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Suppose that some function $f$ is two times differentiable and assume that $f(0)=0$, $f(1)=1$ and $f'(0)=f'(1)=0$. Then there exists an $x \in [0,1]$ such that $|f''(x)| > 4$.

Applying the MVT yields that there exists a $c \in (0,1)$ such that $$ f'(c) = \frac{f(1) - f(0)}{1} = 1. $$ Since $f$ is twice differentiable, we can apply the MVT to the intervals $(0,c)$ and $(c,1)$ to yield $c_1 \in (0,c)$ and $c_2 \in (c,1)$ such that $$ f''(c_1) = \frac{f'(c) - f'(0)}{c} = 1/c, $$ and $$ f''(c_2) = \frac{f'(1) - f'(c)}{1-c} = - \frac{1}{1-c}. $$ But now I'm stuck. How do I use this information?

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Are you sure the inequality in $|f''(x)|>4$ is strict? – wj32 Oct 23 '12 at 6:55
Yes it is strict. – somebody Oct 23 '12 at 7:42
up vote 1 down vote accepted

Here is a laborious proof by contradiction.

Suppose $|f''(x)| \leq 4$ for all $x$.

Define $\phi(x) = f(x)-x$. Clearly $\phi'(x) = f'(x)-1$, $\phi''(x) = f''(x)$.

Furthermore, $\phi(0) = \phi(1) = 0$, $\phi'(0) = \phi'(1) = -1$.

The MVT gives $\phi(x) = \phi(0)+\phi'(0) x + \frac{1}{2}\phi''(c_x) x^2$ for some $c_x \in [0,x]$, which yields the estimate (using $|\phi''(x)| \leq 4$) $\phi(x) \leq -x + 2 x^2$. In particular, $\phi(x) <0$ for $x\in (0,\frac{1}{2})$. A similar analysis, expanded around $x=1$ shows that $\phi(x) >0$ for $x\in (\frac{1}{2}, 1)$. It follows that $\phi(\frac{1}{2}) = 0$.

Now let $\eta(x) = -x + 2 x^2-\phi(x)$. From above, we have $\eta(x)\geq 0$ for $x\in [0,\frac{1}{2}]$. We will show that, in fact, $\eta(x) = 0$. We have $\eta(0) = \eta(\frac{1}{2}) = 0$, hence $\eta$ is maximized at some $\hat{x} \in (0,\frac{1}{2})$. Then using the MVT, we have $\eta(\frac{1}{2}) = \eta(\hat{x}) + \eta'(\hat{x}) (\frac{1}{2}-\hat{x}) + \frac{1}{2}\eta''(c) (\frac{1}{2}-\hat{x})^2$. Since $\eta'(\hat{x})=0$ and $\eta''(c) \geq 0$, this implies that $\eta(\hat{x}) \leq \eta(\frac{1}{2}) = 0$. Hence $\eta(x) = 0$ for $x \in [0,\frac{1}{2}]$ and so $\phi(x) = -x+2x^2$ on this interval.

A similar analysis for the interval $[\frac{1}{2}, 1]$ shows that $\phi(x) = -2x^2+3x-1$ on this interval.

This gives $\phi'(x) = \begin{cases}4x-1, & x \in [0,\frac{1}{2}] \\ 3-4x, & x \in [\frac{1}{2}, 1] \end{cases}$, or $\phi'(x) = 1 -4 | x-\frac{1}{2}|$. Since $\phi'$ is not differentiable at $x=\frac{1}{2}$, $f'$ is not differentiable at $x=\frac{1}{2}$, which is a contradiction.

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Why does the MVT give $\phi(x) = \phi(0)+\phi'(0) x + \frac{1}{2}\phi''(c_x) x^2$ ? I seem to be getting $\phi(x) = \phi(0) + \phi'(x) x$ instead. – somebody Oct 23 '12 at 7:40
@somebody: Maybe Taylor polynomial? – wj32 Oct 23 '12 at 8:29
@wj32: but he said "The MVT gives..." – somebody Oct 23 '12 at 8:34
@somebody: I was just offering a suggestion. After all, Taylor's theorem is really an extension of the MVT. – wj32 Oct 23 '12 at 8:35
@somebody: I should have said Taylor's expansion, but like wj32 wrote, it is basically an extension of the MVT. – copper.hat Oct 23 '12 at 14:35

Applying the mean value theorem to $f'$ at some $x$ yields

$$ f''(c)=\frac{f'(x)-f'(0)}{x-0}=\frac{f'(x)}x $$

for some $c$ with $0\le c\le x$. Thus if $|f''(x)|\le4$ everywhere, then $f'(x)\le4x$ everywhere. But $f'$ is differentiable and hence continuous, so we can apply the fundamental theorem of calculus and the monotonicity of integration to get

$$ f(1/2)=\int_0^{1/2}f'(x)\,\mathrm dx\le\int_0^{1/2}4x\,\mathrm dx=\frac12\;. $$

Now you just need to argue that if $f'(x)\lt4x$ at some point, then by continuity this holds in some neighbourhood of that point, and this would reduce $f(1/2)$ below $1/2$; then an analogous argument for $[1/2,1]$ would lead to a contradiction. That only leaves the possibility

$$ f(x)=\begin{cases}2x^2&x\le1/2\;,\\1-2(1-x)^2&x\gt1/2\;,\end{cases} $$

and this is not twice differentiable at $x=1/2$.

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+1 Very smooth. – copper.hat Oct 23 '12 at 14:46

The idea is this. Consider the two isosceles triangles of height $2$ and with base as the line segment $\overline{0\text{ }1}$. If $f''(x) < 4$ then the graph of $f'(x)$ will never leave the interior of these triangles, because if it does, then it has to enter back again at some point . Then you can apply mean value theorem to arrive at a contradiction because slopes of the sides of the traingles are $\pm 4$ (If its entry is through a different side, it still would have cut the line $y=\pm 4x$ twice). But the maximum positive area under the graph if it stays in the interior of the triangles is $<1$. But it should have been at least $\int _0^1f(x)dx=1$

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Assume that $|f''(x)|\leq 4$ for all $x\in[0,1]$. Using the integral form for the remainder in Taylor's theorem we have $$f\Bigl({1\over2}\Bigr)=\int_0^{1/2} f''(t)\Bigl({1\over2}-t\Bigr)\ dt\leq 4\int_0^{1/2}\tau\ d\tau={1\over2}\ ,$$ with strict inequality if $f''(t)\not\equiv 4$ for $0\leq t\leq{1\over2}$, in particular, if $f''\bigl({1\over2}\bigr)\ne4$. By symmetry we also have $f\bigl({1\over2}\bigr)>{1\over2}$ if $f''\bigl({1\over2}\bigr)\ne -4$.

As no value for $f''\bigl({1\over2}\bigr)$ leads to an admissible value for $f\bigl({1\over2}\bigr)$, such a function cannot exist.

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My solution originally used the integral form, but you need to justify that $f''$ is integrable. – copper.hat Oct 23 '12 at 14:36
+1 It is simpler than I thought, since $f''$ is bounded, $f'$ is Lipschitz, hence absolutely continuous. – copper.hat Oct 23 '12 at 14:49

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