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Scrolling in a book on real analysis I found, as a last exercise, the request to prove the Heisenberg uncertainty principle. The exercise states

Let $f\in \mathcal L_{1}^{2}(\mathbb R)$, such that $\|f\|_{2}=\|\hat f\|_{2}=1$. Prove that $$\left(\int_{\mathbb R}|x|^{2}|f(x)|^{2}\,\mathrm dx\right)\cdot\left(\int_{\mathbb R}|\xi|^{2}|\hat f(\xi)|^{2}\,\mathrm d\xi\right)\geq \frac{1}{(4\pi)^{2}}$$ Hint: suppose that $f\in C_{c}^{\infty}$ and use the following identities: $$\int_{\mathbb R}x\overline{f}(x)f'(x) = -{1\over 2} \|f\|_{2}^{s}$$ Plancherel's identity and Cauchy-Schwarz inequality.

The book states that $$\mathcal L^2_s(\mathbb R^d) := \{f:\mathbb R^d\rightarrow\mathbb C \text{ measurable} \text{ : }(1+|x|^2)^{s\over 2}f\in L^2(\mathbb R^d)\}$$ which calles it weighted $L^2$ spaces.

I was very curious on how to solve this problem because I don't have nearly as much knowledge as it's needed to solve this! So I'm here to ask you if you could give me the solution.

I can't give you my working because, as just stated, I don't know much about real analysis and I was just very curious to see the solution! Probably I'll understand it if I see one but searching on the internet didn't gave me any help.

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  • $\begingroup$ When you copy text from a book, you ought to identify what book it is. $\endgroup$ Sep 21, 2018 at 16:19
  • $\begingroup$ @NateEldredge I'm sorry but probably I used the word book in the wrong way. What I was reading are "notes": in practise a book on real analysis wrote by my professor that he gave to us some time ago $\endgroup$
    – Quiver
    Sep 21, 2018 at 17:19

1 Answer 1

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You have by the given identities

$$ \frac{1}{4} = \left(-\frac{1}{2}\|f\|_2^ 2 \right)^2 = \left(\int_\mathbb{R}{x \overline{f}(x) f'(x) dx} \right)^2 \le \|x \mapsto x \overline{f}(x) \|_2^2 \cdot \|f'\|_2^2 = \|x \mapsto x \overline{f}(x) \|_2^2 \cdot \|\widehat{f'}\|_2^2 $$

The first inequality is Cauchy Schwarz and the last equality is Plancherel Theorem. Using $$ \|\widehat{f'}\|_2 = 2\pi \| x \mapsto x\hat{f}(x)\|_2 $$ we get

$$ \frac{1}{(4 \pi)^2} \leq \|x \mapsto x \overline{f}(x) \|_2^ 2 \| x \mapsto x\hat{f}(x)\|_2^ 2 = \left( \int_\mathbb{R} {|x|}^2{|f(x)|}^2dx \right) \cdot \left( \int_\mathbb{R} {|x|}^2{|\hat{f}(x)|}^2dx \right) $$

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  • $\begingroup$ Just a square typo on the pie. $\endgroup$ Sep 21, 2018 at 13:44
  • $\begingroup$ Oh sry. And thanks for the info. $\endgroup$
    – Christian
    Sep 21, 2018 at 13:45
  • $\begingroup$ Thanks for the solution! The only obscure thing to me now is the notation $$\|x\rightarrow x\overline{f}(x)\|_2$$ what's that all about? $\endgroup$
    – Quiver
    Sep 21, 2018 at 13:47
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    $\begingroup$ It's he $ L^2 $ norm of the mapping $ g(x) = x \overline{f}(x) $. It's just short hand syntax to not explicity define the new map every time. $\endgroup$
    – Christian
    Sep 21, 2018 at 13:50
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    $\begingroup$ @DavideMorgante one has to choose if one wants to be exactly precise in your language, or if one wants to avoid awkward notation. If you look in papers, you might see people write $\| x\bar f\|$ $\endgroup$ Sep 21, 2018 at 13:56

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