Generalization of Theorem (from $ \mathbb R$ to $ \mathbb R^n$)

Question

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)?

:)

Answer 1

\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 $A$.