$ \newcommand{\braket}[1]{\langle #1 \rangle} \newcommand{\abs}[2][]{\left\lvert#2\right\rvert_{\text{#1}}} \newcommand{\ket}[1]{\left\lvert#1 \right.\rangle} \newcommand{\bra}[1]{\langle\left. #1\right\rvert} \newcommand{\braket}[1]{\langle #1 \rangle} \newcommand{\dd}{\text{d}} \newcommand{\dv}[2]{\frac{\dd #1}{\dd #2}} \newcommand{\pdv}[2]{\frac{\partial}{\partial #1}} $
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Title

Let $A\subseteq\mathbb{R}^n$, $\vec{a}\in \text{int}(A)$, $f:A\to\mathbb{R}$, and $B_{\delta}(\vec{a})\in A$ for some $\delta >0$. Suppose that for some $i, j\in\{1,2, \dots, n\}$, the partial derivatives $\frac{\partial f}{\partial x_i}$, $\frac{\delta f}{\delta x_j}$, $\frac{\delta^2 f}{\delta x_j \delta x_i}$, $\frac{\partial^2 f}{\partial x_i \partial x_j}$ exists on some $B_{\delta}(\vec{a})$. If $\frac{\partial ^2 f}{\partial x_j \partial x_i}$ and $\frac{\partial ^2 f}{\partial x_i \partial x_j}$ are both continuous at $\vec{a}$, then $$\begin{align*} \frac{\partial ^2 f}{\partial x_j\partial x_i}(\vec{a})=\frac{\partial^2 f}{\partial x_i\partial x_j}(\vec{a}) \end{align*}$$

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Title

Let $A\subseteq\mathbb{R}^n$, $\vec{a}\in \text{int}(A)$, $f:A\to\mathbb{R}$, and $B_{\delta}(\vec{a})\in A$ for some $\delta >0$. Suppose that for some $i, j\in\{1,2, \dots, n\}$, the partial derivatives $\frac{\partial f}{\partial x_i}$, $\frac{\delta f}{\delta x_j}$, $\frac{\delta^2 f}{\delta x_j \delta x_i}$, $\frac{\partial^2 f}{\partial x_i \partial x_j}$ exists on some $B_{\delta}(\vec{a})$. If $\frac{\partial ^2 f}{\partial x_j \partial x_i}$ and $\frac{\partial ^2 f}{\partial x_i \partial x_j}$ are both continuous at $\vec{a}$, then $$\begin{align*} \frac{\partial ^2 f}{\partial x_j\partial x_i}(\vec{a})=\frac{\partial^2 f}{\partial x_i\partial x_j}(\vec{a}) \end{align*}$$

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proof

Title

Let $A\subseteq\mathbb{R}^n$, $\vec{a}\in \text{int}(A)$, $f:A\to\mathbb{R}$, and $B_{\delta}(\vec{a})\in A$ for some $\delta >0$. Suppose that for some $i, j\in\{1,2, \dots, n\}$, the partial derivatives $\frac{\partial f}{\partial x_i}$, $\frac{\delta f}{\delta x_j}$, $\frac{\delta^2 f}{\delta x_j \delta x_i}$, $\frac{\partial^2 f}{\partial x_i \partial x_j}$ exists on some $B_{\delta}(\vec{a})$. If $\frac{\partial ^2 f}{\partial x_j \partial x_i}$ and $\frac{\partial ^2 f}{\partial x_i \partial x_j}$ are both continuous at $\vec{a}$, then $$\begin{align*} \frac{\partial ^2 f}{\partial x_j\partial x_i}(\vec{a})=\frac{\partial^2 f}{\partial x_i\partial x_j}(\vec{a}) \end{align*}$$

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Title

Let $A\subseteq\mathbb{R}^n$, $\vec{a}\in \text{int}(A)$, $f:A\to\mathbb{R}$, and $B_{\delta}(\vec{a})\in A$ for some $\delta >0$. Suppose that for some $i, j\in\{1,2, \dots, n\}$, the partial derivatives $\frac{\partial f}{\partial x_i}$, $\frac{\delta f}{\delta x_j}$, $\frac{\delta^2 f}{\delta x_j \delta x_i}$, $\frac{\partial^2 f}{\partial x_i \partial x_j}$ exists on some $B_{\delta}(\vec{a})$. If $\frac{\partial ^2 f}{\partial x_j \partial x_i}$ and $\frac{\partial ^2 f}{\partial x_i \partial x_j}$ are both continuous at $\vec{a}$, then $$\begin{align*} \frac{\partial ^2 f}{\partial x_j\partial x_i}(\vec{a})=\frac{\partial^2 f}{\partial x_i\partial x_j}(\vec{a}) \end{align*}$$

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