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