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Let $f \in L^{2}(\mathbb{R})$ be continuously differentiable on $\mathbb{R}$. I am trying to show the following: $( \int |f|^{2} dx)^{2} \leq 4 ( \int |xf(x)|^{2} dx) ( \int |f'|^{2} dx))$.

My first thought is to think about this inequality as $ ||f||^{4}_{2} \leq 4 ||xf(x)||^{2}_{2} ||f'||^{2}_{2} $ and apply Holder's inequality to get $ 4 (\int |xf(x)f'(x)| dx)^{2} \leq 4 ||xf(x)||^{2}_{2} ||f'||^{2}_{2}$ but beyond this I have no intuition, especially what to do with the continuously differentiable assumption. Could someone lend me a hint?

saz
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EarlyGame
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  • I can only give a hint: This sounds like some kind of Sobolev inequality, but I'm not sure if that is what you are looking for. You can get some ideas on these inequalities on wikipedia: https://en.wikipedia.org/wiki/Sobolev_inequality These inequalities always relate the norms of functions to norms of their derivatives. In particular the Nash inequality looks very similar to yours. – KoliG Dec 02 '15 at 15:51
  • This may help: http://math.stackexchange.com/questions/1526513/prove-an-integral-inequality-left-intf2dx-right2-le-4-left-intxfx –  Dec 02 '15 at 18:49

1 Answers1

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Hints:

The claim holds trivially true if the right-hand side equals $+\infty$, therefore we may assume without loss of generality that $$\int x^2 f(x)^2 \, dx < \infty. \tag{1} $$

Let us first assume that $f$ has bounded support.

  1. By the integration by parts formula, we have $$\int_{-R}^R f(x)^2 \, dx = \big[ x f(x)^2 \big]_{x=-R}^R -2 \int_{-R}^R x f(x) f'(x) \, dx$$ for all $R>0$. Since $f$ has bounded support, we can let $R \to \infty$ and obtain $$\int f(x)^2 \, dx = -2 \int x f(x) f'(x) \, dx. \tag{1}$$

For the general case:

  1. Choose a smooth cut-off function $\chi$ such that $0 \leq \chi \leq 1$, $\chi(x) = 1$ for $|x| \leq 1$ and $\chi(x) = 0$ for $|x| \geq 2$. Set $\chi_R(x) := \chi(x/R)$.
  2. Applying step 1 to $f_R(x) := f(x) \chi_R(x)$ gives $$\int \chi_R(x)^2 f(x)^2 \, dx \leq 2 \int \left|x f(x) \frac{d}{dx}(f(x) \chi_R(x)) \right| \, dx. \tag{3}$$
  3. Show that by the product rule and the definition of $\chi_R$, $$\begin{align*}\left| \frac{d}{dx} (f(x) \chi_R(x)) \right| &\leq |f'(x)| + |f(x)| R^{-1} \|\chi'\|_{\infty}. \end{align*}$$
  4. Use the finiteness of the integral $(1)$ and the monotone convergence theorem to let $R \to \infty$ in $(3)$ and conclude $$\int f(x)^2 \, dx \leq 2 \int |x f(x) f'(x)| \, dx.$$
  5. Apply the Cauchy-Schwarz inequality.
saz
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