I read a proof on the Uncertainty Principle (see below) and although the technical part itself is relatively straight forward I still do not understand a certain "w.l.o.g." statement in the proof.
Uncertainty Principle : For $f\in L^2(\mathbb R)$ and $a,b\in\mathbb R$ we have $$\|(x-a)f(x)\|_2\cdot\|(x-b)\hat f(x)\|_2\geq\frac12\|f\|_2^2,$$ where $$\hat f(x):=\frac1{\sqrt{2\pi}}\int_{\mathbb R}f(\xi)e^{-i\xi x}d\xi$$ denotes the Fourier transform of $f$.
The proof asserts that it suffices to show the statement for $a=b=0$. But why? I don't see how to substitute $f$ in $$\|xf(x)\|_2\cdot\|x\hat f(x)\|_2\geq\frac12\|f\|_2^2$$ such that the inequality becomes $$\|(x-a)f(x)\|_2\cdot\|(x-b)\hat f(x)\|_2\geq\frac12\|f\|_2^2$$ The norm $\|\cdot\|_2$ might be invariant under translation, so that I get
$$
\begin{align*}
\|xf(x)\|_2&=\|(x-a)f(x-a)\|_2=\|(x-a)g(x)\|_2,&&g:=T_af\text{ and} \\
\|x\hat f(x)\|_2&=\|(x-b)\hat f(x-b)\|_2=\|(x-b)T_b\hat f(x)\|_2=\|(x-b)\hat h(x)\|_2,&&h:=M_bf,
\end{align*}
$$
where $T_af(x):=f(x-a)$ is the translation operator and $M_bf(x):=f(x)e^{ibx}$ is the modulation operator, but surely $g$ and $h$ aren't equal. How do I get from $a=b=0$ to the general case?
Best Answer
Hint: just apply the result with $a=b=0$ to the function $g(x)=f(x+\alpha)e^{ix\beta}$ for suitable real numbers $\alpha$ and $\beta$.