$let f \in L^2(\mathbb{R}) \text{ and } a,b \in \mathbb{R}$. Then $$ \left(\int_{\mathbb{R}} (x-a)^2 |f(x)|^2 dx \right)^{1/2} \left( \int_{\mathbb{R} }(\xi-b)^{2} |\hat{f}(\xi) |^2 d\xi\right)^{1/2} \geq \frac{1}{2} ||f||_2^2$$
I dont understand the first steps of the proof:
Without loss of generality let $a,b=0 \text{ and } ||f||_2^2 = 1$. Then $$1 = \int_{\mathbb{R}} |f(x)|^2dx = -\int_{\mathbb{R} }x \frac{d}{dx} |f(x)|^2 dx = -\int_{\mathbb{R} }x f'(x)\overline{f(x)} + x\overline{f'(x)}f(x)dx $$
They say, that partial integration was used. I assume it was used in the second equality where the minus appears. But I run into trouble there:
$$\int_{\mathbb{R}} |f(x)|^2dx = \int_{\mathbb{R}} x'|f(x)|^2 dx= x|f(x)|^2 - \int_{\mathbb{R}}x \frac{d}{dx} |f(x)|^2dx$$ So, where does the $x|f(x)|$ hide in the proof?