Sequence of differentiable maps being Cauchy sequence. Let $U\subset E$ (E is normed space finite dimensional or not )be an open convex and $f:U\to F$ where $F$ is a Banach space. I have a sequence of differentiable maps $(f_n)$ such that :


*

*There exist a point $a\in U$ such that $(f_n(a))$ converge

*There exist $x\in U$ and a open neighborhood $V\subset U$ of $x$ such that $$\lim_{n\to \infty}\sup_{y\in V}\Vert(Df_n)(y)-g(y)\Vert=0$$ where $g:U\to \mathcal{L}(E,F)$



Now, given $M>0$ and $\varepsilon>0$, I would like to prove that there exist $N_0\in \Bbb{N}$ such that $n,m\ge N_0$ and $\forall x\in U$ satisfying $\Vert x-a \Vert<M$, we have $$\Vert f_n(x)-f_m(x) \Vert\le \varepsilon.$$

I write
$$
\Vert f_n(x)-f_m(x)\Vert\le\Vert f_n(a)-f_m(a) \Vert+\Vert f_n(x)-f_n(a)-f_m(x)+f_m(a) \Vert,
$$
After that I tried to use the mean value inequality to $f_n-f_m:=h_{m,n}$ which gives us $$\Vert h_{m,n}(x)-h_{m,n}(a)\Vert\le\sup_{y\in [a,x]}\Vert Dh_{m,n}\Vert\Vert x-a\Vert.$$
Now I know that $[a,x]$ is compact ( continuous image of compact set), so I can use the property that every open cover of $[a,x]$ has a finite subcover. Anyway I am stuck because I have ( always) something that depends on $x$. 
Any ideas ?
 A: (This is a somewhat simplified version of the sequence I posted as a comment to the OP.)
Note that we don't have to let conditions 1 and 2 occur anywhere near each other, so let's not.
Start with $\hat{f}_n(x) = x^n$.  This is a standard example of a sequence of functions converging (nonuniformly) to $g \equiv 0$ on $U = (-1,1)$ whose derivatives also do so.  So this sequence satisfies condition 2.  If we offset these $\hat{f}$ vertically, their derivatives remain unchanged so the convergence $D\hat{f}_n \rightarrow g$ holds.
Now, how do we force convergence at $a = -1/2$ without disturbing $D\hat{f}(0)$ (much)?  Multiply through by $\mathrm{e}^{\left(x+\frac{1}{2}\right)^{-2}}$?  No, might modify $\hat{f}(0)$ too much.  Okay, successively narrow the distortion: $f_n(x) = \mathrm{e}^{\left(n\left(x+\frac{1}{2}\right)\right)^{-2}}(x^n - (-1)^n)$.
Checking that we haven't damaged the $Df_n(0)$ condition...  $Df_n(0) = \exp\left( \frac{-1}{n^2 (x+1/2)^2} \right) \hat{f}_n(x)$ and we can make that exponential as close to $1$ as we like on any (small) neighborhood of $x=0$ by taking $n$ large enough, so no damage there.
Now, does this sequence converge to something?  Not even close.  $f_n(0) \rightarrow 1$ for $n$ odd and $\rightarrow -1$ for $n$ even.  In fact, this is approximately the what the functions are doing "everywhere" on $U$.  They all swing toward zero at $x=-1$ (not in $U$) on a shrinking neighborhood, zero on a shrinking neighborhood of $-1/2$, and rising on a shrinking neighborhood of $x=1$.  (The odd $n$s rise from $1$ to $2$ and the evens from $-1$ to $0$.)  Everywhere in between, these functions are very close to $\pm 1$, according to whether $n$ is even or odd.  So $f_n$ and $f_{n+1}$ are further from each other in most norms as $n \rightarrow \infty$.  (In fact, this is a sequence with two limit points.)
Edit:  Another, similar solution:  $f_n(x) = (-1)^n \cos^{2/n}(x)$ where in the exponentiation by $2/n$ we mean to square first, then take the $n^\text{th}$ root of the resulting non-negative number.  This is another sequence trying to converge to the functions $\pm 1$, with derivative $\rightarrow 0$ on neighborhoods of $x=0$ and taking the value $0$ at $x = \pm \pi/2$.  We can think of this on $(-\pi,\pi) \subset \Bbb{R}$ or on the circle.
Notice that you can also simplify your setup:  Since the $f_n(a)$ converge, to $\hat{a}$ say, consider instead the sequence $(f_n-\hat{a})_n$.  This new sequence converges to $0$ at $a$.  Similarly, you can remove $g$ from all the sequence members so that $||Df_n(y)|| \rightarrow 0$ in your condition 2.  This should help you find many more counterexamples more directly.
A: If you really mean in 2. that there exists a specific $x$ such that those conditions hold, then you should label this point $x_0$ or something. If you just call it $x$ and then later use $x$ as a variable, which you seem to do, then confusion and errors can arise.
Assuming you actually mean a specific $x_0$: Let $a=0, x_0 = 1.$ Choose any $f\in C^\infty(\mathbb R)$ with $f(0)=0$ and $f(x) = 1$ for $x>1/2.$ Define $f_n(x) = nf(x).$ Then $f_n(a) = 0$ for all $n,$ and $f_n'\equiv 0$ in $V=(1/2,\infty).$ Thus we have 1. and 2. satisfied, but clearly the $f_n$'s fail to converge uniformly, or even pointwise, in every neighborhood of $a.$ (This seems be the same kind of thing @EricTowers is doing in his answer, although I haven't read it carefully).
Somehow I doubt this is the correct statement of the problem. You should make this clear.
