Convex functions converge uniformly. Then their derivatives too? The following seems true to me:
Question) Let $D$ be an open convex domain in $R^d$, and let $(f_n)$ be a sequence of convex functions which converge uniformly to $f$ in $D$. Then does $(\nabla f_n)$ uniformly converge to $\nabla f$ on any compact subset of $D$?
Since functions might not be differentiable, then this means that  the distance of the set of subdifferentials converge to zero uniformly, that is, $\sup_{x \in K}$ dist$(\partial f_n (x), \partial f(x))$ goes to zero, where $K$ is a compact set in $D$ and $\partial f(x)$ is the subdifferential set of $f$ at $x$.
(But you may just assume that everything is differentiable.)
Is this true, and if so, how to prove it?
As the counterexamples in the answers indicate, the statement above is wrong. But I think the following weaker statement is true:
Question) Let $D$ be an open convex domain in $R^d$, and let $(f_n)$ be a sequence of convex functions which converge uniformly to $f$ in $D$. Then does $(\nabla f_n)$ converge pointwise to $\nabla f$ in $D$? (again in the sense that dist$(\partial f_n (x), \partial f(x)) \to 0$ for each $x \in D$)
And in fact, I think that even pointwise convergence of $f_n$ to $f$ can still imply the pointwise convergence of the derivatives.
Is this true, and if so, how to prove it?
 A: Right, the statement is wrong and there is a very simple counterexample:
$f(x) = |x|$ on $R$, and
$f_n(x) = f(x)$ if $|x| \ge 1/n$ while $f_n(x) = 1/n$ if $|x| \le 1/n$.
Then obviously $f_n$ uniformly converges to $f$, while $\nabla f_n$ does not converge uniformly to $\nabla f$ on any neighborhood of $0$.
But in this example still $\nabla f_n$ converges pointwise to $\nabla f$, so I added a modified question in the main article. Do you think this is true?
A: We will construct counterexample 
(1) Recall the definition : If $g$ is convex, then we have $$d_p g\
v =\lim_{t>0,\ t\rightarrow 0}\ \frac{g(p+tv)-g(p)}{t} $$
(2) We will construct sequence $f_n$ on $\mathbb{R}^2 $ : $$ f
(x,y)=|y| $$
And $f_n$ is obtained by infinite thin cylinder and $f$ More
precisely $$ f_n(x,y)=f_n(x,-y) $$ $$ y>0,\
f_n(x,y)=\frac{\sqrt{2}-\cos\ \theta }{n} =\frac{\sqrt{2}
-\sqrt{1-n^2y^2}}{n} ,\ \frac{ \sin\ \theta }{n} =y $$ for $$ U:=\{
(x,y)| |y|\leq \frac{1}{\sqrt{2}n} \}$$ and
$$f_n|\mathbb{R}^2-U=f$$
Clearly by considering epigraph of $f_n,\ f$ we know that $f_n,\ f$
are convex And uniform convergence is also clear
(3) If $p=(x_0,0)$ and $v=(0,1)$ then $$ d_pf\ v=1,\ d_pf_n\ v=0$$
