Symmetry of second derivative states that for $u=u(x,y)$ if $u_x,u_y$ exists and $u_{xy},u_{yx}$ exists and continuous then $u_{xy}=u_{yx}$.
I proved that statement using the mean value theorem.
While I was looking in Wikipedia there is a section called "Sufficiency of twice-differentiability", if $u(x,y):E^{\text{open set}}\subset \Bbb R^2\to \Bbb R$ and $u_x,u_y,u_{yx}$ exists everywhere and $u_{yx}$ is continuous at a point in $E$ then $u_{xy}$ exists at that point and equal to $u_{yx}$.
My question is, while I proved Symmetry of second derivative I had to assume the continuity of $u_{xy}$ and $u_{yx}$, so how can I prove that this is true without even assuming the existence of one of the second derivative? I'm sitting on this for a long time and I couldn't think on any starting point and would love help with this.
Lemma Let $A:\left( \left( -r,r\right) \setminus\left\{ 0\right\} \right) \times\left( \left( -r,r\right) \setminus\left\{ 0\right\} \right) \rightarrow\mathbb{R}$. Assume that the double limit $\lim_{\left( s,t\right) \rightarrow\left( 0,0\right) }A\left( s,t\right) $ exists in $\mathbb{R}$ and that the limit $\lim_{t\rightarrow0}A\left( s,t\right) $ exists in $\mathbb{R}$ for all $s\in\left( -r,r\right) \setminus\left\{ 0\right\} $. Then the iterated limit $\lim_{s\rightarrow0}\lim_{t\rightarrow 0}A\left( s,t\right) $ exists and $$ \lim_{s\rightarrow0}\lim_{t\rightarrow0}A\left( s,t\right) =\lim_{\left( s,t\right) \rightarrow\left( 0,0\right) }A\left( s,t\right) . $$
Proof Let $\ell=\lim_{\left( s,t\right) \rightarrow\left( 0,0\right) }A\left( s,t\right) $. Then for every $\varepsilon>0$ there exists $\delta =\delta\left( \left( 0,0\right) ,\varepsilon\right) >0$ such that $$ \left\vert A\left( s,t\right) -\ell\right\vert \leq\varepsilon $$ for all $\left( s,t\right) \in\left( \left( -r,r\right) \setminus\left\{ 0\right\} \right) \times\left( \left( -r,r\right) \setminus\left\{ 0\right\} \right) $, with $\sqrt{\left\vert s-0\right\vert ^{2}+\left\vert t-0\right\vert ^{2}}\leq\delta$.
Fix $s\in\left( -\frac{\delta}{2},\frac{\delta}{2}\right) \setminus\left\{ 0\right\} $. Then for all $t\in\left( -\frac{\delta}{2},\frac{\delta} {2}\right) \setminus\left\{ 0\right\} $, $$ \left\vert A\left( s,t\right) -\ell\right\vert \leq\varepsilon $$ and so letting $t\rightarrow0$ in the previous inequality (and using the fact that the limit $\lim_{t\rightarrow0}A\left( s,t\right) $ exists), we get $$ \left\vert \lim_{t\rightarrow0}A\left( s,t\right) -\ell\right\vert \leq\varepsilon $$ for all $s\in\left( -\frac{\delta}{2},\frac{\delta}{2}\right) \setminus \left\{ 0\right\} $. But this implies that there exists $\lim_{s\rightarrow 0}\lim_{t\rightarrow0}A\left( s,t\right) =\ell$.
Proof of the Theorem
Let $\left\vert t\right\vert ,\left\vert s\right\vert <\frac{r}{\sqrt{2}}$. Then the points $\left( x_{0}+s,y_{0}\right) $, $\left( x_{0}+s,y_{0}+t\right) $, and $\left( x_{0},y_{0}+t\right) $ belong to $B\left( \left( x_{0},y_{0}\right) ,r\right) $. Define \begin{align*} A\left( s,t\right) & :=\frac{u\left( x_{0}+s,y_{0}+t\right) -u\left( x_{0}+s,y_{0}\right) -u\left( x_{0},y_{0}+t\right) +u\left( x_{0}% ,y_{0}\right) }{st},\\ g\left( x\right) & :=u\left( x,y_{0}+t\right) -u\left( x,y_{0}\right) . \end{align*} By the mean value theorem $$ A\left( s,t\right) =\frac{g\left( x_{0}+s\right) -g\left( x_{0}\right) }{st}=\frac{g^{\prime}\left( \xi\right) }{t}=\frac{\frac{\partial u}{\partial x}\left( \xi_{t},y_{0}+t\right) -\frac{\partial u}{\partial x}\left( \xi_{t},y_{0}\right) }{t}% $$ where $\xi$ is between $x_{0}$ and $x_{0}+t$. Fix $t$ and consider the function $$ h\left( y\right) :=\frac{\partial u}{\partial x}\left( \xi_{t},y\right) . $$ By the mean value theorem, $$ h\left( b\right) -h\left( a\right) =h^{\prime}\left( c\right) \left( b-a\right) =\frac{\partial^{2}u}{\partial y\partial x}\left( \xi _{t},c\right) \left( b-a\right) $$ for some $c$ between $a$ and $b$. Taking $b=t$ and $a=0$, we get$$ \frac{\partial u}{\partial x}\left( \xi_{t},y_{0}+t\right) -\frac{\partial u}{\partial x}\left( \xi_{t},y_{0}\right) =\frac{\partial^{2}u}{\partial y\partial x}\left( \xi_{t},\eta_{t}\right) t $$ where $\eta_{t}$ is between $y_{0}$ and $y_{0}+t$. Hence,$$ A\left( s,t\right) =\frac{\partial^{2}u}{\partial y\partial x}\left( \xi_{t},\eta_{t}\right) \rightarrow\frac{\partial^{2}u}{\partial y\partial x}\left( x_{0},y_{0}\right) , $$ where we have used the fact that $\left( \xi,\eta\right) \rightarrow\left( x_{0},y_{0}\right) $ as $\left( s,t\right) \rightarrow\left( 0,0\right) $ together with the continuity of $\frac{\partial^{2}u}{\partial y\partial x}$ at $\left( x_{0},y_{0}\right) $. Note that this shows that there exists the limit $$ \lim_{\left( s,t\right) \rightarrow\left( 0,0\right) }A\left( s,t\right) =\frac{\partial^{2}u}{\partial y\partial x}\left( x_{0},y_{0}\right) . $$ On the other hand, for all $s\neq0$, \begin{align*} \lim_{t\rightarrow0}A\left( s,t\right) & =\frac{1}{s}\lim_{t\rightarrow 0}\left[ \frac{u\left( x_{0}+s,y_{0}+t\right) -u\left( x_{0}% +s,y_{0}\right) }{t}-\frac{u\left( x_{0},y_{0}+t\right) -u\left( x_{0},y_{0}\right) }{t}\right] \\ & =\frac{\frac{\partial u}{\partial y}\left( x_{0}+s,y_{0}\right) -\frac{\partial u}{\partial y}\left( x_{0},y_{0}\right) }{s}. \end{align*} Hence, we are in a position to apply the previous lemma to obtain \begin{align*} \frac{\partial^{2}u}{\partial y\partial x}\left( x_{0},y_{0}\right) & =\lim_{\left( s,t\right) \rightarrow\left( 0,0\right) }A\left( s,t\right) =\lim_{s\rightarrow0}\lim_{t\rightarrow0}A\left( s,t\right) \\ & =\lim_{s\rightarrow0}\frac{\frac{\partial u}{\partial y}\left( x_{0}+s,y_{0}\right) -\frac{\partial u}{\partial y}\left( x_{0}% ,y_{0}\right) }{s}=\frac{\partial^{2}u}{\partial x\partial y}\left( x_{0},y_{0}\right) . \end{align*}