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Say you have a real function $h$ that is differentiable at every point and that $h(0)=0$ and $h(2)=0$ and $|h'(x)|\leq 1$ for all $x$. I know how the function looks like but how do you use the mean value theorem to show that $h(1) \leq 1$ and $h(1)$ cannot be $1$?

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Hint: Draw lines of slope $\pm 1$ at $x = 0$ and $x = 2$. By the Mean Value Theorem, no point in the set $\{ (x, y) \, | \, x \in (0, 2), \, y = f(x) \}$ lies outside that parallelogram.

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Hint: The tricky part is showing that $h(1) \neq 1$. Argue that if $h(1) = 1$ then you must have $h'(x) = 1$ for all $0 \leq x < 1$ and $h'(x) = -1$ for all $1 < x \leq 2$. But then how can $h$ be differentiable at $x = 1$?

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Assume $h(1)>1$. Then there exists $\xi\in(0,1)$ with $(1-0)h'(\xi)=h(1)-h(0)$, i.e. $h'(\xi)>1$ contrary to assumption.

Assume $h(1)=1$ and let $h'(1)<1$ then for sufficiently small positive $k$ we have $\frac{h(1-k)-h(1)}{-k}<1$, i.e. $h(1-k)>1-k$, and then for some $\xi\in(0,1-k)$ we find $h'(\xi)>1$ as above. The same works if $h'(1)>-1$ and working on the right half interval. As at least one of the inequalities holds, we are done.

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