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Consider the Hilbert space of functions $f \in L^2(\mathbb R)$ such that $x^2f \in L^2(\mathbb R) $ and $ f'' \in L^2(\mathbb R).$

I am wondering whether it is true that $xf'\in L^2(\mathbb R)$ as well?

It seems natural and perhaps some sort of interpolation should yield the claim but I fail to see how to show this.

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  • $\begingroup$ $L^2$ of $\bf R$, presumably? $\endgroup$
    – Noam D. Elkies
    Commented Jul 6, 2018 at 18:16
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    $\begingroup$ @NoamD.Elkies that's right. $\endgroup$
    – Zorgo
    Commented Jul 6, 2018 at 18:28
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    $\begingroup$ There's probably also a highbrow explanation for this, along the following lines: Your assumptions say that $f$ is in the natural domain of the (harmonic oscillator) operator $Lf=-f''+x^2f$, which can be factored as $L=A^*A$, $A=D+x$, and this should also imply the claim. (The more elementary answers seem more appropriate though.) $\endgroup$
    – Christian Remling
    Commented Jul 7, 2018 at 2:40

2 Answers 2

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By a cutoff function argument, it suffices to assume $f$ is compactly supported, so we can integrate by parts without picking up boundary terms.

Thus $$\int (xf')^2 = \int (x^2f') f' = -\int 2xf'f - \int x^2 f'' f$$ Hence using Cauchy-Schwarz, $$\|xf'\|_2^2 \le \int |2xf f'| + \int |x^2 f f''| \le 2\|xf\|_2 \|f'\|_2 + \|x^2f\|_2 \|f''\|_2.$$ The second term is finite by our assumption. For the first term, note that $$\|xf\|_2^2 = \int x^2 f^2 = \int |x^2f| |f| \le \|x^2f\|_2 \|f\|_2$$ and $$\|f'\|^2 = \int (f')^2 = -\int f f'' \le \int |f f''| \le \|f\|_2 \|f''\|_2.$$

Putting everything together, we have $$\|xf'\|_2^2 \le 2 \|f\|_2 \sqrt{\|f''\|_2 \|x^2f\|_2} + \|x^2f\|_2 \|f''\|_2.$$

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By integration by parts, $$\int (xf')^2 = \int (x^2f') f' = -\int 2xf'f - \int x^2 f'' f.$$ $$-\int 2xf'f = \int f^2=\|f\|^2.$$ By the Cauchy-Schwarz inequality $$\bigg|\int x^2 f'' f\bigg|\le \|x^2f\|\|f''\|.$$ The conclusion is proved.

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    $\begingroup$ Oh yeah, that's much better than mine. (Hint to people as slow as me: in the second line, $2f'f = (f^2)'$.) $\endgroup$
    – Nate Eldredge
    Commented Jul 6, 2018 at 22:21
  • $\begingroup$ @NateEldredge: You are too kind. :-) $\endgroup$
    – Hans
    Commented Jul 7, 2018 at 19:56

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