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My task is to find a function $h:[-1,1] \to \mathbb{R}$ so that

(i) $h(-1) = h(1) = 0$

(ii) $h$ is continuously differentiable on $[-1,1]$

(iii) $h$ is twice differentiable on $(-1,0) \cup (0,1)$

(iv) $|h^{\prime\prime}(x)| < 1$ for all $x \in (-1,0)\cup(0,1)$

(v) $|h(x)| > \frac{1}{2}$ for some $x \in [-1,1]$

The source I have says to use the function

$h(x) = \frac{3}{4}\left(1-x^{4/3}\right)$

which fails to satisfy condition (iv) so it is incorrect. I'm starting to doubt the validity of the problem statement because of this. So my question is does such a function exist? If not, why? Thanks!

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Why don't you play a bit? For instance, imagine $h(x)=\alpha (1-f(x))$, where $f$ is an even function (so that $f(-1)=f(1)$) with $f(1)=0$. You may also try $f(x)=x^q$, and look if there is an exponent $q$ that makes $h$ behave as you wish. – Siminore Jul 14 '12 at 12:58
up vote 3 down vote accepted

Let $h$ satisfy (i)-(iv) and $x_0$ be a point where the maximum of $|h|$ is attained on $[-1,1]\;$. WLOG we can assume that $x_0\ge0$. Then $$ |h(x_0)|=|h(x_0)-h(1)|=\left|\int_{x_0}^1h'(y)\,dy\right|= $$ $$ \left|\int_{x_0}^1\int_{x_0}^yh''(z)\,dz\;dy\right|\le \sup_{(0,1)}|h''|\int_{x_0}^1\int_{x_0}^y\,dz=\frac{(1-x_0)^2}2\sup_{(0,1)}|h''|\le \frac12. $$

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I think there's a $dy$ missing, and it seems you're assuming $h'(x_0)=0$? – joriki Jul 14 '12 at 12:56
@joriki thanks, fixed it. And $h'(x_0)=0$ since $x_0$ is a point where the minimum or the maximum of $h$ on $(-1,1)$ is attained. – Andrew Jul 14 '12 at 13:04
I see, I missed that. – joriki Jul 14 '12 at 13:22
I agree with this! Thanks!! – Suugaku Jul 14 '12 at 14:06

Such a function does exist. Try $h(x) = e^{-x^2b}-\frac{b}{e}$. This function trivially satisfies (i)-(iii). Now you just need to find $b$ such that (iv) and (v) also hold. Assuming $b >0$ we get : $|h^{''}(x)|= |2be^{-x^2b}(2bx^2-1)| \leq 2b |2b-1|$ for $x\in[-1,1]$. So choosing a very small $b$ like $b=1/8$ will make this smaller than $1/2$. For this $b$ we also get $h(0)=1-\frac{1}{8e} > 1/2$. So (iv) and (v) also hold.

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This doesn't satisfy $(i)$. It looks like you confused $\mathrm e^{-x^2b}$ and $\mathrm e^{-x^2}b$. – joriki Jul 14 '12 at 13:29

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