Generalization of Theorem (from $ \mathbb R$ to $ \mathbb R^n$) I've wondering about the generalization of this, in my opinion, beautiful theorem, and I tried to write it but I don't know if it's correct.
Theorem:
Let $f_n:(a,b) \rightarrow \mathbb R$ differentiable functions such that:
1) There exist $g:(a,b) \rightarrow \mathbb R$ such that $f'_n$ $ \rightarrow$ $g$ uniformly over (a,b).
2)There exist $x_0 \in(a,b)$ such that the sequence $\{f_n(x_0)\}$ converges.
Then there exist $f_n:(a,b)  \rightarrow \mathbb R$ such that $f_n \rightarrow f$ uniformly over $(a,b)$, $f$ is differentiable and $f'=g$.
Now my attempt of the generalization is the following:
Let $A\subset \mathbb R^n$, open, convex, bounded and $f_n:A  \rightarrow \mathbb R^m$ differentiable.
1)There exist h: $\mathbb R^n\rightarrow \mathbb R^m$ such that $T_n \rightarrow H$ where $T_n$ is the differential of $f_n$.
2)There exist $x_0 \in $ A such that the sequence $\{f_n(x_0)\}$ converges. 
Then exist $f:A  \rightarrow \mathbb R^m$ such that $f_n \rightarrow f$ uniformly over A, $f$ is differentiable and $T_n$=h.
My questions are: Is my generalization correct? is it wrong? 
If it's correct, could have been written in a more elegant way?
Also which would be the the proof (for the generalization theorem)?
:)
 A: \begin{align*}
&  \frac{f(x)-f(x_{0})-H(x_{0})\cdot(x-x_{0})}{\Vert x-x_{0\Vert}}%
=\frac{f(x)-f(x_{0})-[f_{n}(x)-f_{n}(x_{0})]}{\Vert x-x_{0\Vert}}\\
&  +\frac{f_{n}(x)-f_{n}(x_{0})-\nabla f_{n}(x_{0})\cdot(x-x_{0})}{\Vert
x-x_{0\Vert}}+\frac{(\nabla f_{n}(x_{0})-H(x_{0}))\cdot(x-x_{0})}{\Vert
x-x_{0\Vert}}\\
&  =:I+II+III.
\end{align*}
Since $A$ is convex, by applying the mean value theorem to the function
$$
g_{n,m}(t)=f_{m}(tx+(1-t)x_{0})-f_{n}(tx+(1-t)x_{0}),\quad t\in\lbrack0,1]
$$
there is $t_{0}$ such that
\begin{align*}
&  f_{m}(x)-f_{m}(x_{0})-[f_{n}(x)-f_{n}(x_{0})]=g_{n,m}(1)-g_{n,m}(0)\\
&  =g_{n,m}^{\prime}(t_{0})=(\nabla f_{m}(z_{0})-\nabla f_{n}(z_{0}%
))\cdot(x-x_{0}),
\end{align*}
where $z_{0}=t_{0}x+(1-t_{0})x_{0}$. By uniform convergence of the gradients,
$$
\Vert\nabla f_{m}(z)-\nabla f_{n}(z)\Vert\leq\Vert\nabla f_{m}(z)-H(z)\Vert
+\Vert\nabla f_{n}(z)-H(z)\Vert\leq2\varepsilon
$$
for all $n,m\geq n_{\varepsilon}$ and all $z\in A$. Hence, by Cauchy's
inequality
\begin{align*}
\left\vert \frac{f_{m}(x)-f_{m}(x_{0})-[f_{n}(x)-f_{n}(x_{0})]}{\Vert
x-x_{0\Vert}}\right\vert  & =\left\vert \frac{(\nabla f_{m}(z_{0})-\nabla
f_{n}(z_{0}))\cdot(x-x_{0})}{\Vert x-x_{0\Vert}}\right\vert \\
& \leq\Vert\nabla f_{m}(z_{0})-\nabla f_{n}(z_{0})\Vert\leq2\varepsilon.
\end{align*}
Since $A$ is bounded, this inequality implies that
\begin{align*}
\vert f_{m}(x)-f_{n}(x)\vert\le|f_{m}(x_{0})-f_{n}(x_{0})|+\Vert
x-x_{0}\Vert  2\varepsilon\le |f_{m}(x_{0})-f_{n}(x_{0})|+2M\varepsilon.
\end{align*}
and so $\{f_n\}$ is a uniform Cauchy sequence and so it converges uniformly to a function $f$. Letting $m\rightarrow\infty$ we get $$
\left\vert \frac{f(x)-f(x_{0})-[f_{n}(x)-f_{n}(x_{0})]}{\Vert x-x_{0\Vert}%
}\right\vert \leq2\varepsilon
$$
for all $n\geq n_{\varepsilon}$. This takes care of $I$. Taking
$n=n_{\varepsilon}$ and using the fact that $f_{n_{\varepsilon}}$ is
differentiable at $x_{0}$ we get that
$$
\left\vert \frac{f_{n_{\varepsilon}}(x)-f_{n_{\varepsilon}}(x_{0})-\nabla
f_{n_{\varepsilon}}(x_{0})\cdot(x-x_{0})}{\Vert x-x_{0\Vert}}\right\vert
\leq\varepsilon
$$
for all $x\in A$ with $0<\Vert x-x_{0}\Vert\leq\delta_{\varepsilon}$. This
takes care of $II$. 
Lastly, by Cauchy's inequality
$$
\left\vert \frac{(\nabla f_{n}(x_{0})-H(x_{0}))\cdot(x-x_{0})}{\Vert
x-x_{0\Vert}}\right\vert \leq\Vert\nabla f_{n}(x_{0})-H(x_{0})\Vert
\leq\varepsilon
$$
for all $n\geq n_{\varepsilon}$. In conclusion we have that for all $x\in A$
with $0<\Vert x-x_{0}\Vert\leq\delta_{\varepsilon}$,
$$
\left\vert \frac{f(x)-f(x_{0})-H(x_{0})\cdot(x-x_{0})}{\Vert x-x_{0\Vert}%
}\right\vert \leq4\varepsilon
$$
which implies that $f$ is differentiable at $x_{0}$ with $\nabla
f(x_{0})=H(x_{0})$. By repeating the proof with $x_0$ replaced by any other point, we get that $f$ is differentiable in $A$.
