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Let $T$ be a bounded linear operator on $L^2(\mathbb R)$. So, let us now assume that $T$ commutes with the translations $\tau_x$. How do I now show that $T$ is given by a convolution with respect to a distribution?

By the way, I know I can probably find the proof somewhere in one of Stein's books, but I would like to prove it myself without knowing what it should be but I'm struggling a bit. So I would like some hints. Especially I would like a method of deriving the result without knowing what it should be. If that is not possible, an intuitive argument why it should be true would be nice as well.

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2 Answers

up vote 2 down vote accepted

Some suggestions:

Even though the delta function is not in $L^2$, heuristically what does your condition on $T$ say when applied to the delta function? Now, given the answer to the previous question and the fact that every function is an average of translated delta functions, what does one expect $Tf$ to be for an arbitrary $L^2$ function $f(x)$?

You can also do this on the Fourier transform side.. i.e. consider $G = F \circ T \circ F^{-1}$ as Plop suggested and then look at how $G$ behaves on a given $\delta(x - a)$. Then again use the idea that an $L^2$ function is an average of translated delta functions.

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Okay, thank you. What could be a pointer to even consider the Dirac delta? –  Jonas Teuwen Nov 28 '10 at 18:24
    
Look at the case functions and convolutions on the unit circle, where the same theorem holds. This can be proven basically by writing $f(x)$ as the sum of its Fourier series, examining how $F$ behaves on each $c_n e^{int}$ and adding. (Of course I'm simplifying things.) So for functions on $R$ you might try finding a continuous analogue to this. –  Zarrax Nov 28 '10 at 18:58
    
I'm trying to parse what you said. If we feed the delta distribution to the operator $T$ (or $T \tau_x$ or $\tau_x T$), then we need to know what the adjoint operator is of $T$ to see what it gives. What do I miss? –  Jonas Teuwen Nov 28 '10 at 21:30
    
In other words, for a fixed $y$, $T(\delta(x - y))$ is the shift by $y$ of $T(\delta)$. Now use that a function $f(x)$ can be expressed as an average of delta functions via $f(x) = \int_Rf(y)\delta(x - y)\,dy$. –  Zarrax Nov 28 '10 at 21:35
    
Ah, that is what you mean, yes, I see. Thank you. However, does this hint how to prove the theorem? It still doesn't look clear to me how to approach it. –  Jonas Teuwen Nov 28 '10 at 22:09
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Maybe you should look at $F \circ T \circ F^{-1}$ where $F$ denotes Fourier transform, so that $\tau_x$ becomes multiplication by $t \mapsto e^{itx}$ and convolution becomes multiplication by something as well.

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