Local version of the estimate $\sup |f'| \le 2 \sqrt{\sup |f| \cdot \sup|f''|}$

Question

It is well known that if $f : \mathbb{R} \to \mathbb{R}$ belongs to $C^2(\mathbb{R})$ and $\sup_\mathbb{R} |f| \le P$, $\sup_\mathbb{R}|f''| \le Q$, then

$$\sup_{\mathbb{R}} |f'| \le 2 \sqrt{PQ}.$$

I would like to know whether a local version of this estimate holds.

For instance, if $f \in C^2[0,1]$ (with the one-sided derivatives existing at the endpoints) does there exist some constant $K > 0$ so that for all $0 < a < b < 1$

$$\sup_{(a,b)} |f'|\le K \sqrt{\sup_{[0,1]} |f|\cdot \sup_{[0,1]} |f''|}?$$

By essentially repeating the proof of the global case (via Taylor's formula), I can show

$$ |f'(x)| \le K (\sup_{[0,1]} |f| + \sup_{[0,1]} |f''|), \qquad x \in [0,1].$$

I have also tried multiplying $f$ by a cutoff function identically one on $(a,b)$ and supported in $(0,1)$ and then applying the global case. But using that strategy I'm not able to make $K$ independent of $a$ and $b$.

Hints or solutions are greatly appreciated!

Answer 1

If $f'$ is never zero, then the local version is false. Take $f(x)=mx+q$. Then the right-hand side is zero and the left-hand side is $m$.

If $f'$ has a zero, then the inequality is true. The only proof I know is complicated. You can find it in Lemma 7.38 in Sobolev book. There might be a simpler proof since there they are considering different $L^p$ spaces for $f$, $f'$ and $f''$, while you are considering the same space for all three.

If $f'$ is never zero, then I think that the best inequality you can hope for is
$$ \sup_{(a,b)}|f^{\prime}(x)|\leq c\ell^{-1}\sup_{(a,b)}|f(x)|+c\sqrt {\sup_{(a,b)}|f(x)|\sup_{(a,b)}|f^{\prime\prime}(x)|}% $$ for every $0<\ell\leq(b-a)$, where $c>0$ does not depend on $a$ and $b$. It's a special case of Theorem 7.41 in the same book. Again the proof is complicated because it treats different spaces. This inequality is sharp: If you consider again the case $f(x)=mx$ and $a=0$ you get $$\sup_{(a,b)}|f(x)|=mb$$ and so you need to divide by something less than $b$ if you want to get $\sup_{(a,b)}|f^{\prime}(x)|=m$.

I would be very interested in finding simpler proofs.

Answer 2

I find that the best $K$ is no more than $\sqrt{2}$ ,if the domain is not less than $2\sqrt{\dfrac{2\sup|f|}{\sup|f''|}}$.$$$$Here is a solution with Tyler's expansion: $$ \forall x\in\mathbb{R},\forall h>0,\\\ f(x+h)=f(x)+hf'(x)+\frac{h^2}{2}f''(\eta_1)\text{ ,where $\eta_1\in(x,x+h)$}\\ f(x-h)=f(x)-hf'(x)+\frac{h^2}{2}f''(\eta_2)\text{ ,where $\eta_2\in(x-h,x)$}\\ \Rightarrow f(x+h)-f(x-h)=2hf'(x)+\frac{h^2}{2}[f''(\eta_1)+f''(\eta_2)]\\ \text{Hence, }|2hf'(x)|\leq|f(x+h)-f(x-h)|+|\frac{h^2}{2}[f''(\eta_1)+f''(\eta_2)]|\leq\\ 2\sup|f|+h^2\sup |f''| $$ We deduce that: $$2\sup|f'|\leq\frac{2\sup|f|}{h}+h\sup|f''|$$ Since $\sup|f^{(k)}|$ is a constant. We can let $$h=\sqrt{\dfrac{2\sup|f|}{\sup|f''|}}$$ We deduce that $$\boxed{\sup |f'| \le \sqrt{2}\cdot \sqrt{\sup |f| \cdot \sup|f''|}}$$When the domain is less than $2\sqrt{\dfrac{2\sup|f|}{\sup|f''|}}$, suppose the domain to be $[a,b]$, where $b>a$. Then we have: $$\boxed{\sup |f'| \le \frac{2\sup|f|}{b-a}+\frac{(b-a)\sup|f''|}{4}}$$