5

Let $T: L^2(\mathbb R) \to L^2(\mathbb R)$ be an operator that commutes with all translations $f(x) \mapsto f(x-y)$. Why does it follow that the Fourier transform of $Tf$ is then given by multiplication with a (bounded) function? In Bounded linear operators that commute with translation the user robjohn gives a seemingly easy to follow proof. But I have some problems understanding the calculation: let $\psi$ be the Gaussian such that $\hat \psi = \psi$. Then

\begin{align} \psi(\xi)\;(Tf)^\wedge(\xi) &=(\psi*Tf)^\wedge(\xi)\\ &=\left(\int_{\mathbb{R}^n}\psi(x-y)\;Tf(y)\;\mathrm{d}y\right)^\wedge(\xi)\\ &=\left(\int_{\mathbb{R}^n}\psi(y)\;Tf(x-y)\;\mathrm{d}y\right)^\wedge(\xi)\\ &=\left(\int_{\mathbb{R}^n}T^{\;*}\psi(y)\;f(x-y)\;\mathrm{d}y\right)^\wedge(\xi)\\ &=(T^{\;*}\psi)^\wedge(\xi)\hat{f}(\xi) \end{align}

1) Where exactly is the assumption that $T$ commutes with translations, is used?

In the literature i only found a more general proof, e.g., in Stein: Introduction to Fourier analysis on Euclidean spaces, Theorem 3.16, but I am interested in a simple proof in the above basic setting.

Also on more conceptual level I would like to understand how one can see this as the fact that Fourier transform diagonalizes translation invariant operators.

Community
  • 1
Mekanik
  • 1,861

2 Answers2

4

Regarding (1), the notation $Tf(x-y)$ could mean at least two different things: The function $Tf$ evaluated at $x-y$ or $T$ applied to the function $f(x-y)$. If you look closely you see you need one interpretation to make one equality above work and the other interpretation to make another equality above work. But the two are the same, precisely because $T$ commutes with translations.

Regarding (2): Yes, there's a change of variable. No, there's no minus sign missing.

Consider Lebesgue measure on the line. It's a fact that $$\int g(t)\,dt=\int g(-t)\,dt.$$Think about it - this is obvious on "area under the curve" grounds!

That's because in the present context $\int$ means $\int_{\mathbb R}$. Now, in a calculus course $\int g(t)\,dt$ would not mean the integral over the line, it would denote an antiderivative. Yes, if $G'(t)=g(t)$ then $-G'(-t)$ is an antiderivative for $g(-t)$; that's the minus sign you think is missing. But if we're thinking of things this way then that minus sign is cancelled by the minus sign in $$\int_{\infty}^{-\infty}=-\int_{-\infty}^\infty.$$

Regarding your last question, the point is that if you look at it right then a diagonal matrix is a multiplication operator!. Say $D$ is an $n\times n$ diagonal matrix with $\lambda_j$ on the diagonal. Then $$Dx=(\lambda_1x_1,\dots,\lambda_nx_n).$$Applying $D$ multiplies the components of $x$ by the $\lambda_j$.

David C. Ullrich
  • 92,839
  • thank you for the clear and elaborate explanations! regarding minus sign, i dont know how I could not see that... – Mekanik Jul 08 '15 at 18:51
3

Where exactly is the assumption that T commutes with translations, is used?

When you write that

$$\int_\mathbb{R} \psi(y) (Tf)(x-y) dy = \int_\mathbb{R} (T^*\psi)(y) f(x-y) dy$$

You means that

$$\langle \psi , \tau_x(Tf) \rangle = \langle \psi , T(\tau_x f) \rangle= \langle T^*\psi(y), \tau_x f \rangle$$

And this is exactly the commutativity of $T$ and $\tau_x$

Tryss
  • 14,490
  • 20
  • 35