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For each $\epsilon>0$ there exists a $C_\epsilon>0$ with $$\|f'\|_\infty\leq \epsilon \|f''\|_\infty + C_\epsilon \|f\|_\infty$$ for all $f\in C^2_b(\Bbb R)=\{f\in C^2(\Bbb R): f,f'\text{ and }f'' \text{ are bounded}\}$.

I tried arguing by contradiction, obtaining a bad sequence of functions, but I didn't manage to generate the contradiction.

It seems to me that we should use the Mean Value Theorem. E.g. we can write $f(x+\epsilon)-f(x)=f'(c_1)\epsilon$ and $f'(x+\epsilon)-f'(x)=f''(c_2)\epsilon$ for certain $x<c_i<x+\epsilon$. But also toying around with this hasn't led me far.

I appreciate your help.

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We can assume that $f$ is real-valued.

Replacing $f$ by $cf$, we can save typing by assuming that $||f''||_\infty= 1$.

So $$|f'(t+h)-f'(t)|\le h\quad(h>0);$$in particular $$f'(t+h)\ge f'(t)-h\quad(h>0).$$

So $$f(t+h)=f(t)+\int_0^hf'(t+s)\,ds\ge f(t)+\int_0^h(f'(t)-s)\,dt =f(t)+hf'(t)-\frac12h^2.$$Assuming that $f'(t)\ge0$ this shows that $$|f(t+h)-f(t)|\ge h|f'(t)|-\frac12h^2.$$Considering $-f$ in place of $f$ we see that last inequality also holds if $f'(t)<0$.

Now $|a-b|\ge c$ implies that $|a|\ge c/2$ or $|b|\ge c/2$, so the previous inequality shows that for every $t\in \Bbb R$ and $h>0$ we have $$||f||_\infty\ge\frac12h|f'(t)|-\frac14h^2.$$Hence, recalling that $||f''||_\infty=1$, we have $$||f||_\infty\ge\frac12h||f'||_\infty-\frac14h^2||f''||_\infty,$$or $$||f'||_\infty\le \frac2h||f||_\infty+\frac h2||f''||_\infty.$$Let $h=2\epsilon$ an this becomes the inequality you want.


Or let $h=2(||f||_\infty/||f''||_\infty)^{1/2}$ and you get the Landua inequality $$||f'||_\infty\le 2(||f||_\infty||f''||_\infty)^{1/2}.$$

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