I am working through Stein and Shakarchi's Fourier Analysis and am stuck on Exercise 22 of Chapter 5, which I quote below. Preliminary notation: $\mathcal{S}$ is the Schwartz space of functions on $\mathbb{R}$; $\hat{f}$ is the Fourier transform of $f$ (the unitary version, i.e. $\hat{f}(\xi)=\int_\mathbb{R} f(x)e^{-2\pi ix\xi}\,dx$).

Chapter 5, Exercise 22 The heuristic assertion stated before Theorem 4.1 can be made precise as follows. If $F$ is a function on $\mathbb{R}$, then we say that the preponderance of its mass is contained in an interval $I$ (centered at the origin) if $$\int_I x^2|F(x)|^2\,dx \geq \frac12 \int_\mathbb{R} x^2|F(x)|^2\,dx. \tag{16}\label{eq}$$ Now suppose $f\in\mathcal{S}$, and \eqref{eq} holds with $F=f$ and $I=I_1$; also with $F=\hat{f}$ and $I=I_2$. Then if $L_j$ denotes the length of $I_j$, we have $$L_1L_2\geq\frac{1}{2\pi}.$$ A similar conclusion holds if the intervals are not necessarily centered at the origin.

For context, here is Theorem 4.1 (the uncertainty principle) along with the heuristic assertion referred to in the exercise:

The mathematical thrust of the [uncertainty] principle can be formulated in terms of a relation between a function and its Fourier transform. The basic underlying law, formulated in its vaguest and most general form, states that a function and its Fourier transform cannot both be essentially localized. Somewhat more precisely, if the "preponderance" of the mass of a function is concentrated in an interval of length $L$, then the preponderance of the mass of its Fourier transform cannot lie in an interval of length essentially smaller than $L^{-1}$. The exact statement is as follows.
Theorem 4.1 Suppose $\psi$ is a function in $\mathcal{S}(\mathbb{R})$ which satisfies the normalizing condition $\int_{-\infty}^\infty |\psi(x)|^2\,dx=1$. Then $$\left(\int_{-\infty}^\infty x^2|\psi(x)|^2\,dx\right)\left(\int_{-\infty}^\infty \xi^2|\hat{\psi}(\xi)|^2\,d\xi\right) \geq \frac{1}{16\pi^2},$$ and equality holds if and only if $\psi(x)=Ae^{-Bx^2}$ where $B>0$ and $|A|^2=\sqrt{2B/\pi}$. In fact, we have $$\left(\int_{-\infty}^\infty (x-x_0)^2|\psi(x)|^2\,dx\right)\left(\int_{-\infty}^\infty (\xi-\xi_0)^2|\hat{\psi}(\xi)|^2\,d\xi\right) \geq \frac{1}{16\pi^2},$$ for every $x_0,\xi_0\in\mathbb{R}$.

OK, back to the exercise. In trying to get a handle on it, I have noted three things:

  • By scaling $f$ by a constant, I can assume that $f$ is normalized in some way; for example, that $\int_\mathbb{R} |f(x)|^2\,dx=1$ or perhaps $\int_\mathbb{R} x^2|f(x)|^2\,dx = 1$.

  • Let $\delta>0$. For intervals centered at the origin: If a preponderance of mass of $f(x)$ is contained in $I_1$, then a preponderance of mass of $f(\delta x)$ is contained in $\delta^{-1}I_1$. Also, if a preponderance of mass of $\hat{f}(\xi)$ is contained in $I_2$, then a preponderance of mass of $\delta^{-1}\hat{f}(\delta^{-1}\xi)$ (which is the Fourier transform of $f(\delta x)$) is contained in $\delta I_2$. Since $(\delta^{-1}L_1)(\delta L_2)=L_1L_2$, I can, for example, choose $\delta$ so that $\delta^{-1}L_1=\delta L_2$ and prove the statement for $f(\delta x)$ and its Fourier transform. In other words, I can assume that $I_1=I_2$.

  • I expected the case of equality to be when $f$ is a Gaussian, because this is the case of equality for the uncertainty principle. However, using Mathematica I have found that this is not true.

Unfortunately, these three observations are as far as I have been able to get on this exercise. I think my basic problem is that I don't see how to express the quantities $L_1$ and $L_2$ in terms of other known quantities.


First note that Theorem 4.1 holds for $\psi\in L^2$ by a simple approximation argument.

Now assume $f\in L^2$, $||f||_2=1$. Then, since $|x|\le L_1/2$ for $x\in I_1$, $$\left(\frac{L_1}{2}\right)^2\ge \left(\frac{L_1}{2}\right)^2\int_{I_1}|f|^2\ge\int_{I_1}x^2|f(x)|^2\ge\frac12\int_{\Bbb R}x^2|f(x)|^2.$$Similarly $$\left(\frac{L_2}{2}\right)^2\ge\frac12\int\xi^2|\hat f(\xi)|^2,$$so 4.1 shows that $$\frac{(L_1L_2)^2}{16}\ge\left(\frac14\right)\frac1{16\pi^2}.$$

Now for the "similarly for intervals not centered at the origin" bit: Seems to me that if $I$ is an interval centered at $a$ we want to say the definition of "the preponderance of the mass is on $I$" is $$\int_I(x-a)^2|f(x)|^2\ge\frac12\int_{\Bbb R}(x-a)^2|f(x)|^2.$$Similarly for $\hat f$. And now we should note that applying 4.1 with $e^{2\pi ibx}f(x-a)$ (or something like that, I may not have the minus signs straight) in place of $f$ shows that $$\int(x-a)^2|f(x)|^2\int(\xi-b)^2|\hat f(\xi)|^2\ge\frac1{16\pi^2}.$$

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