5
$\begingroup$

Landau inequality is about the bounds of derivatives of a real(complex)-valued function defined on some interval of the real line (I heard this from a lecture). I learned the simplest case is: $$ \| f' \|_{L^\infty (-\infty,+\infty)}^2 \leq 4 \|f\| \cdot \|f''\| ,$$ the later norms are the same as the first one. I tried several functions $f$, and they are indeed the case, yet I don't have any idea how to get the general result (or the proof).

Can someone be kind enough to give me some hints on this? Besides, does this inequality have similar results on finite intervals? Can someone give me some references on this? Thank you very much!

$\endgroup$
3
$\begingroup$

I just tried to prove the very elementary one, the one that you typed. I hope the proof is correct. PS: the wikipedia page doesn't give any hint of the proof.

For a function $f\in C^2(-\infty, \infty)$, we have the following Taylor expansion: $$ f(t)=f(r)+(t-r)f'(r)+\int_r^t (t-s)f''(s)ds$$

So $$ f'(r)=\frac{f(t)-f(r)}{t-r}-\frac{1}{t-r}\int_r^t (t-s)f''(s)ds$$

Take absolute value on both sides, $$ |f'(r)|\leq \left|\frac{f(t)-f(r)}{t-r}\right|+\frac{1}{|t-r|}\left|\int_r^t (t-s)f''(s)ds\right|\leq \frac{1}{|t-r|}|f(t)-f(r)|+\frac{|t-r|}{2}\|f''\|_\text{sup}\leq \frac{2\|f\|_\text{sup}}{|t-r|}+\frac{\|f''\|_\text{sup}}{2}|t-r|$$

Let $s=|t-r|$. Since we assume $f\in C^2(-\infty, \infty)$, so the above inequality is true for all $s$: $$ |f'(r)|\leq \frac{1}{s}2\|f\|+\frac{\|f''\|}{2}s $$

So it's still true even we take the infimum over $s$: $$ |f'(r)|\leq \inf_{s}\left(\frac{1}{s}2\|f\|+\frac{\|f''\|}{2}s\right) $$

Now let's discuss the infimum of the function $$ g(s)=\frac{C}{s}+Ds $$ for some constants $C$ and $D$. Since $s$ can be taken from $-\infty$ to $\infty$, this function can achieve its infimum when $s=\sqrt{\frac{C}{D}}$ and its infimum is $2\sqrt{CD}$.

So for any $r$, we have $$ |f'(r)|\leq 2\sqrt{2\|f\|_\text{sup}\times \frac{\|f''\|_\text{sup}}{2}}=2\sqrt{\|f\|\cdot \|f''\|} $$

Therefore, $\|f'\|_\text{sup}\leq 2\sqrt{ \|f\|\cdot \|f''\|}$ and take the square on both sides: $$ \|f'\|^2\leq 4\|f\|\|f''\|$$

More general cases can see reference here: http://ajmaa.org/RGMIA/papers/v6n2/inequalities.pdf

$\endgroup$
1
$\begingroup$

Wikipedia calls it the Landau-Kolmogorov inequality. The link should get you started.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.