Symmetry of second derivative - Sufficiency of twice-differentiability 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.
 A: 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*}
