kantorovich. Next we recall Kantorovich’s theorem on the approximation of a derivative almost everywhere and state Proposition 4 (obtained not so long ago). All these results, except the last one, pertain to the approximation of univariate functions on $[0,1]$, whereas Proposition 4 (see below) is concerned with the approximation of a bivariate function (and its derivatives); however, this can help the reader get up to speed quickly. The result on the approximation of derivatives (Proposition 2) was announced in Chlodovsky’s and Kantorovich’s talks at the First All-Union Congress of Mathematicians.1 However, this result has never been fully published by any of them.2 To these authors’ knowledge, the first published proof of Proposition 2 (with a mention of the names of Chlodovsky and Kantorovich) belongs to Lorentz3 [13]. Below results taken from other papers are formulated as propositions. For readily accessible results, we refer the reader to [13], Theorem 1, [2], Theorem 2.1.1, and also4 to [5], Theorem 15.1, and [5], Theorems 10.1 and 10.2. Proposition 1 (Bernstein [1]). If $f$ is a continuous function on $[0,1]$, then the sequence of polynomials $B_n(f,x)$ converges uniformly to $f(x)$ on $[0,1]$ as $n \to \infty$. Proposition 2 (Lorentz [13], Theorem 1). If the $r$th derivative $f^{(r)}(x_0)$ of a bounded function $f$ exists at a point $x_0$, then In [9], to study the approximation properties of mere Borel measurable functions, rather than regular ones, it was proposed to modify the polynomials by replacing the integrable function $f(x)$ on $[0,1]$ by So the derivative of a polynomial defined for $F(x)$ assumes the form
$$
\begin{equation*}
\begin{aligned} \, P_n(f,x) &:=\frac{d}{dx}B_{n+1}(F,x) \\ &\,=\sum_{k=0}^{n}\binom{n}{k}x^{k} {(1-x)}^{n-k}(n+1)\int_{k/(n+1)}^{(k+1)/(n+1)}f(t)\,dt. \end{aligned}
\end{equation*}
\notag
$$
Proposition 3 (Kantorovich [9]). The convergence holds at each point $x \in (0,1)$ where $f(x)$ is equal to the derivative of the integral $ \displaystyle F(x)=\int_{0}^{x}f(t)\,dt$ (that is, almost everywhere). For an exposition of this result also see Theorem 2.1.1 in [2] with a reference to Kantorovich. It is also worth pointing out that [16] deals with the convergence of Bernstein polynomials for Riemann integrable functions at points of discontinuity; however, this problem is not what we are interested in here. The one-dimensional analogues of the modified multivariate polynomials $\widetilde B_n$ (see (3.1) below) would be closer in spirit to Kantorovich’s modification mentioned above, however, the type of convergence in the main theorem that follows is slightly different. In addition, it is worth recalling that in the multivariate case there is no analogue of the one-dimensional Lebesgue differentiation theorem, and so the result of [9] cannot be automatically carried over to the multivariate setting. We also note that in our paper we consider the problem of the approximation of the second-order mixed derivative, which is a different task. For a function of two variables one of the two classical variants of Bernstein polynomials is as follows:
$$
\begin{equation*}
{B}_{n}(f,x_1,x_2)=\sum_{k_1, k_2=0}^{n} f \biggl(\frac{k_1}{n},\frac{k_2}{n}\biggr) \binom{n}{k_1} \binom{n}{k_2} x_{1}^{k_1} (1-x_1)^{n-k_1} x_{2}^{k_2}(1-x_2)^{n-k_2},
\end{equation*}
\notag
$$
where $0\leqslant x_1, x_2 \leqslant 1$, and similarly, it can be defined on the $d$-dimensional cube $0\leqslant x_1, \dots, x_d \leqslant 1$. In what follows we use this form. (In the second classical variant, given a polynomial, a function is approximated on a simplex, rather than on a square/cube; of course, both variants are the same in the one-dimensional case.) In [12], given a function $f\in C^m(\mathbb{R}^d)$, all the derivatives of order $m$ of multivariate Bernstein polynomials were shown to converge uniformly on the $d$-dimensional cube to the corresponding derivatives of $f$, namely, $ {B}_{n}^{(m)}(f,x) \rightrightarrows f^{(m)}(x)$ as $n \to \infty$ (see Theorem 4 in [12]); a similar result will also be proved for multi-indices $\mathbf k=(k_1,\dots,k_d)$, namely, ${B}_{n}^{(\mathbf k)}(f,x) \rightrightarrows f^{(\mathbf k)}(x)$ as $n \to \infty$ under the assumption that only the derivative corresponding to this multi-index exists and is continuous (without assuming that this also holds for all derivatives of this order). That this fact is of independent interest is demonstrated by Tolstov’s theorem (see Proposition 12 below), which does not assert convergence, but rather the equality of mixed derivatives in each order (however, it is required in its statement that all derivatives of order 2 exist). The same assumption on all derivatives of some order is typical for some other versions of this result. Note, however, that this assumption is not imposed in Proposition 4 or the theorem. Proposition 4 (see [12], Theorem 4). (1) If $m>0$ and $f \in C^{m}({\mathbb{R}}^d)$, then where $K^{d}$ is the $d$-dimensional cube. (2) If ${\mathbf k}=(k_1,\dots,k_d)$ is a multi-index and $f \in C^{\mathbf k}(\mathbb{R}^d)$, that is, the derivative corresponding to the multi-index ${\mathbf k}$ is defined and continuous on $\mathbb R^d$ (it is not assumed that all derivatives of order $|{\mathbf k}|$ exist), then where $K^{d}$ is the $d$-dimensional cube.In comparison to this result, in the main theorem that follows we succeeded in eliminating the assumption that the mixed second-order derivative of $f$ are continuous on $[0,1]^2:=K \, (\equiv K^2)$, and, under the condition $\dfrac{\partial^2 f}{\partial x_1 \,\partial x_2}\in L_1(K)$ and some others, for a modified Bernstein–Stancu polynomial depending on some random vector the second-order mixed derivative of this polynomial with respect to the variables $x_1$ and $ x_2$ is shown to converge to the second-order mixed derivative of the function $f$ in the $L_1$-norm with respect to $(x_1,x_2)\times\omega$ relative to the direct product of the two-dimensional Lebesgue measure on $K$ and a probability measure $\mathsf P$ on a measure space $(\Omega, \mathcal F, \mathsf P)$ (for the precise definitions, see § 3). Here $(x_1,x_2) \in K$ and $\omega \in \Omega$. It should be noted that this convergence is slightly different from the one in the above result from [12]. Proposition 4 is required below, in the proof of the main theorem of § 5; we formulated this result above for easier references and the convenience of the reader. In what follows we consider only bivariate functions, because we are concerned here with second-order mixed derivatives, to which consideration of a larger number of variables is not relevant. § 3. The main resultTo formulate the main result we consider two independent random variables5 $\alpha_i \in U[0,1]$, $i=1,2$, on a probability space $(\Omega, \mathcal F, \mathsf P)$. We also set $\alpha= (\alpha_1, \alpha_2)$. The Bernstein–Stancu type polynomials, which we study in our paper, have the form
$$
\begin{equation}
\begin{aligned} \, \notag \widetilde {B}_{\alpha,n}(f,x_1,x_2) &=\sum_{k_1, k_2=0}^{n} f\biggl(\frac{k_1+\alpha_1}{n},\frac{k_2+\alpha_2}{n}\biggr) \\ &\qquad \times \binom{n}{k_1} \binom{n}{k_2} x_{1}^{k_1} (1-x_1)^{n-k_1} x_{2}^{k_2} (1-x_2)^{n-k_2}. \end{aligned}
\end{equation}
\tag{3.1}
$$
We also extend $f(x_1,x_2)$ by zero outside $K=[0,1]^2$. So polynomials are defined on the square, because we are interested in the two-dimensional version of these polynomials; of course, they can also similarly be defined on the $d$-dimensional cube. As noted above, similar polynomials for nonrandom parameters $\alpha_i$ were proposed by Stancu [3]. Theorem. Let the Borel function $f(x_1,x_2)$ be bounded on $[0,1]^2=K$. Assume that the (classical) derivatives $\dfrac{\partial{f(x_1,x_2)}}{\partial{x_1}}$ and $ \dfrac{\partial{f(x_1,x_2)}}{\partial{x_2}} $ exist and lie in $L_1(K)$ and the second-order mixed derivative exists and lies in $L_1(K)$. Then Corollary. If, under the hypotheses of the theorem, the mixed derivative $f_{x_2x_1}$ exists and lies in $L_1(K)$, then convergence as in (3.2) holds and for almost all $(x_1,x_2)$, This result is immediate from (3.2) because
$$
\begin{equation*}
\dfrac{\partial^2 \widetilde{B}_{\alpha,n}(f,x_1,x_2)}{\partial x_1 \,\partial x_2}= \dfrac{\partial^2 \widetilde{B}_{\alpha,n}(f,x_1,x_2)}{\partial x_2\, \partial x_1 }.
\end{equation*}
\notag
$$
This equality is verified, for example, by an appeal to any of Propositions 5–8 (see § 6 below).
§ 4. Auxiliary resultsAs far as we know, the following lemma is well known in the theory of integration. However, we were unable to find an accurate reference, so following an advice of some colleagues, we present it with a proof, of course, without claiming the authorship. Lemma 1. 1. Let $g\in L_1([0,1])$, and let $g$ be extended by zero outside $[0,1]$. Then 2. Let $g\in L_1([0,1]^2)$, and let $g$ be extended by zero outside the square $K=[0,1]^2$. Then Proof. 1. Let $\delta>0$, and let there exist $h\in C([0,1])$ such that Then Here the last integral tends to zero as $t\to 0$ since $h(x+t)$ converges pointwise to $h(x)$ and since the function $h$ is bounded by the Lebesgue dominated convergence theorem.
2. Again, let $\delta>0$, and let $h\in C(K)$ be such that Note that in this case both $g$ and $h$ are defined on $K$. Then we have
$$
\begin{equation*}
\begin{aligned} \, &\iint_K |g(x_1+t_1, x_2+t_2)-g(x_1,x_2)|\,dx_1\,dx_2 \\ &\qquad \leqslant \iint_K |g(x_1+t_1, x_2+t_2)-h(x_1+t_1,x_2+t_2)|\,dx_1\,dx_2 \\ &\qquad\qquad +\iint_K |g(x_1, x_2)-h(x_1,x_2)|\,dx_1\,dx_2 \\ &\qquad\qquad +\iint_K |h(x_1+t_1, x_2+t_2)-h(x_1,x_2)|\,dx_1\,dx_2 \\ &\qquad \leqslant 2\delta+\iint_K |h(x_1+t_1, x_2+t_2)-h(x_1,x_2)|\,dx_1\,dx_2. \end{aligned}
\end{equation*}
\notag
$$
As in the one-dimensional case, since $h$ is (uniformly) continuous and bounded on $K$ the last integral tends to zero as $t_1,t_2\to 0$ by the Lebesgue dominated convergence theorem.
This proves Lemma 1. In what follows we deal with functions given by rather complicated expressions (of course, the corresponding derivatives are even more complicated), and so it is convenient to use the shorthand notation
$$
\begin{equation*}
\begin{gathered} \, \Delta_{(x_1)}f(x_1,x_2):=f\biggl(x_1+\frac{1}{n}, x_{2}\biggr)-f(x_1, x_2), \\ \Delta_{(x_2)}f(x_1,x_2):=f\biggl(x_1, x_{2}+\frac{1}{n}\biggr)-f(x_1, x_2), \\ f_{x_i}:=\frac{\partial{f(x_1,x_2)}}{\partial{x_i}}\quad\text{and} \quad f_{x_i,x_j}:=\frac{\partial^2{f(x_1,x_2)}}{\partial{x_i}\,\partial{x_j}}= \frac{\partial{}}{\partial{x_i}} \biggl(\frac{\partial{f(x_1,x_2)}}{\partial{x_j}} \biggr), \quad i,j=1, 2, \end{gathered}
\end{equation*}
\notag
$$
which reduces significantly the amount of mathematics, even though the procedure is still quite long. Here the subscript $x_i$ in $\Delta_{(x_i)}$ means only that the difference is taken with respect to the variable $x_i$; note also that there may be no actual dependence on this variable, at least, if it is not indicated as an argument of the function to which the corresponding convolution operator is applied. Lemma 2. 1. Let $f$ be a bounded Borel function on $K$ and for all $x_1$ and $x_2$ let the classical first-order derivative $\dfrac{\partial f(x_1, x_2)}{\partial x_1} \in L_1(K)$ exist. Then The same result also holds for the derivative $f_{x_2}$ with respect to the other variable, provided that this derivative $\dfrac{\partial f(x_1, x_2)}{\partial x_2} $ exists and lies in $L_1(K)$. 2. For all $x_1$ and $x_2$ let the classical mixed derivative $ \dfrac{\partial^2 f(x_1, x_2)}{\partial x_1\, \partial x_2} $ exist and lie in $L_1(K)$ (in particular, the function $f_{x_2}(x_1, x_2)$ is bounded for each $x_2$). Then
$$
\begin{equation}
\int_{0}^{1}dx_1 \int_{0}^{1}dx_2\, \biggl| n \biggl(f_{x_2}\biggl(x_1+\frac{1}{n}, x_2\biggr)-f_{x_2}(x_1, x_2) \biggr) -\frac{\partial^2 f(x_1, x_2)}{\partial x_1\, \partial x_2}\biggr|\to 0, \qquad n \to \infty.
\end{equation}
\tag{4.3}
$$
In other words,
$$
\begin{equation*}
\int_{0}^{1}dx_1 \int_{0}^{1}dx_2\, \bigl|n\Delta_{(x_1)}f_{x_2}(x_1,x_2)-f_{x_1 x_2}(x_1,x_2)\bigr| \to 0, \qquad n \to \infty.
\end{equation*}
\notag
$$
In addition,
$$
\begin{equation}
\Delta_{(x_1)}f_{x_2}(x_1,x_2)=\frac{\partial }{\partial x_2}\Delta_{(x_1)}f(x_1,x_2)
\end{equation}
\tag{4.4}
$$
and
$$
\begin{equation}
\int_{0}^{1}dx_1 \int_{0}^{1}dx_2\, \bigl|n^2\Delta_{(x_2)}\Delta_{(x_1)}f(x_1,x_2)-n\Delta_{(x_1)}f_{x_2}(x_1,x_2)\bigr| \to 0, \qquad n \to \infty.
\end{equation}
\tag{4.5}
$$
Similarly, if for all $x_1$ and $ x_2$ the classical mixed derivative $\dfrac{\partial^2 f(x_1, x_2)}{\partial x_2\, \partial x_1} $ exists and lies in $L_1(K)$ (and, in particular, the function $f_{x_1}(x_1, x_2)$ is bounded for each $x_1$), then, as $n \to \infty$, and also Proof. We will repeatedly make use of auxiliary functions $g\in C(K)$ to approximate the function $f$, its derivatives, and also $F^\varepsilon$ for some integral differences. In general, these functions will be different for different $f$, $f_{x_1}$, $f_{x_2}$ and $f_{x_1x_2}$ and different differences.
1. We verify (4.1). To do this we approximate $f$ by smooth functions ${f^\varepsilon \mkern-2mu\!\in\! C^\infty\mkern-1mu(\mathbb R^2\mkern-1mu)}$. For definiteness we assume that $f^{\varepsilon}$ approximates $f$ by convolutions with the same nonnegative kernel $\varphi \in C_0^{\infty}(\mathbb R)$ (where $0$ means that the support in each of the two variables is compact), where $\displaystyle \int_{-\infty}^{\infty}\varphi(x)\,dx=1$ and $ \varphi^{\varepsilon}(x)=\dfrac{1}{\varepsilon}\varphi\biggl(\dfrac{x}{\varepsilon}\biggr)$. It is known that
$$
\begin{equation*}
f^{\varepsilon}(x_1, x_2)=\int dy_1\int dy_2\, \bigl[f(x_1-y_1, x_2-y_2) \varphi^{\varepsilon}(y_1)\varphi^{\varepsilon}(y_2)\bigr] \in C^{\infty}.
\end{equation*}
\notag
$$
In what follows products $\varphi^{\varepsilon}(y_1)\varphi^{\varepsilon}(y_2)$ are sometimes denoted briefly by
$$
\begin{equation*}
\varphi^{\varepsilon}(y_1)\varphi^{\varepsilon}(y_2)=: \overline \varphi^{\varepsilon}(y_1,y_2) , \qquad f^\varepsilon=f*\overline \varphi^\varepsilon.
\end{equation*}
\notag
$$
We claim that under the hypotheses of this assertion of the lemma the derivative with respect to $x_1$ commutes with integration, that is,
$$
\begin{equation}
\partial_{x_1}f^{\varepsilon}(x_1, x_2)\equiv f_{x_1}^{\varepsilon}(x_1, x_2)=\int dy_1\int dy_2\, \bigl[f_{x_1}(x_1-y_1, x_2-y_2) \varphi^{\varepsilon}(y_1)\varphi^{\varepsilon}(y_2)\bigr].
\end{equation}
\tag{4.8}
$$
To prove this it suffices to show that, for all $0\leqslant x_1 \leqslant x'_1 \leqslant 1$ and each $x_2$,
$$
\begin{equation*}
f^{\varepsilon}(x'_1, x_2)-f^{\varepsilon}(x_1, x_2) =\int_{x_1}^{x'_1} dz \int dy_1\int dy_2 \, \bigl[f_{z}(z-y_1, x_2-y_2) \varphi^{\varepsilon}(y_1)\varphi^{\varepsilon}(y_2)\bigr].
\end{equation*}
\notag
$$
Applying Fubini’s theorem and using the fundamental theorem of calculus, we have
$$
\begin{equation*}
\begin{aligned} \, &f^{\varepsilon}(x_1, x_2) +\int_{x_1}^{x'_1} dz \int dy_1\int dy_2 \, \bigl[f_{z}(z-y_1, x_2-y_2) \varphi^{\varepsilon}(y_1)\varphi^{\varepsilon}(y_2)\bigr] \\ &\qquad =\int dy_1\int dy_2 \, \bigl[f(x_1-y_1, x_2-y_2) \varphi^{\varepsilon}(y_1)\varphi^{\varepsilon}(y_2)\bigr] \\ &\qquad\qquad +\int dy_1\int dy_2 \int_{x_1}^{x'_1} dz \, \bigl[f_{z}(z-y_1, x_2-y_2) \varphi^{\varepsilon}(y_1)\varphi^{\varepsilon}(y_2)\bigr] \\ &\qquad =\iint \biggl[f(x_1-y_1, x_2-y_2)+\int_{x_1}^{x'_1}\, dz f_{z}(z-y_1, x_2-y_2) \biggr] \varphi^{\varepsilon}(y_1)\varphi^{\varepsilon}(y_2)\,dy_1\, dy_2 \\ &\qquad =\iint f(x'_1-y_1, x_2-y_2) \varphi^{\varepsilon}(y_1)\varphi^{\varepsilon}(y_2)\,dy_1 \, dy_2=f^{\varepsilon}(x'_1, x_2) \end{aligned}
\end{equation*}
\notag
$$
provided that all integrals here are absolutely convergent (which is the case by the hypotheses of the lemma). This proves the claim.
We next employ the auxiliary function $f^\varepsilon$ to estimate the upper limit as $n\to\infty$ of the (nonnegative) integral on the left in (4.1). This integral depends on $n$, but not on $\varepsilon$. We claim that this limit is majorized by some (also nonnegative) expression of $\varepsilon$, which, in turn, tends to zero as $\varepsilon \to 0$. In other words, in estimating auxiliary expressions depending on $n$ and $\varepsilon$ we will be concerned with repeated limits of the form $\lim_{\varepsilon\to 0} \limsup_{n\to\infty}$. (If the integrands were not nonnegative, then, of course, we would also have to deal with limits of the form $\lim_{\varepsilon\to 0} \liminf_{n\to\infty}$; however, this is not the case here.) We have
$$
\begin{equation*}
\begin{aligned} \, &\int_{0}^{1}dx_1 \int_{0}^{1}dx_2\, \biggl| n \biggl(f\biggl(x_1+\frac{1}{n}, x_2\biggr)-f(x_1, x_2) \biggr) -f_{x_1}(x_1, x_2)\biggr| \\ &\qquad \leqslant\int_{0}^{1}dx_1 \int_{0}^{1}dx_2\, \bigl|f^{\varepsilon}_{x_1}(x_1, x_2)-f_{x_1}(x_1, x_2)\bigr| \\ &\qquad\qquad +\int_{0}^{1}dx_1 \int_{0}^{1}dx_2\, \biggl|n \biggl(f^{\varepsilon}\biggl(x_1+\frac{1}{n}, x_2\biggr)-f^{\varepsilon}(x_1, x_2) \biggr)-f^{\varepsilon}_{x_1}(x_1, x_2)\biggr| \\ &\qquad\qquad +\int_{0}^{1}dx_1 \int_{0}^{1}dx_2 \, \biggl|n \biggl(f\biggl(x_1+\frac{1}{n}, x_2\biggr) -f(x_1, x_2) \biggr) \\ &\qquad\qquad\qquad -n \biggl(f^{\varepsilon}\biggl(x_1+\frac{1}{n}, x_2\biggr)-f^{\varepsilon}(x_1, x_2) \biggr)\biggr| \\ &\qquad=:J_{1}^{\varepsilon}+J_{2}^{n,\varepsilon}+J_{3}^{n,\varepsilon}. \end{aligned}
\end{equation*}
\notag
$$
We claim that
$$
\begin{equation}
J_{1}^{\varepsilon}=\int_{0}^{1}dx_1 \int_{0}^{1}dx_2\, \bigl|f^{\varepsilon}_{x_1}(x_1, x_2)-f_{x_1}(x_1,x_2)\bigr|\to 0, \qquad \varepsilon \to 0.
\end{equation}
\tag{4.9}
$$
In fact, the space $C(K)$ is dense in $L_1(K)$, and so for each $\delta>0$ there exists a function $g\in C(K)$ such that By the above we also have
$$
\begin{equation*}
\|f^\varepsilon_{x_1}-g^\varepsilon\|_{L_1(K)} \leqslant \delta,
\end{equation*}
\notag
$$
where $g^\varepsilon=g*\overline\varphi^\varepsilon $. Next, we have
$$
\begin{equation*}
\begin{aligned} \, &\int_{0}^{1}dx_1 \int_{0}^{1}dx_2 \, \bigl|f^{\varepsilon}_{x_1}(x_1, x_2) -f_{x_1}(x_1,x_2)\bigr| \\ &\qquad \leqslant \int_{0}^{1}dx_1 \int_{0}^{1}dx_2 \, \bigl |f^{\varepsilon}_{x_1} (x_1, x_2)- g^\varepsilon(x_1,x_2)\bigr| \\ &\qquad\qquad +\int_{0}^{1}dx_1 \int_{0}^{1}dx_2 \, \bigl|f_{x_1} (x_1, x_2)-g(x_1,x_2)\bigr| \\ &\qquad\qquad +\int_{0}^{1}dx_1 \int_{0}^{1}dx_2 \, \bigl|g^{\varepsilon} (x_1, x_2)-g(x_1,x_2)\bigr| \\ &\qquad \leqslant \int_{0}^{1}dx_1 \int_{0}^{1}dx_2\, \bigl|g^{\varepsilon}(x_1, x_2)-g(x_1,x_2)\bigr|+2\delta. \end{aligned}
\end{equation*}
\notag
$$
Here, since $g$ is uniformly continuous and $\varphi$ has a compact support, the integrand $g^{\varepsilon}(x_1, x_2)-g(x_1,x_2)$ on the right-hand side tends to zero as $\varepsilon\to 0$ for all ${(x_1,x_2)\in K}$. Now the required convergence (4.9) follows, since $\delta>0$ is arbitrary and the integral $J^\varepsilon_1$ is independent of $\delta$.
Consider the second integral. For fixed $\varepsilon$, as $ n \to \infty$, we have
$$
\begin{equation}
J_{2}^{n,\varepsilon}=\int_{0}^{1}dx_1 \int_{0}^{1}dx_2\, \biggl|n \biggl(f^{\varepsilon}\biggl( x_1+\frac{1}{n}, x_2\biggr)-f^{\varepsilon}(x_1, x_2) \biggr) - f^{\varepsilon}_{x_1}(x_1, x_2)\biggr| \to0,
\end{equation}
\tag{4.10}
$$
where we have used the definition of the derivative of the continuous function $f^{\varepsilon}$ and employed Lebesgue’s dominated convergence theorem; in addition, the finite differences $n \biggl(f^{\varepsilon}\biggl(x_1+\dfrac{1}{n}, x_2\biggr)-f^{\varepsilon}(x_1, x_2) \biggr)$ are bounded by the mean value theorem or by the fundamental theorem of calculus.
Consider $J_{3}^{n,\varepsilon}$. To curtail the arguments, we transform the integrand in $J_{3}^{n,\varepsilon}$ first. We have
$$
\begin{equation*}
\begin{aligned} \, & n\biggl|(f^{\varepsilon}-f)\biggl(x_1+\frac{1}{n}, x_2\biggr) +(f-f^{\varepsilon})(x_1, x_2) \biggr| \\ &\qquad =n\biggl|(f*\varphi^{\varepsilon}-f)\biggl(x_1+\frac{1}{n}, x_2\biggr) +(f-f*\varphi^{\varepsilon})(x_1, x_2) \biggr| \\ &\qquad =\biggl|\int_0^1 dz \, (f_{x_1}*\varphi^{\varepsilon}) \biggl(x_1+\frac{z}{n}, x_2\biggr)-\int_0^1 dz \, f_{x_1} \biggl(x_1+\frac{z}{n}, x_2\biggr) \biggr|. \end{aligned}
\end{equation*}
\notag
$$
Using Fubini’s theorem and the fundamental theorem of calculus we can write the integral $J_{3}^{n,\varepsilon}$ as
$$
\begin{equation*}
\begin{aligned} \, &J_{3}^{n,\varepsilon}=\int_0^1 dx_1 \int_0^1 dx_2 \, \biggl|\int_{-\infty}^{\infty} dy_1 \int_{-\infty}^{\infty} dy_2 \, \biggl[\biggl(f\biggl(x_1+\frac{1}{n}-y_1, x_2-y_2\biggr) \\ &\qquad\qquad -f(x_1-y_1, x_2-y_2)\biggr) \varphi^{\varepsilon}(y_1)\varphi^{\varepsilon}(y_2)\biggr] -\biggl[f\biggl(x_1+\frac{1}{n}, x_2\biggr)- f(x_1, x_2)\biggr]\biggr| \\ &\qquad =\int_0^1 dx_1 \int_0^1 dx_2 \, \biggl| \int_{-\infty}^{\infty} dy_1 \int_{-\infty}^{\infty} dy_2 \, \biggl(f\biggl(x_1+\frac{1}{n}-y_1, x_2-y_2\biggr)\varphi^{\varepsilon}(y_1)\varphi^{\varepsilon}(y_2) \\ &\qquad\qquad -f\biggl(x_1+\frac{1}{n}, x_2\biggr)\biggr) -\bigl(f(x_1-y_1, x_2-y_2)\varphi^{\varepsilon}(y_1)\varphi^{\varepsilon}(y_2)-f(x_1, x_2)\bigr)\biggr|. \end{aligned}
\end{equation*}
\notag
$$
We claim that this expression tends to zero as $n \to \infty$ for any fixed $\varepsilon>0$. In fact, consider the function
$$
\begin{equation*}
F^\varepsilon(x_1,x_2):=\int_{-\infty}^{\infty} dy_1 \int_{-\infty}^{\infty} dy_2 \, f(x_1-y_1, x_2-y_2)\varphi^{\varepsilon}(y_1)\varphi^{\varepsilon}(y_2)-f(x_1, x_2).
\end{equation*}
\notag
$$
It is bounded, and therefore it lies in $L_1(K)$. We also have
$$
\begin{equation*}
J_{3}^{n,\varepsilon}= \int_0^1 dx_1 \int_0^1 dx_2 \, \biggl(F^\varepsilon\biggl(x_1+\frac1{n},x_2\biggr)- F^\varepsilon(x_1,x_2)\biggr).
\end{equation*}
\notag
$$
By Lemma 1 this expression tends to zero as $n\to \infty$. Hence
$$
\begin{equation*}
\lim_{\varepsilon\to 0} \lim_{n\to\infty} J_{3}^{n,\varepsilon}=0,
\end{equation*}
\notag
$$
as required. Now (4.1) follows from (4.9) and (4.10).
Convergence (4.2), which is quite similar, follows by interchanging $x_1$ and $x_2$. 2. Let us prove (4.3). Proceeding as in the similar claim for the first derivatives, we will show that, under the hypotheses of this assertion of the lemma, the second mixed derivative $f$ also commutes with the operation of convolution with the function $\varphi^\varepsilon$,
$$
\begin{equation}
(f^\varepsilon)_{x_1x_2}(x_1,x_2)=(f_{x_1x_2}*\varphi^\varepsilon)(x_1,x_2).
\end{equation}
\tag{4.11}
$$
Next we use the fact that the second-order mixed derivative can be interchanged with the double integral in the definition of the convolution, provided that all required integrals are absolutely convergent. Indeed, by (4.8) above we have
$$
\begin{equation*}
\begin{aligned} \, (f^\varepsilon)_{x_1x_2}(x_1,x_2) &=\frac{\partial}{\partial x_2}\frac{\partial}{\partial x_1} \int_0^1 \biggl(\int_0^1 f(x_1-y_1,x_2-y_2)\varphi^\varepsilon(y_2) \,dy_2\biggr)\varphi^\varepsilon(y_1) \,dy_1 \\ &= \frac{\partial}{\partial x_2} \int_0^1 \frac{\partial}{\partial x_1} \biggl(\int_0^1 f(x_1-y_1,x_2-y_2)\varphi^\varepsilon(y_2)\, dy_2\biggr)\varphi^\varepsilon(y_1) \,dy_1 \end{aligned}
\end{equation*}
\notag
$$
(all integrals here are absolutely convergent by the hypotheses of the lemma). Then for the inner integral the variable $x_1$ is a parameter, and so, since the right-hand side is absolutely convergent,
$$
\begin{equation*}
\begin{aligned} \, & \frac{\partial}{\partial x_2} \int_0^1 \frac{\partial}{\partial x_1}\biggl(\int_0^1 f(x_1-y_1,x_2-y_2)\varphi^\varepsilon(y_2) \,dy_2\biggr)\varphi^\varepsilon(y_1) \,dy_1 \\ &\qquad =\frac{\partial}{\partial x_2} \int_0^1 \biggl(\int_0^1 f_{x_1}(x_1-y_1,x_2-y_2)\varphi^\varepsilon(y_2)\, dy_2\biggr)\varphi^\varepsilon(y_1) \,dy_1. \end{aligned}
\end{equation*}
\notag
$$
Now by Fubini’s theorem
$$
\begin{equation*}
\begin{aligned} \, &\frac{\partial}{\partial x_2} \int_0^1 \biggl(\int_0^1 f_{x_1}(x_1-y_1,x_2-y_2)\varphi^\varepsilon(y_2) \,dy_2\biggr)\varphi^\varepsilon(y_1) \,dy_1 \\ &\qquad = \frac{\partial}{\partial x_2} \int_0^1 \biggl(\int_0^1 f_{x_1}(x_1-y_1,x_2-y_2)\varphi^\varepsilon(y_1)\, dy_1\biggr)\varphi^\varepsilon(y_2) \,dy_2, \end{aligned}
\end{equation*}
\notag
$$
and so, using (4.8) and differentiating with respect to the parameter (which is now $x_2$) under the integral sign, we find that
$$
\begin{equation*}
\begin{aligned} \, &\frac{\partial}{\partial x_2} \int_0^1 \biggl(\int_0^1 f_{x_1}(x_1-y_1,x_2-y_2)\varphi^\varepsilon(y_1) \,dy_1\biggr)\varphi^\varepsilon(y_2) \,dy_2 \\ &\qquad =\int_0^1 \frac{\partial}{\partial x_2} \biggl(\int_0^1 f_{x_1}(x_1-y_1,x_2-y_2)\varphi^\varepsilon(y_1)\, dy_1\biggr)\varphi^\varepsilon(y_2) \,dy_2 \\ &\qquad =\int_0^1 \biggl(\int_0^1 f_{x_1x_2}(x_1-y_1,x_2-y_2)\varphi^\varepsilon(y_1)\, dy_1\biggr)\varphi^\varepsilon(y_2) \,dy_2, \end{aligned}
\end{equation*}
\notag
$$
as claimed for (4.11). So we obtain
$$
\begin{equation*}
(f_{x_1x_2})^{\varepsilon}(x_1, x_2) \equiv (f_{x_1x_2}*\overline\varphi^\varepsilon)(x_1, x_2).
\end{equation*}
\notag
$$
In what follows, each of these expressions is denoted for brevity by $f_{x_1x_2}^{\varepsilon}(x_1, x_2)$.
We now have
$$
\begin{equation*}
\begin{aligned} \, &\int_{0}^{1}dx_1 \int_{0}^{1}dx_2 \, \biggl| n \biggl(f_{x_2}\biggl(x_1+\frac{1}{n}, x_2\biggr) -f_{x_2}(x_1, x_2) \biggr)-f_{x_1x_2}(x_1, x_2)\biggr| \\ &\qquad \leqslant\int_{0}^{1}dx_1 \int_{0}^{1}dx_2 \, \bigl|f^{\varepsilon}_{x_1x_2}(x_1, x_2)-f_{x_1x_2}(x_1, x_2)\bigr| \\ &\qquad\qquad +\int_{0}^{1}dx_1 \int_{0}^{1}dx_2 \, \biggl|n \biggl(f^{\varepsilon}_{x_2}\biggl( x_1+\frac{1}{n}, x_2\biggr)-f^{\varepsilon}_{x_2}(x_1, x_2) \biggr)- f^{\varepsilon}_{x_1x_2}(x_1, x_2)\biggr| \\ &\qquad\qquad +\int_{0}^{1}dx_1 \int_{0}^{1}dx_2 \, \biggl|n \biggl(f_{x_2}\biggl(x_1+\frac{1}{n}, x_2\biggr) -f_{x_2}(x_1, x_2) \biggr) \\ &\qquad\qquad\qquad -n \biggl(f^{\varepsilon}_{x_2}\biggl(x_1+\frac{1}{n}, x_2\biggr)-f^{\varepsilon}_{x_2}(x_1, x_2) \biggr)\biggr| \\ &\qquad =J_{4}^{\varepsilon}+J_{5}^{n,\varepsilon}+J_{6}^{n,\varepsilon}. \end{aligned}
\end{equation*}
\notag
$$
Let us prove that each of these integrals converges to zero in the sense of the limits $\lim_{\varepsilon\to 0} \lim_{n\to\infty}(\dots)$. First we claim that
$$
\begin{equation}
J_{4}^{\varepsilon}=\int_{0}^{1}dx_1 \int_{0}^{1}dx_2 \, \bigl|f^{\varepsilon}_{x_1x_2}(x_1, x_2)-f_{x_1x_2}(x_1, x_2)\bigr|\to0, \qquad \varepsilon \to 0.
\end{equation}
\tag{4.12}
$$
Indeed, for some $\delta>0$ let $g\in C(K)$ be a function such that Let us show by applying elementary inequalities, the above property (4.11) and Fubini’s theorem that
$$
\begin{equation*}
\|f^\varepsilon_{x_1x_2}-g^\varepsilon \|_{L_1(K)} \leqslant \delta,
\end{equation*}
\notag
$$
where $g^\varepsilon=g * \overline\varphi^\varepsilon$. Indeed,
$$
\begin{equation*}
\begin{aligned} \, \|f^\varepsilon_{x_1x_2}-g^\varepsilon \|_{L_1(K)} &= \iint_{K} |f^\varepsilon_{x_1x_2}(x_1,x_2)-g^\varepsilon(x_1,x_2)| \,dx_1\,dx_2 \\ &= \iint_{K} \biggl|\iint_{K} (f_{x_1x_2}(x_1-y_1,x_2-y_2) \\ &\qquad - g(x_1-y_1,x_2-y_2))\overline\varphi^\varepsilon(y_1,y_2)\,dy_1\,dy_2\biggr| \,dx_1\,dx_2 \\ &\leqslant \iint_{K} \biggl(\iint_{K} |f_{x_1x_2}(x_1-y_1,x_2-y_2) \\ &\qquad - g(x_1-y_1,x_2-y_2)|\overline\varphi^\varepsilon(y_1,y_2)\, dx_1\,dx_2\biggr)\, dy_1\,dy_2 \\ &\leqslant \iint_{K}(\|f_{x_1x_2}-g \|_{L_1(K)}) \overline\varphi^\varepsilon(y_1,y_2) \,dy_1\,dy_2 \\ &\leqslant \delta \iint_{K} \overline\varphi^\varepsilon(y_1,y_2)=\delta. \end{aligned}
\end{equation*}
\notag
$$
Now, for each $\delta>0$,
$$
\begin{equation*}
\begin{aligned} \, &\int_{0}^{1}dx_1 \int_{0}^{1}dx_2 \, \bigl|f^{\varepsilon}_{x_1x_2}(x_1, x_2)-f_{x_1x_2}(x_1, x_2)\bigr| \\ &\qquad \leqslant \int_{0}^{1}dx_1 \int_{0}^{1}dx_2 \, \bigl|f^{\varepsilon}_{x_1x_2}(x_1, x_2)- g^\varepsilon(x_1, x_2)\bigr| \\ &\qquad\qquad +\int_{0}^{1}dx_1 \int_{0}^{1}dx_2 \, \bigl|f_{x_1x_2}(x_1, x_2)- g(x_1, x_2)\bigr| \\ &\qquad\qquad +\int_{0}^{1}dx_1 \int_{0}^{1}dx_2 \, \bigl|g^{\varepsilon}(x_1, x_2)- g(x_1, x_2)\bigr| \\ &\qquad \leqslant \int_{0}^{1}dx_1 \int_{0}^{1}dx_2 \, \bigl|g^{\varepsilon}(x_1, x_2)- g(x_1, x_2)\bigr|+2\delta. \end{aligned}
\end{equation*}
\notag
$$
Here the integral on the right in the last inequality tends to zero as $\varepsilon \to 0$ because $g$ is uniformly continuous. The integral $J_{4}^{\varepsilon}$ is independent of $\delta$, and so we have the required convergence (4.12) to zero.
Consider the term $J_{5}^{n,\varepsilon}$. By the definition of the derivative and Lebesgue’s dominated convergence theorem, as $n \to \infty$, we have
$$
\begin{equation*}
J_{5}^{n,\varepsilon}=\int_{0}^{1}dx_1 \int_{0}^{1}dx_2\biggl|n \biggl( f^{\varepsilon}_{x_2}\biggl(x_1+\frac{1}{n}, x_2\biggr)-f^{\varepsilon}_{x_2}(x_1, x_2) \biggr) - f^{\varepsilon}_{x_1x_2}(x_1, x_2)\biggr| \to0
\end{equation*}
\notag
$$
for the bounded smooth function $f^{\varepsilon}$; here the finite differences
$$
\begin{equation*}
n \biggl(f^{\varepsilon}_{x_2}\biggl(x_1+\dfrac{1}{n}, x_2\biggr)-f^{\varepsilon}_{x_2}(x_1, x_2) \biggr)
\end{equation*}
\notag
$$
are bounded by the mean value theorem or by the fundamental theorem of calculus, similarly to the corresponding result for $J_{2}^{n,\varepsilon}$. Hence
$$
\begin{equation*}
\lim_{\varepsilon\to 0} \limsup_{n\to\infty} J_{5}^{n,\varepsilon}=0.
\end{equation*}
\notag
$$
Consider the integral $J_{6}^{n,\varepsilon}$. Using the fundamental theorem of calculus and since the differentiation operator (with respect to $x_1$) and convolution commute, we transform the integrand in $J_{6}^{n,\varepsilon}$ as follows:
$$
\begin{equation*}
\begin{aligned} \, &n \biggl|(f_{x_2}^{\varepsilon}-f_{x_2})\biggl(x_1+\frac{1}{n}, x_2\biggr)-(f_{x_2}^{\varepsilon}-f_{x_2})(x_1, x_2)\biggr| \\ &\qquad =n \biggl|(f_{x_2}*\varphi^{\varepsilon}-f_{x_2}) \biggl(x_1+\frac{1}{n}, x_2\biggr)+(f_{x_2}-f_{x_2}*\varphi^{\varepsilon}) (x_1, x_2)\biggr| \\ &\qquad =\biggl|\int_0^1 dz\, (f_{x_2 x_1}*\varphi^{\varepsilon})\biggl(x_1+\frac{z}{n},x_2\biggr)-\int_0^1 dz f_{x_2 x_1}\biggl(x_1+\frac{z}{n}, x_2\biggl)\biggr|. \end{aligned}
\end{equation*}
\notag
$$
As a result, we have
$$
\begin{equation*}
\begin{aligned} \, J_{6}^{n,\varepsilon} &=\int_0^1 dx_1 \int_0^1 dx_2 \, \biggl|\int_{-\infty}^{\infty} dy_1 \int_{-\infty}^{\infty} dy_2 \, \biggl(f_{x_2}\biggl(x_1+\frac{1}{n}-y_1, x_2-y_2\biggr) \\ &\quad - f_{x_2}(x_1-y_1, x_2-y_2)\biggr)\varphi^{\varepsilon}(y_1)\varphi^{\varepsilon}(y_2)- \biggl(f_{x_2}\biggl(x_1+\frac{1}{n}, x_2\biggr)- f_{x_2}(x_1, x_2)\biggr)\biggr| \\ &=\int_0^1 dx_1 \int_0^1 dx_2 \, \biggl| \int_{-\infty}^{\infty} dy_1 \int_{-\infty}^{\infty} dy_2 \, \biggl(f_{x_2}\biggl(x_1+\frac{1}{n}-y_1, x_2-y_2\biggr)\varphi^{\varepsilon}(y_1)\varphi^{\varepsilon}(y_2) \\ &\quad -f_{x_2}\biggl(x_1+\frac{1}{n}, x_2\biggl)\biggr)- \bigl(f_{x_2}(x_1-y_1, x_2-y_2)\varphi^{\varepsilon}(y_1)\varphi^{\varepsilon}(y_2)-f_{x_2}(x_1, x_2)\bigr)\biggr|. \end{aligned}
\end{equation*}
\notag
$$
Setting
$$
\begin{equation*}
F^\varepsilon(x_1,x_2):=\int_{-\infty}^{\infty} dy_1 \int_{-\infty}^{\infty} dy_2 \, \bigl( f_{x_2}(x_1-y_1, x_2-y_2) \varphi^{\varepsilon}(y_1)\varphi^{\varepsilon}(y_2)-f_{x_2}(x_1, x_2)\bigr),
\end{equation*}
\notag
$$
we can write the integral $J_{6}^{n,\varepsilon}$ as
$$
\begin{equation*}
J_{6}^{n,\varepsilon}=\int_0^1 dx_1 \int_0^1 dx_2 \, \biggl| F^\varepsilon\biggl(x_1+\frac1{n},x_2\biggr)- F^\varepsilon(x_1,x_2)\biggr|.
\end{equation*}
\notag
$$
The function $F^\varepsilon$ is bounded, and therefore $\lim_{n\to\infty} J_{6}^{n,\varepsilon}=0$. As a result,
$$
\begin{equation*}
\lim_{\varepsilon \to 0}\lim_{n\to\infty} J_{6}^{n,\varepsilon}=0,
\end{equation*}
\notag
$$
as required.
3. Let us verify (4.5). Applying the fundamental theorem of calculus in the form for $a=1/n$ we find that
$$
\begin{equation*}
\begin{aligned} \, &n\Delta_{(x_1)} f_{x_2}(x_1, x_2) = n\biggl(f_{x_2}\biggl(x_1+\frac1{n}, x_2\biggr) - f_{x_2}(x_1, x_2)\biggr) \\ &\qquad= \int_0^1 f_{x_2 x_1}\biggl(x_1+\frac{t}{n}, x_2\biggr)\, dt \end{aligned}
\end{equation*}
\notag
$$
and
$$
\begin{equation*}
\begin{aligned} \, &n^2\Delta_{(x_2)}\Delta_{(x_1)}f(x_1, x_2) \\ &\qquad =n^2\biggl[f\biggl(x_1+\frac1{n}, x_2+\frac1{n}\biggr)-f\biggl(x_1+\frac1{n}, x_2\biggl) -f\biggl(x_1, x_2+\frac1{n}r\biggr)+f(x_1, x_2)\biggr] \\ &\qquad = n \int_0^1 f_{x_2}\biggl(x_1+\frac1{n}, x_2+\frac{t}{n}\biggr)\, ds-n \int_0^1 f_{x_2}\biggl(x_1, x_2+\frac{t}{n}\biggr)\, dt \\ &\qquad = n \int_0^1\biggl[f_{x_2}\biggl(x_1+\frac1{n}, x_2+\frac{t}{n}\biggr) -f_{x_2}\biggl(x_1, x_2+\frac{t}{n}\biggr)\biggr] \,dt \\ &\qquad = \iint_{K} f_{x_2 x_1}\biggl(x_1+\frac{s_1}{n}, x_2+\frac{s_2}{n}\biggr)\, ds_1\, ds_2. \end{aligned}
\end{equation*}
\notag
$$
Therefore,
$$
\begin{equation*}
\begin{aligned} \, &\int_0^1 dx_1 \int_0^1 dx_2 \, \bigl|n\Delta_{(x_1)} f_{x_2}(x_1, x_2)- n^2\Delta_{(x_2)}\Delta_{(x_1)}f(x_1, x_2)\bigr| \\ &\qquad =\int_0^1 dx_1 \int_0^1 dx_2 \, \biggl|\int_0^1 f_{x_1x_2}\biggl(x_1+\frac{t_1}{n}, x_2\biggr)\,dt_1 \\ &\qquad\qquad- \int_0^1 dt_1 \int_0^1 f_{x_1x_2 }\biggl(x_1 + \frac{t_1}{n}, x_2 + \frac{t_2}{n}\biggr)\,dt_2\biggr| \\ &\qquad \leqslant \int_0^1 dx_1 \int_0^1 dx_2 \, \int_0^1 dt_1 \int_0^1 dt_2 \, \biggl|f_{x_1x_2}\biggl(x_1+\frac{t_1}{n}, x_2\biggr) \\ &\qquad\qquad-f_{x_1x_2}\biggl(x_1+\frac{t_1}{n}, x_2+\frac{t_2}{n}\biggl)\biggr| \\ &\qquad \leqslant \int_0^1 dx_1 \int_0^1 dx_2 \int_0^1 dt_1 \int_0^1 dt_2 \, \biggl|f_{x_1x_2}\biggl(x_1+\frac{t_1}{n}, x_2+\frac{t_2}{n}\biggr)-f_{x_1x_2}(x_1, x_2)\biggr| \\ &\qquad\qquad +\int_0^1 dx_1 \int_0^1 dx_2 \int_0^1 dt_1 \int_0^1 dt_2 \, \biggl|f_{x_1x_2}(x_1, x_2) -f_{x_1x_2}\biggl(x_1+\frac{t_1}{n}, x_2\biggr)\biggr|. \end{aligned}
\end{equation*}
\notag
$$
By Lemma 1, both terms on the right in the last inequality tend to zero as $n\to\infty$. Hence
$$
\begin{equation*}
\int_{0}^{1}dx_1 \int_{0}^{1}dx_2 \, \bigl| n^2\Delta_{(x_2)}\Delta_{(x_1)}f(x_1,x_2) - n\Delta_{(x_1)}f_{x_2}(x_1,x_2)\bigr| \to 0.
\end{equation*}
\notag
$$
This proof of Lemma 2 completes by noting that (4.6) is equivalent to (4.3), and (4.7) is equivalent to (4.5). § 5. Proof of the theoremFor brevity, we set
$$
\begin{equation*}
\Pi_{k_1,k_2}(x_1,x_2):=\binom{n-1}{k_1} \binom{n-1}{k_2} x_{1}^{k_1}{(1-x_1)}^{n-k_1-1} x_{2}^{k_2} {(1-x_2)}^{n-k_2-1}.
\end{equation*}
\notag
$$
The expression on the right appears naturally by repeated differentiation of the polynomials $\widetilde B_{\alpha,n}$. Note that, by the binomial formula, for all $x_1$ and $x_2$,
$$
\begin{equation}
\sum_{0\leqslant k_1, k_2 \leqslant n-1}\Pi_{k_1,k_2}(x_1,x_2)=1.
\end{equation}
\tag{5.1}
$$
We claim that
$$
\begin{equation}
\begin{aligned} \, &\mathsf E \int_{0}^{1} dx_1 \int_{0}^{1} dx_2 \, \biggl| \frac{\partial^2 \widetilde {B}_{\alpha,n}(f,x_1,x_2)}{\partial x_1\, \partial x_2}-\frac{\partial^2 (f,x_1,x_2)}{\partial x_1 \,\partial x_2}\biggr| \nonumber \\ &\qquad ={\mathsf E} \int_{0}^{1} dx_1 \int_{0}^{1} dx_2\, \Biggl| \sum_{\substack{0 \leqslant k_1 < n\\ 0 \leqslant k_2 < n}} n^2 \biggl[ \Delta_{(x_1)} \Delta_{(x_2)}f\biggl(\frac{k_1+\alpha_1}{n},\frac{k_2+\alpha_2}{n}\biggr) \nonumber \\ &\qquad\qquad -\Delta_{(x_1)} \Delta_{(x_2)}f(x_1,x_2)\biggr] \Pi_{k_1,k_2}(x_1,x_2)\Biggr|+o(1) \nonumber \\ &=: I^n+o(1), \qquad n\to\infty, \end{aligned}
\end{equation}
\tag{5.2}
$$
by (4.3) and (4.5) in Lemma 2. Indeed, differentiating $\widetilde {B}_{\alpha,n}$ with respect to $x_1$, we have
$$
\begin{equation*}
\begin{aligned} \, &\frac{\partial \widetilde {B}_{\alpha,n}(f,x_1,x_2)}{\partial x_1} \\ &\qquad= \frac{\partial}{\partial x_1} \sum_{k_1, k_2=0}^{n} f\biggl( \frac{k_1}{n}+\frac{\alpha_1}{n},\frac{k_2}{n}+\frac{\alpha_2}{n}\biggr) \\ &\qquad\qquad\qquad \times \binom{n}{k_1} \binom{n}{k_2} x_{1}^{k_1} {(1-x_1)}^{n-k_1} x_{2}^{k_2} {(1-x_2)}^{n-k_2} \\ &\qquad =\sum_{\substack{0 < k_1 \leqslant n\\ 0 \leqslant k_2 \leqslant n}} f\biggl(\frac{k_1}{n}+\frac{\alpha_1}{n},\frac{k_2}{n}+\frac{\alpha_2}{n}\biggr) \\ &\qquad\qquad\qquad \times \binom{n}{k_1} \binom{n}{k_2} k_{1} x_{1}^{k_1-1} {(1-x_1)}^{n-k_1} x_{2}^{k_2} {(1-x_2)}^{n-k_2} \\ &\qquad\qquad- \sum_{\substack{0 \leqslant k_1 < n\\ 0 \leqslant k_2 \leqslant n}} f\biggl(\frac{k_1}{n}+\frac{\alpha_1}{n},\frac{k_2}{n}+\frac{\alpha_2}{n}\biggr) \\ &\qquad\qquad\qquad \times \binom{n}{k_1} \binom{n}{k_2} x_{1}^{k_1} (n-k_1) {(1-x_1)}^{n-k_1-1} x_{2}^{k_2} {(1-x_2)}^{n-k_2} \\ &\qquad =\sum_{\substack{0 \leqslant k_1 < n\\ 0 \leqslant k_2 \leqslant n}} f\biggl(\frac{k_1+\alpha_1+1}{n},\frac{k_2+\alpha_2}{n}\biggr) \\ &\qquad\qquad\qquad \times \binom{n}{k_1+1} \binom{n}{k_2} (k_{1}+1) x_{1}^{k_1} {(1-x_1)}^{n-k_1-1} x_{2}^{k_2} {(1-x_2)}^{n-k_2} \\ &\qquad\qquad - \sum_{\substack{0 \leqslant k_1 < n\\ 0 \leqslant k_2 \leqslant n}} f\biggl(\frac{k_1+\alpha_1}{n},\frac{k_2+\alpha_2}{n}\biggr) \\ &\qquad\qquad\qquad \times \binom{n}{k_1} \binom{n}{k_2} x_{1}^{k_1} (n-k_1) {(1-x_1)}^{n-k_1-1} x_{2}^{k_2} {(1-x_2)}^{n-k_2} \\ &\qquad =\sum_{\substack{0 \leqslant k_1 < n\\ 0 \leqslant k_2 \leqslant n}} n \Delta_{(x_1)} f\biggl(\frac{k_1+\alpha_1}{n},\frac{k_2+\alpha_2}{n}\biggr) \\ &\qquad\qquad\qquad \times \binom{n-1}{k_1} \binom{n}{k_2} x_{1}^{k_1} {(1-x_1)}^{n-k_1-1} x_{2}^{k_2} {(1-x_2)}^{n-k_2}. \end{aligned}
\end{equation*}
\notag
$$
Differentiating this expression with respect to $x_2$ we obtain
$$
\begin{equation*}
\begin{aligned} \, &\frac{\partial^2 \widetilde {B}_{\alpha,n}(f,x_1,x_2)}{\partial x_1 \,\partial x_2} \\ &\qquad =\frac{\partial^2}{\partial x_1 \, \partial x_2} \sum_{k_1, k_2=0}^{n} f\biggl( \frac{k_1+\alpha_1}{n},\frac{k_2+\alpha_2}{n}\biggr) \\ &\qquad\qquad\qquad \times \binom{n}{k_1} \binom{n}{k_2} x_{1}^{k_1} {(1-x_1)}^{n-k_1} x_{2}^{k_2} {(1-x_2)}^{n-k_2} \\ &\qquad =\frac{\partial}{\partial x_2} \sum_{\substack{0 \leqslant k_1 < n\\ 0 \leqslant k_2 \leqslant n}} n \Delta_{(x_1)} f\biggl(\frac{k_1+\alpha_1}{n},\frac{k_2+\alpha_2}{n}\biggr) \\ &\qquad\qquad\qquad \times \binom{n-1}{k_1} \binom{n}{k_2} x_{1}^{k_1} {(1-x_1)}^{n-k_1-1} x_{2}^{k_2} {(1-x_2)}^{n-k_2} \\ &\qquad =\sum_{\substack{0 < k_1 \leqslant n\\ 0 \leqslant k_2 < n}} n \Delta_{(x_1)} f\biggl(\frac{k_1+\alpha_1}{n},\frac{k_2+\alpha_2}{n}\biggr) \\ &\qquad\qquad\qquad \times \binom{n-1}{k_1} \binom{n}{k_2} x_{1}^{k_1} {(1-x_1)}^{n-k_1-1} k_2 x_{2}^{k_2-1} {(1-x_2)}^{n-k_2} \\ &\qquad\qquad - \sum_{\substack{0 \leqslant k_1 < n\\ 0 < k_2 \leqslant n}} n \Delta_{(x_1)} f\biggl(\frac{k_1+\alpha_1}{n},\frac{k_2+\alpha_2}{n}\biggr) \\ &\qquad\qquad\qquad \times \binom{n-1}{k_1} \binom{n}{k_2} x_{1}^{k_1} {(1-x_1)}^{n-k_1-1} x_{2}^{k_2}(n-k_2) {(1-x_2)}^{n-k_2-1} \\ &\qquad =\sum_{\substack{0 \leqslant k_1 < n\\ 0 \leqslant k_2 < n}} f\biggl(\frac{k_1+\alpha_1 }{n},\frac{k_2+\alpha_2+1}{n}\biggr) \\ &\qquad\qquad\qquad \times \binom{n-1}{k_1} \binom{n}{k_2+1} x_{1}^{k_1} {(1-x_1)}^{n-k_1-1} x_{2}^{k_2} {(1-x_2)}^{n-k_2-1} \\ &\qquad\qquad - \sum_{\substack{0 \leqslant k_1 < n\\ 0 < k_2 \leqslant n}} n \Delta_{(x_1)} f\biggl(\frac{k_1+\alpha_1}{n},\frac{k_2+\alpha_2}{n}\biggr) \\ &\qquad\qquad\qquad \times\binom{n-1}{k_1} \binom{n}{k_2} x_{1}^{k_1} {(1-x_1)}^{n-k_1-1} x_{2}^{k_2}(n-k_2) {(1-x_2)}^{n-k_2-1} \\ &\qquad =\sum_{\substack{0 \leqslant k_1 < n\\ 0 \leqslant k_2 < n}} n^2 \biggl(\Delta_{(x_1)} f\biggl(\frac{k_1+\alpha_1}{n},\frac{k_2+\alpha_2+1}{n}\biggr) -\Delta_{(x_1)} f\biggl(\frac{k_1+\alpha_1}{n},\frac{k_2+\alpha_2}{n}\biggr) \biggr) \\ &\qquad\qquad\qquad \times \binom{n-1}{k_1} \binom{n-1}{k_2} x_{1}^{k_1} {(1-x_1)}^{n-k_1-1} x_{2}^{k_2} {(1-x_2)}^{n-k_2-1} \\ &\qquad =\sum_{\substack{0 \leqslant k_1 < n\\ 0 \leqslant k_2 < n}} n^2 \Delta_{(x_1)} \Delta_{(x_2)}f \biggl( \frac{k_1+\alpha_1}{n},\frac{k_2+\alpha_2}{n}\biggr) \\ &\qquad\qquad\qquad \times \binom{n-1}{k_1} \binom{n-1}{k_2} x_{1}^{k_1} {(1-x_1)}^{n-k_1-1} x_{2}^{k_2} {(1-x_2)}^{n-k_2-1}. \end{aligned}
\end{equation*}
\notag
$$
Now formula (5.2) follows from the above representation for the mixed derivative $\dfrac{\partial^2 \widetilde {B}_{\alpha,n}(f,x_1,x_2)}{\partial x_1 \,\partial x_2}$ and from Lemma 2. Let us estimate the (nonnegative) integral $I^n$ on the right in (5.2) from above. To do this we use the smoothed function $f^\varepsilon=f*\overline \varphi^\varepsilon$ from the proof of Lemma 2. We have
$$
\begin{equation*}
\begin{aligned} \, I^n &\leqslant n^2 \mathsf E \int_{0}^{1} dx_1 \int_{0}^{1} dx_2 \, \Biggl| \sum_{\substack{0 \leqslant k_1 < n\\ 0 \leqslant k_2 < n}} \biggl(\Delta_{(x_1)}\Delta_{(x_2)}f^{\varepsilon}\biggl( \frac{k_1+\alpha_1}{n},\frac{k_2+\alpha_2}{n}\biggr) \\ &\qquad\qquad -\Delta_{(x_1)}\Delta_{(x_2)} f^{\varepsilon}(x_1,x_2) \biggr)\Pi_{k_1,k_2}(x_1,x_2)\Biggr| \\ &\qquad +n^2 \mathsf E \int_{0}^{1} dx_1 \int_{0}^{1} dx_2\, \Biggl|\sum_{\substack{0 \leqslant k_1 < n\\ 0 \leqslant k_2 < n}} \bigl(\Delta_{(x_1)}\Delta_{(x_2)}f^{\varepsilon}(x_1,x_2) \\ &\qquad\qquad-\Delta_{(x_1)} \Delta_{(x_2)}f(x_1,x_2 )\bigr) \Pi_{k_1,k_2}(x_1,x_2)\Biggr| \\ &\qquad+n^2 \mathsf E \int_{0}^{1} dx_1 \int_{0}^{1} dx_2\, \Biggl|\sum_{\substack{0 \leqslant k_1 < n\\ 0 \leqslant k_2 < n}} \biggl(\Delta_{(x_1)}\Delta_{(x_2)}f^{\varepsilon}\biggl( \frac{k_1+\alpha_1}{n},\frac{k_2+\alpha_2}{n}\biggr) \\ &\qquad\qquad-\Delta_{(x_1)}\Delta_{(x_2)} f\biggl(\frac{k_1+\alpha_1}{n},\frac{k_2+\alpha_2}{n} \biggr)\biggr)\Pi_{k_1,k_2}(x_1,x_2)\Biggr| \\ &=: I_1^{n,\varepsilon}+I_2^{n,\varepsilon}+I_3^{n,\varepsilon}. \end{aligned}
\end{equation*}
\notag
$$
All three integrals $I_k^{n,\varepsilon}$, $1\leqslant k\leqslant 3$, are nonnegative, and the second integral $I_2^{n,\varepsilon}$ does not contain the random variables $\alpha_1$ and $\alpha_2$. Hence the expectation sign before the double integral in $I_2^{n,\varepsilon}$ can be dropped. We claim that
$$
\begin{equation*}
\lim_{\varepsilon\to 0} \limsup_{n\to\infty} I_k^{n,\varepsilon}=0, \qquad k=1,2,3.
\end{equation*}
\notag
$$
It will be convenient to consider $I_1^{n,\varepsilon}$ last; see the end of the proof. Let us estimate the integral $I_2^{n,\varepsilon}$. To this end we represent the estimate as a sum of three integrals. By (5.1) we have
$$
\begin{equation*}
\begin{aligned} \, I_2^{n,\varepsilon} &=\int_0^1 dx_1 \int_0^1 dx_2 \, \bigl|n^2\Delta_{(x_1)}\Delta_{(x_2)}f^{\varepsilon}(x_1, x_2) -n^2\Delta_{(x_1)}\Delta_{(x_2)}f(x_1, x_2)\bigr| \\ &\leqslant \int_0^1 dx_1 \int_0^1 dx_2 \, \bigl|n^2 \Delta_{(x_1)}\Delta_{(x_2)}f^{\varepsilon}(x_1, x_2)-f_{x_1 x_2}^{\varepsilon}(x_1, x_2)\bigr| \\ &\qquad +\int_0^1 dx_1 \int_0^1 dx_2 \, \bigl| f_{x_1 x_2}^{\varepsilon}(x_1, x_2) -f_{x_1 x_2}(x_1, x_2)\bigr| \\ &\qquad +\int_0^1 dx_1 \int_0^1 dx_2 \, \bigl|f_{x_1 x_2}(x_1, x_2)-n^2\Delta_{(x_1)}\Delta_{(x_2)}f(x_1, x_2)\bigr| \\ &=:I_{21}^{n,\varepsilon}+I_{22}^{\varepsilon}+I_{23}^{n}. \end{aligned}
\end{equation*}
\notag
$$
Consider the integral $I_{22}^{\varepsilon}$. Since $f_{x_1x_2} \in L_1(K)$, this integral converges to zero as $\varepsilon \to 0$ by Lemma 1. The integral $I_{21}^{n,\varepsilon}$ is estimated as
$$
\begin{equation*}
\begin{aligned} \, 0 &\leqslant I_{21}^{n,\varepsilon} \leqslant \int_0^1 dx_1 \int_0^1 dx_2 \, \bigl|n^2 \Delta_{(x_1)}\Delta_{(x_2)}f^{\varepsilon}(x_1, x_2)- n \Delta_{(x _1)}f_{x_2}^{\varepsilon}(x_1, x_2)\bigr| \\ &\qquad +\int_0^1 dx_1 \int_0^1 dx_2 \, \bigl|n \Delta_{(x _1)}f_{x_2}^{\varepsilon}(x_1, x_2)-f_{x_1 x_2}^{\varepsilon}(x_1, x_2)\bigr| \\ &=: I_{21a}^{n,\varepsilon}+I_{21b}^{n,\varepsilon}. \end{aligned}
\end{equation*}
\notag
$$
Now $I_{{21}b}^{n,\varepsilon} \to 0$ as $n\to\infty$ by (4.3) in Lemma 2. A similar analysis shows that $\lim_{n\to\infty} I_{21a}^{n,\varepsilon}=0$, where we use formula (4.7) from Lemma 2. Consider the integral $I_{23}^{n}$. Let us estimate this expression from above by the sum of two integrals. We have $\Delta_{(x_2)}\Delta_{(x_1)}f(x_1, x_2)=\Delta_{(x_1)}\Delta_{(x_2)}f(x_1, x_2)$, and so
$$
\begin{equation*}
\begin{aligned} \, &\int_0^1 dx_1 \int_0^1 dx_2 \, \bigl|f_{x_1 x_2}(x_1, x_2)-n^2\Delta_{(x_1)}\Delta_{(x_2)}f(x_1, x_2)\bigr| \\ &\qquad \leqslant\int_0^1 dx_1 \int_0^1 dx_2 \, \bigl|n \Delta_{(x_1)} f_{x_2}(x_1, x_2)-n^2\Delta_{(x_2)}\Delta_{(x_1)}f(x_1, x_2)\bigr| \\ &\qquad\qquad +\int_0^1 dx_1 \int_0^1 dx_2 \, \bigl|f_{x_1 x_2}(x_1, x_2)-n \Delta_{(x_1)} f_{x_2}(x_1, x_2)\bigr| \\ &\qquad=I_{23a}^{n}+I_{23b}^{n}. \end{aligned}
\end{equation*}
\notag
$$
For $I_{23a}^{n}$ we have $I_{23a}^{n} \to0$ as $n \to \infty$ by (4.5) in Lemma 2. For $I_{23b}^{n}$ we have $I_{23b}^{n} \to0$ as $n \to \infty$ by (4.3) in Lemma 2. So, for $I_2^{n, \varepsilon}$,
$$
\begin{equation}
0\leqslant \lim_{\varepsilon\to 0} \limsup_{n\to\infty} I_2^{n,\varepsilon}\leqslant \lim_{\varepsilon\to 0} \limsup_{n\to\infty} (I_{21}^{n, \varepsilon}+I_{22}^{\varepsilon}+I_{23}^{n})=0.
\end{equation}
\tag{5.3}
$$
Now consider $I_3^{n, \varepsilon}$. In what follows we require the following well-known property of the beta function:
$$
\begin{equation*}
\mathcal{B}(x, y) =\int_{0}^{1} t^{x-1}(1-t)^{y-1}\, dt=\frac{\Gamma(x)\Gamma(y)}{\Gamma(x+y)}.
\end{equation*}
\notag
$$
Let us transform $I_3^{n, \varepsilon}$ by using the well-known combinatorial formulae expressing the gamma function of an integer argument in terms of factorials. As a result, we obtain an expression (for estimation of $I_3^{n, \varepsilon}$) that is analogous to $I_2^{n, \varepsilon}$. We have
$$
\begin{equation*}
\begin{aligned} \, &I_3^{n, \varepsilon} =n^2 \mathsf E \int_0^1 dx_1 \int_0^1 \Biggl|\Biggl[ \sum_{\substack{0 \leqslant k_1 < n\\ 0 \leqslant k_2 < n}} \Delta_{(x_1)}\Delta_{(x_2)}f^{\varepsilon}\biggl(\frac{k_1+\alpha_1}{n}, \frac{k_2+\alpha_2}{n}\biggr) \\ &\qquad -\Delta_{(x_1)}\Delta_{(x_2)}f\biggl(\frac{k_1+\alpha_1}{n}, \frac{k_2+\alpha_2}{n}\biggr) \Biggr] \\ &\qquad\qquad \times x^{k_1}(1-x)^{n-k_1-1}x^{k_2}(1-x)^{n-k_2-1}\binom{n-1}{k_{1}}\binom{n-1}{k_{2}} \Biggr|\,dx_2 \\ &\leqslant n^2 \mathsf E\sum_{\substack{0 \leqslant k_1 < n\\ 0 \leqslant k_2 < n}} \int_0^1 dx_1 \int_0^1 \biggl| \Delta_{(x_1)}\Delta_{(x_2)}f^{\varepsilon} \biggl(\frac{k_1+\alpha_1}{n},\frac{k_2+\alpha_2}{n}\biggr) \\ &\qquad -\Delta_{(x_1)}\Delta_{(x_2)}f\biggl(\frac{k_1+\alpha_1}{n}, \frac{k_2+\alpha_2}{n}\biggr)\biggr| \\ &\qquad\qquad \times x^{k_1}(1-x)^{n-k_1-1}x^{k_2}(1-x)^{n-k_2-1} \binom{n-1}{k_{1}}\binom{n-1}{k_{2}}\,dx_2 \\ &=n^2 \mathsf E\! \sum_{\substack{0 \leqslant k_1 <n\\ 0 \leqslant k_2< n}} \biggl| \Delta_{(x_1)}\Delta_{(x_2)}f^{\varepsilon}\biggl(\frac{k_1{\kern1pt}{+}\,\alpha_1}{n}, \frac{k_2{\kern1pt}{+}\,\alpha_2}{n}\biggr) {\kern1pt}{-}\,\Delta_{(x_1)}\Delta_{(x_2)}f\biggl(\frac{k_1{\kern1pt}{+}\,\alpha_1}{n}, \frac{k_2{\kern1pt}{+}\,\alpha_2}{n}\biggr)\biggr| \\ &\qquad\qquad \times\int_0^1 dx_1 \int_0^1 \biggl[ x^{k_1}(1-x)^{n-k_1-1}x^{k_2}(1-x)^{n-k_2-1}\binom{n-1}{k_{1}} \binom{n-1}{k_{2}}\,dx_2 \biggr] \\ &=\mathsf E\! \sum_{\substack{0 \leqslant k_1 < n\\ 0 \leqslant k_2 < n}} \biggl| \Delta_{(x_1)}\Delta_{(x_2)}f^{\varepsilon} \biggl(\frac{k_1+\alpha_1}{n},\frac{k_2+\alpha_2}{n}\biggr)- \Delta_{(x_1)}\Delta_{(x_2)}f\biggl(\frac{k_1+\alpha_1}{n}, \frac{k_2+\alpha_2}{n}\biggr) \biggr| \\ &\qquad\qquad \times n^2 \binom{n-1}{k_{1}}\binom{n-1}{k_{2}} \mathcal{B}(k_1+1, n-k_1)\mathcal{B}(k_2+1, n-k_2) \\ &=\mathsf E\! \sum_{\substack{0 \leqslant k_1 < n\\ 0 \leqslant k_2 < n}} \biggl| \Delta_{(x_1)}\Delta_{(x_2)}f^{\varepsilon} \biggl(\frac{k_1+\alpha_1}{n},\frac{k_2+\alpha_2}{n}\biggr)- \Delta_{(x_1)}\Delta_{(x_2)}f\biggl(\frac{k_1+\alpha_1}{n}, \frac{k_2+\alpha_2}{n}\biggr) \biggr| \\ &\qquad\qquad \times n^2\binom{n-1}{k_{1}}\binom{n-1}{k_{2}}\frac{\Gamma(k_1+1)\Gamma(n-k_1)}{\Gamma(n+1)} \frac{\Gamma(k_2+1)\Gamma(n-k_2)}{\Gamma(n+1)} \\ &=\mathsf E\! \sum_{\substack{0 \leqslant k_1 < n\\ 0 \leqslant k_2 < n}} \biggl| \Delta_{(x_1)}\Delta_{(x_2)}f^{\varepsilon} \biggl(\frac{k_1+\alpha_1}{n},\frac{k_2+\alpha_2}{n}\biggr)- \Delta_{(x_1)}\Delta_{(x_2)}f\biggl(\frac{k_1+\alpha_1}{n}, \frac{k_2+\alpha_2}{n}\biggr) \biggr| \\ &\qquad\qquad \times\binom{n-1}{k_{1}}\binom{n-1}{k_{2}}\frac{k_1!\,(n-k_1-1)!}{(n-1)!} \frac{k_2!\,(n-k_2-1)!}{(n-1)!} \\ &=\sum_{\substack{0 \leqslant k_1 < n\\ 0 \leqslant k_2 < n}} {\mathsf E} \biggl| \Delta_{(x_1)}\Delta_{(x_2)}f^{\varepsilon} \biggl(\frac{k_1\,{+}\,\alpha_1}{n},\frac{k_2\,{+}\,\alpha_2}{n}\biggr)\,{-}\, \Delta_{(x_1)}\Delta_{(x_2)}f\biggl(\frac{k_1\,{+}\,\alpha_1}{n}, \frac{k_2\,{+}\,\alpha_2}{n}\biggr) \biggr|. \end{aligned}
\end{equation*}
\notag
$$
We consider separately each term of this double sum. Writing the expectation as an integral (by the properties of a uniform distribution on $[0,1]$) and changing the variables by the formula $a_i/n=a'_i$, $i=1,2$, we find that for $0\leqslant k_1,k_2<n$ we have the equalities
$$
\begin{equation*}
\begin{aligned} \, & {\mathsf E} \biggl| \Delta_{(x_1)}\Delta_{(x_2)}f^{\varepsilon} \biggl(\frac{k_1+\alpha_1}{n},\frac{k_2+\alpha_2}{n}\biggr)- \Delta_{(x_1)}\Delta_{(x_2)}f\biggl(\frac{k_1+\alpha_1}{n}, \frac{k_2+\alpha_2}{n}\biggr) \biggr| \\ &\quad =\int_{0}^{1} d{a_1} \int_{0}^{1} d{a_2}\, \biggl|\Delta_{(x_1)}\Delta_{(x_2)}f^{\varepsilon}\biggl( \frac{a_1+k_1}{n},\frac{a_2+k_2}{n}\biggr) \\ &\quad\qquad -\Delta_{(x_1)} \Delta_{(x_2)}f\biggl(\frac{a_1+k_1}{n},\frac{a_2+k_2}{n} \biggr) \biggr| \\ &\quad =n^2 \int_{k_1/n}^{(k_1+1)/n} d{a'_{1}} \int_{k_2/n}^{(k_2+1)/n} d{a'_{2}}\, \bigl|\Delta_{(x_1)}\Delta_{(x_2)}f^{\varepsilon}(a'_{1},a'_{2}) -\Delta_{(x_1)}\Delta_{(x_2)} f(a'_{1},a'_{2}) \bigr| \end{aligned}
\end{equation*}
\notag
$$
(for the evaluation of the expectation the random variables $\alpha_1$ and $\alpha_2$ are replaced by the variables of integration $a_1$ and $a_2$, respectively). Now summing over $0\leqslant k_1,k_2<n$ we have
$$
\begin{equation}
\begin{aligned} \, &\sum_{\substack{0 \leqslant k_1 < n\\ 0 \leqslant k_2 < n}} \int_{0}^{1} d{a_1} \int_{0}^{1} \biggl| \Delta_{(x_1)}\Delta_{(x_2)}f^{\varepsilon} \biggl(\frac{k_1+a_1}{n},\frac{k_2+a_2}{n}\biggr) \nonumber \\ &\qquad\qquad -\Delta_{(x_1)}\Delta_{(x_2)} f\biggl(\frac{k_1+a_1}{n},\frac{k_2+a_2}{n} \biggr) d{\alpha_2} \bigg| \nonumber \\ &\qquad =n^2 \int_{0}^{1}da_{1} \int_{0}^{1}da_{2} \, \bigl| \Delta_{(x_1)}\Delta_{(x_2)}f^{\varepsilon}(a_{1},a_{2}) -\Delta_{(x_1)}\Delta_{(x_2)} f(a_{1},a_{2}) \bigr|. \end{aligned}
\end{equation}
\tag{5.4}
$$
From the last expression we see that the convergence
$$
\begin{equation}
\lim_{\varepsilon\to 0} \limsup_{n\to\infty}I_3^{n, \varepsilon}=0
\end{equation}
\tag{5.5}
$$
is proved by precisely the same argument as in the proof of the convergence of $I_2^{n, \varepsilon}$ to zero, with $x_i$ replaced by $a_i$, $i=1,2$. Now the convergence for an arbitrary small $\varepsilon>0$, which was fixed above for the integrals $I_2^{n, \varepsilon}$ and $I_3^{n, \varepsilon}$, follows from Proposition 4 for $d=2$ and multi-index $k=(1,1)$, since $f^\varepsilon$ is smooth.So, using (5.3), (5.5) and (5.6) we have
$$
\begin{equation*}
\lim_{\varepsilon\to 0} \limsup_{n\to\infty}( I_1^{n, \varepsilon}+I_2^{n, \varepsilon}+I_3^{n, \varepsilon})=0.
\end{equation*}
\notag
$$
Hence which proves the required equality
$$
\begin{equation*}
\lim_{n\to\infty} \mathsf E \int_0^1 dx_1 \int_0^1 dx_2 \, \biggl| \frac{\partial^2 \widetilde{B}_{\alpha,n}(f,x_1,x_2)}{\partial x_1 \,\partial x_2}- \frac{\partial^2 f(x_1,x_2)}{\partial x_1 \,\partial x_2}\biggr|=0.
\end{equation*}
\notag
$$
This completes the proof of the theorem. § 6. Some comments on mixed derivativesWe present some theorems on mixed derivatives, which are formally different but close in spirit to the corollary in § 3. We also state a related result, due to Stepanov, on the first-order derivative. Proposition 5 (see [17], Theorem 8.2.3). Let the function $f\colon G \mapsto \mathbb R$ have the partial derivatives $f_{xy}$ and $f_{yx}$ in a domain $G$. Then these derivatives are equal at each point $(x_0,y_0)\in G$ where they are continuous: A useful extension will be given in Proposition 10, which is Exercise 8.4.2b in [17]. In that result, it is not assumed a priori that both partial derivatives exist at a point. Instead, it is assumed that only one of these derivatives exists; the existence of the other follows from the assertion of the proposition. In the following classical theorem it is also assumed that the second differential exists at a point. Proposition 6 (Young [18]). Let $f\colon D \subset \mathbb{R}^{2} \to \mathbb{R}$ and $(x^0_1, x^0_2) \in D$. Assume that the first-order partial derivatives $ \partial f/\partial x_1$ and $\partial f/\partial x_2$ of $f$ exist in some neighbourhood of $(x^0_1, x^0_2)$ and are differentiable at $(x^0_1, x^0_2)$. Then the second-order mixed derivatives of $f$ at $(x^0_1, x^0_2)$ are equal: Young’s original (very long) proof of 1908 is quite obscure, as it sometimes happens to pioneering results. For a modern elegant and short argument, see § 8.12.3 of [19]. For completeness we state this result. Proposition 7 (see [19], § 8.12.3). Let $G$ be an open subset of $\mathbb R^n$. If a map $f$ from $G$ to a Banach space $F$ is twice differentiable at $x_0$, then the partial derivatives $D_i f$ are differentiable at $x_0$ and Proposition 8 (Aksoy and Martelli [20]). The following assertions are equivalent. (1) Let $g \in C(U, \mathbb R^2)$ and $[a, b] * [c, d] \subset U$. Then
$$
\begin{equation*}
\int_a^b \int_c^d g(x, y)\,dy\,dx=\int_c^d \int_a^b g(x, y)\,dx\,dy.
\end{equation*}
\notag
$$
(2) Let $f \in C^1(U, \mathbb R^2)$, and let the mixed derivative $f_{xy} \in C(U, R^2)$ exist. Then the derivative $f_{yx}$ also exists and Proposition 9 (see [21] and [22], Theorem 3.1.9). Let $E\subset G \subset \mathbb{R}^{m}$ be a measurable set, where $G$ is open, let $f\colon G \to \mathbb{R}^{n}$ be a locally bounded measurable function, and let $\limsup_{x\to a}\dfrac{|f(x)-f(a)|}{|x-a|} < \infty$ for almost all points $a\in E$. Then $f$ is differentiable at $L_m$-almost all points of $E$, where $L_m$ is the m-dimensional Lebesgue measure. Of course, this proposition also applies to the second derivatives. However, the problem of the equality of mixed derivatives is not considered in this statement. Proposition 10 (see [17], Exercise 8.4.2b). Let the function $f$ have partial derivatives $f_x$ and $f_y$ in some neighbourhood of a point $(x_0,y_0)$. Assume that the mixed derivative $f_{xy}$ (or $f_{yx}$) exists in $U$ and is continuous at $(x_0,y_0)$. Then the derivative $f_{yx}$ ($f_{xy}$, respectively) also exists at this point and Proposition 11 (Tolstov [23], Theorem 7). Let $f(x_1, x_2)$ be a linearly continuous measurable function (that is, $f$ is continuous in each variable) in a domain $G$ and, as concerns its partial derivatives $\dfrac{\partial f(x_1,x_2)}{\partial x_1}$ and $\dfrac{\partial f(x_1,x_2)}{\partial x_2}$, let each derivative with respect to each variable be finite on a set $E$ of positive two-dimensional measure. Then the mixed derivatives $\dfrac{\partial^2 f(x_1,x_2)}{\partial x_1 \,\partial x_2}$ and $\dfrac{\partial^2 f(x_1,x_2)}{\partial x_2 \,\partial x_1}$ exist and agree almost everywhere on $E$. Proposition 12 (Tolstov [23], Theorem 8). Let $f(x_1, x_2)$ be a linearly continuous function, and let the derivatives $\dfrac{\partial^2 f(x_1, x_2)}{\partial x^2_1}$, $\dfrac{\partial^2 f(x_1, x_2)}{\partial x_1 \,\partial x_2}$, $\dfrac{\partial^2 f(x_1, x_2)}{\partial x_2 \,\partial x_1}$ and $\dfrac{\partial^2 f(x_1, x_2)}{\partial x^2_2}$ exist in a domain $G$. Then almost everywhere on $G$. Now the problem of the equality of two mixed derivatives can be solved quite easily (modulo Schwarz’s or Young’s theorems; see Propositions 5 and 6 above) in the case of Sobolev derivatives. For the convenience of the reader, we recall that a function $g$ is the second-order mixed Sobolev derivative of $f$ with respect to the variables $x$ and $y$ in $L_p(G)$ if there exists a sequence of smooth functions $f^n \in C^\infty(G)$ such that andThe following elementary result is a consequence of the fact that $f_{xy}^\varepsilon = f_{yx}^\varepsilon$ everywhere if $f^\varepsilon$ is a smooth function. Proposition 13 (on mixed Sobolev derivatives). Let the $L_p(G)$-function $f$, $p>0$, have a mixed Sobolev derivative $f_{xy}\in L_p(G)$. Then the Sobolev derivative $f_{yx}\in L_p(G)$ also exists and the functions $f_{yx}$ and $f_{xy}$ are equal almost everywhere on $G$, that is, $\|f_{yx}-f_{xy}\|_{L_p(G)}=0$. AcknowledgementThe authors are grateful to an anonymous referee for several helpful comments and suggestions. 
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