Consider the most classical form of the Mean Value Theorem: given a positive continuous function $f\in C([0,2])$ and a continuous function $g\in C([0,2])$, there exist $c\in(\frac{1}{2},\frac{3}{2})$ such that
$$
\int\_{\frac{1}{2}}^{\frac{3}{2}} f(t)g(t) dt = 
g(c)\int\_{\frac{1}{2}}^{\frac{3}{2}} f(t) dt.
$$
Of course, this $c$ need not be unique. Now, consider instead 
$$
\int\_{\frac{1}{2}+x}^{\frac{3}{2}+x} f(t)g(t) dt = 
g(c(x))\int\_{\frac{1}{2}+x}^{\frac{3}{2}+x} f(t) dt.
$$
with $x\in(0,\frac{1}{2})$. Is it possible to choose $c(x)$ so that it is continuous? 
And if $g\in C^{0,\alpha}([0,2])$, can $c$ be $C^{0,\alpha}([0,\frac{1}{2}])$ as well? 



Regularity for the Mean Value Theorem
No. Take $f \equiv 1$ and define $g(x)$ as follows:


$$ g(x) = \begin{cases} 100 x & x \leq 1/2 \\
50 - (2x - 1) & 1/2 < x \leq 1 \\
49 + (2x - 2) & 1 < x< \leq 3/2 \\
50 - 50(2x - 3) \end{cases} $$


Since $g$ is piecewise linear and continuous, it is Lipschitz continuous. 


We compute 


$$ \int\_{1/2}^{3/2} g(t) \mathrm{d} t = 49.5 $$


which is attained at $c \in \{3/4,5/4\}$. 


It is easy to see, by graphing the function $g(x)$, that 
$$ \int\_{1/2 + x}^{3/2 + x} g(t) \mathrm{d}t $$
is decreasing as a function of $x$. Which means that regardless of which choice of $c \in \{3/4,5/4\}$ your curve $c(x)$, as $x$ increases, moves toward $1/2$. However, you see that 
$$ \int\_1^2 g(t) \mathrm{d}t = 149/4 < 49 $$
so it is impossible that $c(x)$ is continuous. 


