An exercise concerning a uniform control of |f'(0)| by uniform norms of f and f''

Question

This is an exercise in Microlocal Analysis for Differential Operators :An Introduction written by Alain Grigis and Johannes Sjostrand.

(1)Let $f \in C^2[0,\varepsilon]$ for some $\varepsilon >0$ with $f(0)\ge0,f'(0)\ge0$. Try to get a lower bound on $\sup_{t \in [0,\varepsilon]}f(t)$ using also $||f''||_{L^\infty}$.

(2)Show that for $\forall f \in C^2([-\varepsilon,\varepsilon])$, there exists a $C_\varepsilon$ independent from $f$ such that $$|f'(0)| \le C_\varepsilon (||f||_{L^\infty}^{1/2}||f''||_{L^\infty}^{1/2}+||f||_{L^\infty}).$$

At first sight one may seek for an inequality looks like $$ \sup_{t \in [0,\varepsilon]} f(t) \ge C_\varepsilon ||f''||_{L^\infty}.$$ But an example of setting $f(x)=-x^3$ says that this is impossible. So I think that we need to consider first the second problem and find out what kinds of estimate is needed.

Hours of attempts are made. Here are my considerations. Put for simplicity $M_0 = ||f||_{L^\infty},M_2=||f''||_{L^\infty},R=M_0/M_2$. We will see that a lower bound of $R$ is vital in this problem. Suppose first $f(0) \ge 0,f'(0) \ge 0$. Once this situation is established, for general $f$ we just apply the result to $g(x) = f(x)-f(0)$ or $g(x) = -(f(x)-f(0))$ to conclude the proof.

Write by Taylor's expansion $$f(h)=f(0)+hf'(0)+\frac{h^2}{2}f''(\xi),$$ $$f(-k)=f(0)-kf'(0)+\frac{k^2}{2}f''(\eta).$$ Subtracting, we obtain $$(h+k)f'(0)=f(h)-f(-k)+\frac{h^2}{2}f''(\xi)-\frac{k^2}{2}f''(\eta)$$ and so $$|f'(0)| \le \frac{2}{h+k}M_0+\frac{h^2+k^2}{2(h+k)}M_2.$$

If we can choose $h,k$ such that $$\frac{h^2+k^2}{2(h+k)}=A_\varepsilon M_0^{1/2}M_2^{-1/2}$$ $$\frac{2}{h+k}=B_\varepsilon,$$ Then the problem is solved, where $A$ and $B$ are chosen so that $h$ and $k$ lie in $[-\varepsilon,\varepsilon]$.

After some calculation, we find that the equation $k$ shall satisfies is $$k^2-2B^{-1}k=4AR^{1/2}B^{-2}-4B^{-2}.$$ It's easy to see that we need a lower bound on $R$ to ensure that there exist $A$ making this equation has a solution. But the lower bound of $R$ is exactly the problem in (1).

Thanks a lot for reading so far.

Answer 1

If we let $h=k$ in your derivation, then it follows $$|f'(0)|\leq\frac1hM_0+\frac h2M_2.$$ Note that this inequality holds for all $h\in(0,\varepsilon)$, hence

• If $M_2=0$, choose $h=\varepsilon/2$ then $C_\varepsilon=2/\varepsilon$,
• If $M_2>0$ and $\sqrt{2M_0/M_2}<\varepsilon$, choose $h=\sqrt{2M_0/M_2}$ then $C_\varepsilon=\sqrt2$,
• If $M_2>0$ and $\sqrt{2M_0/M_2}\geq\varepsilon$, choose $h=\varepsilon/2$ then $$\frac1hM_0+\frac h2M_2=\frac2\varepsilon M_0+\frac\varepsilon4M_2\leq\frac2\varepsilon M_0+\frac14\sqrt{2M_0M_2}.$$

To summarise, we can choose $C_\varepsilon=\max(2/\varepsilon,\sqrt2)$.

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This $C_\varepsilon$ is just an example, you can of course make it better with the same idea.