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Question:

Find a example function $f$,such $f\in C^2(R)$,and such $$\left(\sup_{x\in R}|f'(x)|\right)^2=2\sup_{x\in R}|f(x)|\cdot\sup_{x\in R}|f''(x)|$$

This problem is from when I prove this inequality

$$\left(\sup_{x\in R}|f'(x)|\right)^2\le 2\sup_{x\in R}|f(x)|\cdot\sup_{x\in R}|f''(x)|$$

poof: this inequality I have$$f(x+y)=f(x)+f'(x)y+f''(\xi)\dfrac{y^2}{2}$$ $$f(x-y)=f(x)-f'(x)y+f''(\eta)\dfrac{y^2}{2}$$ then $$f(x+y)-f(x-y)=2f'(x)y+\left(f''(\xi)-f''(\eta)\right)\dfrac{y^2}{2}$$ then $$2y|f'(x)|=|f(x+y)-f(x-y)-(f''(\xi)-f''(\eta))\dfrac{y^2}{2}|\le 2a+by^2$$ so $$\sup_{x\in R}|f'(x)|\le\dfrac{a}{y}+\dfrac{by}{2}$$ By Done

But this problem ask the constant $2$ is best.so I think we must find a example,I can't find the function such $f\in C^2(R)$

math110
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1 Answers1

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For an integer $n\geq3$ we consider $f_n(x)$ defined as follows: $$f_n(x)=\left\{\matrix{\frac{1}{2n}- \frac{1}{2} &\hbox{if}& x <-1\cr \frac{n}{6}(x+1)^3-\frac{1}{2}+\frac{1}{2n}&\hbox{if}&-1< x <\frac{1}{n}-1 \cr \frac{x^2}{2}+\left(1-\frac{1}{2n}\right)x+\frac{1}{6n^2} &\hbox{if}& \frac{1}{n}-1< x <-\frac{1}{n}\cr -\frac{n x^3}{6}-\frac{x}{n}+x &\hbox{if}& -\frac{1}{n}<x<\frac{1}{n}\cr -\frac{x^2}{2}+\left(1-\frac{1}{2n}\right)x-\frac{1}{6n^2} &\hbox{if}&\frac{1}{n}<x <1-\frac{1}{n} \cr \frac{n}{6}(x-1)^3+\frac{1}{2}-\frac{1}{2n}&\hbox{if}& 1-\frac{1}{n}<x<1\cr \frac{1}{2}-\frac{1}{2n}&\hbox{if}& x>1\cr }\right. $$ the functions $f_n$, $f'_n$ and $f''_n$ are depicted in the next figure:

$\hspace{2cm}$

It is straightforward to check that $$\eqalign{\sup_{x\in\mathbb{R}}|f_n(x)|&=\frac{1}{2}-\frac{1}{2n}\cr \sup_{x\in\mathbb{R}}|f'_n(x)|&=f'_n(0)=1-\frac{1}{n}\cr \sup_{x\in\mathbb{R}}|f''_n(x)|&=1 }$$ So if $c$ is a constant that satisfies $$\left(\sup_{\mathbb{R}}|f'|\right)^2\leq c\sup_{\mathbb{R}}|f|\cdot \sup_{\mathbb{R}}|f''|$$ for every bounded function $f:\mathbb{R}\to\mathbb{R}$ having bounded second derivative, then, applying this to $f_n$ we see that $$\left(1-\frac{1}{n}\right)^2\leq c\left(\frac{1}{2}-\frac{1}{2n}\right)$$ Letting $n$ tend to $+\infty$ we get $c\geq 2$. And we know that $c=2$ works, so $2$ is the best constant.

Remark 1. The sequence of functions $(f_n)_n$ does converge to a function that has a continuous derivative, but not a continuous second derivative.

Remark 2. Proving that $c=2$ is the best constant does not necessarily mean the we have to find an example that realizes the equality.

Omran Kouba
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