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Let $f:\mathbb{R}\rightarrow\mathbb{R}$ be an everywhere twice continuously differentiable function, and suppose $f(1)=f'(1)=f(-1)=f'(-1)=0$. What is the maximum possible value for $f(0)$ given $|f''(x)|\leq 1$ for all $x\in\mathbb{R}$?

Thanks for the help!

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  • $\begingroup$ Hint: use limited development up to second order in points $x=-1$ and $x=1$. $\endgroup$ – TZakrevskiy Feb 14 '14 at 14:45
  • $\begingroup$ Could you elaborate on what "limited development" means? I am not familiar with that term. $\endgroup$ – user88849 Feb 14 '14 at 14:50
  • $\begingroup$ $f(x+y) = f(x)+f'(x)y+f''(\xi)y^2/2$ where $\xi\in (\min(x,x+y),\max(x,x+y))$. $\endgroup$ – TZakrevskiy Feb 14 '14 at 14:55
  • $\begingroup$ I can see how that immediately gives $f(0)\leq\frac{1}{2}$, so if I can realize that with an example, the proof would be complete. I do not see a way to lower this to anything like 1/4, but then the trouble is that I cannot find such a function. Mostly I've tried playing around with quartic polynomials and the like, but obviously these would have to be modified away from $[-1,1]$ to control the second derivative on all of $\mathbb R$. $\endgroup$ – user88849 Feb 14 '14 at 15:07
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By continuity, $f$ assumes its maximum on $[-1,1]$ at some point $a$. Wlog. $a\ge 0$. By the condition, $f(1-h)\le \frac12h^2$. From $f'(a)=0$, also by the condition $f(a+h)\ge f(a)-\frac12h^2$. This implies $$ f(a)-\frac{(1-a)^2}8\le f\left(a+\frac{1-a}2\right)= f\left(1-\frac{1-a}2\right)\le \frac{(1-a)^2}8$$ hence $$f(0)\le f(a)\le \frac{(1-a)^2}{4}\le \frac14.$$ This is sharp as we see from $$ f(x)=\begin{cases}\frac12(1+x)^2&\text{if }x\le-\frac12\\ \frac14-\frac12x^2&\text{if }-\frac12\le x\le \frac12\\ \frac12(1-x)^2&\text{if }x\ge \frac12\end{cases}$$ which is twice dfferentiable except at $\pm\frac12$, but can be approximated arbitrarily well with a smoother function:

For $0<\epsilon<\frac12$ let $$ g(x)=\begin{cases}1&\text{if }x<-\frac12-\epsilon\text{ or } x>\frac12+\epsilon\\ \frac1\epsilon(x+\frac12)&\text{if }-\frac12-\epsilon\le x\le-\frac12+\epsilon\\ -1&\text{if }-\frac12+\epsilon<x<\frac12+\epsilon\\ \frac1\epsilon(\frac12-x)&\text{if }+\frac12-\epsilon\le x\le+\frac12+\epsilon,\end{cases}$$ then $G(x)=\int_{-1}^x g(t)\,\mathrm dt$ and $F(x)=\int_{-1}^x G(t)\,\mathrm dt$. Then $F(x)$ fulfills the conditions and almost coincides with $f(x)$. In fact $|G(x)-f'(x)|\le \int_{-\epsilon}^\epsilon \frac t\epsilon\,\mathrm dt =\epsilon$ and therefore $|F(0)-f(0)|\le \epsilon$ (because $F(-1)=f(-1)=0$).

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