Estimate of the derivative of a function that is supremum of a function of two variables I tried to show the following estimation and am not sure, if it holds:

Given $U\subseteq\mathbb R$ and $f:\mathbb R\times \mathbb R\to\mathbb R$. Its derivative with respect to the second argument exists and is denoted by $f_b(a,b)$:
$\displaystyle\frac{d}{db}\left(\sup_{a\in U}f(a,b)\right)\ge \inf_{a\in U}f_b(a,b)$

My proof:
There exists a sequence $a_n$ in $U$ such that
$\displaystyle\lim_{n\to\infty}f(a_n,b)=\sup_{a\in U}f(a,b)$
$\forall n\in\mathbb N$ the following holds:
$\sup_{a\in U}f(a,b+h)-\sup_{a\in U}f(a,b)\ge f(a_n,b+h)-\sup_{a\in U}f(a,b)$
Using the differentiability of $f$:
$=f(a_n,b)+h f_b(a_n,b)-\sup_{a\in U}f(a,b)+\mathcal O(h^2)$
This is always greater than
$\displaystyle\ge f(a_n,b)+h\;\inf_{a\in U} f_b(a,b)-\sup_{a\in U}f(a,b)+\mathcal O(h^2)$
As this holds for an arbitrary $n$ and thus $f(a_n,n)$ comes arbitrarily close to $\sup_{a\in U}f(a,b)$ it holds:
$\sup_{a\in U}f(a,b+h)-\sup_{a\in U}f(a,b)\ge h\;\inf_{a\in U} f_b(a,b)+\mathcal O(h^2)$
Deviding by $h$ and performing the limit yields the assertion.

Before estimating $\displaystyle\frac{d}{db}\left(\sup_{a\in U}f(a,b)\right)$ one should probably ask if it exists. For example, if $f(a,b)=ab$ and $U=(-1,1)$, then $\sup_{a\in U}f(a,b)=|b|$, which is not differentiable at $0$. 
But even assuming the differentiability of the envelope, the result is false as stated. For example, the zero function on $(-1,1)$ is the supremum of functions $\min(0, a^{-1}((b-a)^2-a^2))$ over $0<a<1$. At $b=0$ each of these has $b$-derivative equal to $1/2$; thus, the infimum of $b$-derivative is $1/2$  while the derivative of supremum is $0$.
The problem with the proof is hidden by the $\mathcal{O}$ symbol, the uniformity of which with respect to $n$ was not discussed. One needs the assumption of uniform differentiability to make it work.
