5
$\begingroup$

Does there exist a constant $C>0$ such that for all $f \in H^3(\mathbb R)$ $$\int_{\mathbb R} \vert x f''(x) \vert^2 \ dx \le C \int_{\mathbb R} \vert f'''(x) \vert^2 + \vert x^3f(x) \vert^2 + \vert f(x) \vert^2 \ dx? $$

The question comes from the fact that it is very easy to see that

$$\int_{\mathbb R} \vert x f'(x) \vert^2 \ dx \le C \int_{\mathbb R} \vert f''(x) \vert^2 + \vert x^2f(x) \vert^2 + \vert f(x) \vert^2 \ dx, $$

but the proof in that case (integrating by parts and using the Cauchy-Schwarz inequality) does not obviously carry over to the case of the first line.

$\endgroup$
  • 1
    $\begingroup$ might want to try scaling argument first to see if its even possible... take $ f $ smooth and compactly supported and then assume inequality holds and put $ f_\lambda(x)= f(\lambda x)$ and vary $ \lambda>0$ and see if you get contradiction... $\endgroup$ – Math604 Dec 2 at 22:42
  • $\begingroup$ What is $H^2(R)$? Hardy space? $\endgroup$ – Alexandre Eremenko Dec 2 at 23:42
  • $\begingroup$ @AlexandreEremenko: I presume it's Sobolev space, but it probably really should be $H^3$. $\endgroup$ – Nate Eldredge Dec 2 at 23:42
  • $\begingroup$ @NateEldredge correct, sorry for the typo. It is the Sobolev space. $\endgroup$ – Tokoyo Dec 2 at 23:44
18
$\begingroup$

This really belongs to MSE rather than to MO, but I'm too lazy to initiate the moving process, so I'll just answer.

There may be more intelligent ways to do it, but you can also integrate by parts and get what you want, say, for smooth functions with compact support, after which you should carefully pass to the limit to extend it to the corresponding Sobolev class.

Let's just solve a more general problem. For non-negative integer $a,b,c$ and an infinitely smooth compactly supported real-valued $f$, denote
$$ I(a,b,c)=\int_{-\infty}^\infty x^af^{(b)}(x)f^{(c)}(x)dx. $$

Claim: If $r,R>0$ are integers and $b,c\le R$, $\frac ar+\frac{b+c-2R}{R}\le 0$, then $$|I(a,b,c)|\le C[I(0,0,0)+I(2r,0,0)+I(0,R,R)]$$

Indeed, there are only finitely many triples $(a,b,c)$ with the above property (admissible triples). Let $M$ be the maximum of $|I(a,b,c)|$ over the admissible triples. The integration by parts formula yields $$ |I(a,b,c)|\le a|I(a-1,b-1,c)|+|I(a,b-1,c+1)| $$ as long as $b>0$ and $c<R$. Note that the triples on the right are still admissible (if $a=0$, the first term just disappears) and $b$ gets smaller every time we do this trick. Thus, after finitely many operations, we arrive at a bound including only "extremal admissible triples" where either $b=0$ or $c=R$. Let $m$ be the maximum of $|I(a,b,c)|$ over the extremal admissible triples. We see that $M\le C_1m$ for some $C_1>0$.

If the extremal admissible triple is of the kind $(a,b,R)$, then we can use Cauchy-Schwarz to write $$ |I(a,b,R)|\le \delta I(2a,b,b)+\delta^{-1}I(0,R,R) $$ with any $\delta>0$ we want. Note that the triple $(2a,b,b)$ is still admissible if $(a,b,R)$ is. We'll choose $\delta=(2C_1)^{-1}$. Then, if $m$ is attained on a triple of this type, we have $$ M\le C_1m\le C_1[\delta M+\delta^{-1}I(0,R,R)]=\frac M2+2C_1^2I(0,R,R) $$ whence $M\le 4C_1^2I(0,R,R)$.

Let us now consider extremal admissible triples of the kind $(a,0,c)$. If $a\le r$, then we can use Cauchy-Schwarz in the form $$ |I(a,0,c)|\le \delta I(0,c,c)+\delta^{-1}I(2a,0,0) $$ and, with the same choice of $\delta$ and same argument, get $$ M\le 4C_1^2I(2a,0,0)\le 4C_1^2[I(0,0,0)+I(2r,0,0)] $$ (the crucial point is that $(0,c,c)$ is admissible).

If $a>r$, then we can use Cauchy-Schwarz in the form $$ |I(a,0,c)|\le \delta I(2(a-r),c,c)+\delta^{-1}I(2r,0,0) $$ and, with the same choice of $\delta$ and same argument, get $$ M\le 4C_1^2I(2r,0,0) $$ (the crucial point is that $(2(a-r),c,c)$ is now admissible).

Either way, we get what we claimed. The inequality the OP asked about corresponds to the case $r=R=3$, $a=b=c=2$.

$\endgroup$
  • 15
    $\begingroup$ Too lazy to click 'flag', so go the far easier route of writing this long and detailed answer? :-) $\endgroup$ – LSpice Dec 3 at 1:48
  • 2
    $\begingroup$ @LSpice Not to click "flag" but to vote to close with 3 clicks and to write the same answer :-) $\endgroup$ – fedja Dec 3 at 12:24
15
$\begingroup$

The inequality is true. One integration by parts and standard Cauchy-Schwarz gives $$\int |xf''|^2 \leq C\left(\int |f'''|^2 + x^4|f'|^2 + |f'|^2\right).$$ The second term can be handled by integrating by parts once and applying Cauchy-Schwarz: $$\int x^4|f'|^2 \leq \delta \int x^2|f''|^2 + C_{\delta}\left(\int(x^2+x^6)f^2\right),$$ and $x^2+x^6$ is smaller than a fixed constant times $(1+x^6)$. Finally, the third term in the first inequality can be handled using the Fourier transform and Parseval's identity: $$\int |f'|^2 = \int |\xi|^2|\hat{f}|^2 \leq \int(1+|\xi|^6)|\hat{f}|^2 = \int (|f|^2 + |f'''|^2).$$

$\endgroup$

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.