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I have trouble with the following task:

Let $g \in C^1(\mathbb{R})$ with $g$, $xg$, $g' \in L^2(\mathbb{R})$ and $\int_\mathbb{R} g^2 d \lambda = 1$.

(a) Show that $\int_\mathbb{R} x^2g^2(x) dx \cdot \int_\mathbb{R} \alpha^2|Fg|^2(\alpha) d \alpha \geq \frac{\pi}{2}$, where $Fg$ is the Fourier transformation of g

(b) For which $g$ holds (a) equality?

I already got the hint to consider $\int_{[a, b]} g^2 \cdot 1 d \lambda$, but that led me only to $\int_{[a, b]} g^2 \cdot 1 d \lambda = bg(b)^2 - ag(a)^2 - 2 \int_{[a, b]} xg'(x) dx$

Yet, I don't see how it relates to the inequality, which is to be proven. Any hint? For (b) I'm completely lost; any hint how to start?

Thanks a lot!

NamedFormula
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  • Note that $g \in L^2(\mathbb{R})$ implies that $g\to 0$ as $x\to \pm \infty$. So, integrating by parts: $\int g^2 = -2 \int x g'(x) g(x)dx$. Using Cauchy-Schwarz on the right hand side you get something like $|g'|{L^2}|gx|{L^2}$. Then, maybe use $|g'|{L^2}=|F(g')|{L^2}$ to finish – Matt Dec 20 '16 at 16:55
  • @Matt: How does $g $ being square integrable imply that $g(x) \to 0$ as $|x| \to \infty$? – copper.hat Dec 20 '16 at 16:59
  • @copper.hat: Indeed, this is a good question considering my claim is not true! There should be a work around... – Matt Dec 20 '16 at 17:08
  • @copper.hat Since $g\in C^1(\mathbb{R})$, we have $g$ is differentiable with bounded derivative, so $g$ is uniformly continuous. Using boundedness of $g$ we can then establish $g^2$ is uniformly continuous. I believe this is sufficient to establish decay at infinity. – Matt Dec 20 '16 at 17:23
  • Look at http://math.stackexchange.com/q/253844/27978. – copper.hat Dec 20 '16 at 18:08
  • @Matt: It doesn't matter here, but $g$ can be smooth, integrable and have unbounded derivative. – copper.hat Dec 20 '16 at 18:09

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