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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?

Cubi73
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1 Answers1

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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 $\beta$.

Kavi Rama Murthy
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    Ahh, now I get it. The one thing I was missing the whole time was $|f(x)e^{ig(x)}|_p=|f|_p$ for real-valued $g$. Thank you :) – Cubi73 Mar 22 '19 at 11:36
  • I see that $$\int_{\mathbb R} x^2 | g(x) |^2 dx = \int_{\mathbb R} (x - \alpha) | f(x) |^2 dx$$ but I struggle to compute $\hat{g}$. I've gotten this far: \begin{align} \sqrt{2 \pi}\cdot\hat{g}(\xi) & = \int_{\mathbb R} f(x + \alpha) e^{- i x (\xi - \beta)} dx= \int_{\mathbb R} f(y) e^{- i (y - \alpha)(\xi - \beta)} dy \ & = e^{i\alpha (\xi - \beta)} \int_{\mathbb R} f(y) e^{- i y(\xi - \beta)} dy \end{align} How can I continue? – ViktorStein Mar 03 '20 at 00:46
  • I tried to continue like this $$\int_{\mathbb{R}} f(y) e^{-i y (\xi - \beta)} dy = \int_{\mathbb{R}} f(y) e^{i y \beta} e^{-i y \xi} dy = \sqrt{2 \pi }\hat{f}(\xi - \beta).$$ But this yields $$\int_{\mathbb{R}} \xi^2 | \hat{g}(\xi) |^2 d \xi = \int_{\mathbb{R}} \xi^2 | \hat{f}(\xi - \beta) |^2 d\xi = \int_{\mathbb{R}} (\xi + \beta)^2 | \hat{f}(\xi) |^2 d\xi,$$ right? – ViktorStein Mar 03 '20 at 01:36
  • So did you mean to write $g(x) = f(x + \alpha) e^{-i x \beta}$ or are my calculations wrong? – ViktorStein Mar 03 '20 at 01:44
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    @ViktorGlombik I think you are confusing $a,b$ with $\alpha,\beta$. Your equation after "I've gotten this far" shows that $\hat g(\xi)=e^{i\alpha(\xi-\beta)}\hat f(\xi-\beta)$. It immediately follows that $|\hat g(\xi)|=|\hat f(\xi-\beta)|$, which subsequently implies $|x\hat g(x)|_2=|x\hat f(x-\beta)|_2=|(x+\beta)\hat f(x)|_2$. This means you need to choose $\beta:=-b$. Similarly, we obtain $|g(x)|=|f(x+\alpha)|$ from the definition of $g$ and therefore $|xg(x)|_2=|(x-\alpha)f(x)|_2$, so you need to choose $\alpha:=a$. – Cubi73 Mar 03 '20 at 02:04
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    @ViktorGlombik But as you have guessed, one could also define $g(x)=f(x+a)e^{-ixb}$ (with $a$ and $b$ instead of $\alpha$ and $\beta$). – Cubi73 Mar 03 '20 at 02:12