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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?

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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.

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  • $\begingroup$ Thanks! Silly question, sorry. A function which is flat on a sub-interval probably does it as well. $\endgroup$ – username Oct 11 '13 at 11:11

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