3

Let $L$ be a linear translationally invariant operator $L:\{f:\mathbb{Z} \rightarrow \mathbb{C}\} \rightarrow \{f:\mathbb{Z} \rightarrow \mathbb{C}\}$, i.e. $$LT=TL,$$ where $T:\{f:\mathbb{Z} \rightarrow \mathbb{C}\} \rightarrow \{f:\mathbb{Z} \rightarrow \mathbb{C}\}$ is the translation operator defined by $(Tf)(x) := f(x+1)$ for every $f$.

I'd like to see a proof for that $L$ can be represented by some stencil, i.e., there exist numbers $\alpha_k \in \mathbb{C}$, such that $$(Lf)(x)= \sum_k \alpha_k f(x+k) $$ is true for every $f$ and $x$. I guess a proof could be an elegant one-liner, but somehow I don't see how to do that.

Edit: I think that this statement is very similar to the theorem that the translation invariant operators on $L^2$ are exactly the multiplier operators.

Nils
  • 133
  • 9
  • Interesting question. Is there any particular reason you believe that this is true? – Ben Grossmann May 22 '17 at 15:45
  • Note that this is not necessarily true unless we include infinite sums. For instance, an operator like $$ (Lf)(x) = \sum_{k=1}^\infty 2^{-k} f(x+k) $$ is not expressible with a finite sum. – Ben Grossmann May 22 '17 at 15:48
  • It would be easiest, I think, if you restricted the set of sequences to $$ \ell^2(\Bbb Z) = {f:\Bbb Z \to \Bbb C : \sum_{k \in \Bbb Z}|f(k)|^2 < \infty} $$ in which case you'd be considering the commutant of the unitary operator $T$ over a Hilbert space – Ben Grossmann May 22 '17 at 16:00
  • I'm actually OK with Infinite sums. Somehow, this is the way I think about translationally invariant operators. I think I have a proof using Fourier-transformation, even though it is not that short. Yes, for that purpose I should restict $f$ to $\ell^2$ – Nils May 22 '17 at 16:02
  • It might help if you gave the outline of your proof – Ben Grossmann May 22 '17 at 16:29
  • It would be very easy to show that this were true if you were looking at functions $f: \Bbb Z_n \to \Bbb C$, where $\Bbb Z_n$ are the integers modulo $n$. – Ben Grossmann May 22 '17 at 16:31

1 Answers1

0

The following "proof" has a couple of errors:

Let $P=[0,2\pi]$ and $F$ be the Fourier transformation, i.e., $$F:\ell^2(\mathbb{Z}) = \{f:\mathbb{Z} \rightarrow \mathbb{C} : \sum_{k\in\mathbb{Z}}|f(k)|^2 < \infty\} \rightarrow \{\hat{f}(k):P\rightarrow \mathbb{C} :\int_P |\hat{f}(k)|^2 dk < \infty \}\}$$ with $$(Ff)(k) = \hat{f}(k) = \sum_{x\in\mathbb{Z}} f(x) e^{-i k x}$$ and $$(F^{-1}\hat{f})(x) = f(x) = \frac{1}{|P|}\int_P \hat{f}(k) e^{i k x}.$$

The eigenfunctions of $T$ are $e^{i k \cdot}$ since $Te^{ik x} = e^{ik(x+1)} = e^{ik} e^{ikx}$ for every $x$. Since $TL=LT$ the eigenfunctions of $L$ are the same, i.e., there exist numbers $m_k\in\mathbb{C}$, such that $Le^{ikx}=m_ke^{ikx}$ for every $x$.

We additionaly have that \begin{align}\widehat{Tf}(k) &= (F f(\cdot + 1) )(k)\\ &= \sum_{x\in\mathbb{Z}} f(x+1) e^{-i k x} \\ &= \sum_{x\in\mathbb{Z}} f(x) e^{-i k (x-1)} = e^{ik} \hat{f}(k).\end{align}

Using $f(x)=\frac{1}{|P|}\int_P \hat{f}(k) e^{i k x} dk$ we finally find

\begin{align} (Lf)(x) &= \frac{1}{|P|}\int_P m_k \hat{f}(k) e^{i k x} dk \\ &= \frac{1}{|P|}\int_P m_k \hat{f}(k) e^{i k x} dk \\ &= \frac{1}{|P|}\int_P m_k \widehat{T^x f}(k) dk \\ &= \frac{1}{|P|}\int_P m_k (FT^x f(\cdot))(k) dk \\ &= \frac{1}{|P|}\int_P m_k (F f(\cdot + x))(k) dk \\ &= \frac{1}{|P|}\int_P m_k \sum_{\ell\in\mathbb{Z}} f(x+\ell) e^{-i k \ell} dk \\ &= \sum_{\ell\in\mathbb{Z}} f(x+\ell)( \frac{1}{|P|}\int_P m_k e^{-i k \ell} dk )\\ &:= \sum_{\ell\in\mathbb{Z}} f(x+\ell) \alpha_k. \end{align}

Nils
  • 133
  • 9
  • Since $TL = LT$, the eigenfunctions of $L$ are the same: this strikes me as a very suspicious step. Even for finite dimensional operators (i.e. the $f$ over $\Bbb Z_n$ case), this amounts to the assumption that $L$ is diagonalizable. – Ben Grossmann May 22 '17 at 16:55
  • Hmmm. Do you have a counter example? Can $L$ be singular if it commutes with $T$?(Despite $L=0$) – Nils May 22 '17 at 17:07
  • Of course $L$ can be singular (e.g. $L = 0$), the issue is whether $L$ can be defective. In the finite dimensional case, it turns out that every matrix which commutes with $T$ is necessarily normal. – Ben Grossmann May 22 '17 at 17:11
  • For a non-trivial example, note that $T - I$ is singular but commutes with $T$. – Ben Grossmann May 22 '17 at 17:13
  • Thank you very much. I actually don't see the problem with the singularity. $T-I$ has the eigenfunctions $e^{ik \cdot}$ and the eigenvalues $e^{ik}-1$. And this operator can be represented as $f(x) \mapsto f(x+1)-f(x)$. What can go wrong? – Nils May 23 '17 at 08:45
  • @Omnomnomnom: I think I see at least a couple of problems in the infinite case which we cannot bypass so easily. You said, that it is easy to show in the finite case (functions $f:\mathbb{Z}_n \rightarrow \mathbb{C}$.). Do you see a simpler proof than the one I stated here? – Nils May 26 '17 at 12:54
  • actually, your proof works very well in the finite dimensional case. Effectively, we're trying to prove that the matrices commuting with $T$ are the circulant matrices. In this setting, we do have the luxury of saying that if $TL = LT$ (since $T$ is diagonalizable with no repeating eigenvalues), $L$ must be diagonalizable with the same eigenbasis (namely, the DFT basis). – Ben Grossmann May 26 '17 at 13:55
  • Okay thanks, I think I understand the finite case. So a proof in the infinite setting has to look fundamentally different, since the translation operator does not even have eigenfunctions... – Nils May 26 '17 at 14:14
  • right, but it is a unitary operator on a Hilbert space, and the spectrum of the operator is all complex numbers with magnitude $1$, so there's that. – Ben Grossmann May 26 '17 at 14:30
  • More on the spectrum of your operator here. – Ben Grossmann May 26 '17 at 14:34
  • It might also be useful to consider exercise 4.2.2 of Pedersen's "Analysis Now". Notably, there is a one to one correspondence between the continuous $\Bbb C$-valued functions on the unit circle in the complex plane and the "stencil" operators. In particular: to show that an operator $L$ is a "stencil operator", it suffices to state that $L = f(T)$ for some continuous function $T$, where $f(T)$ is defined via functional calculus. – Ben Grossmann May 26 '17 at 14:47
  • Thanks for the interesting notes. I'm going to take a look at that exercise asap. (You meant 4.4.2 instead of 4.2.2). – Nils May 26 '17 at 17:57
  • Yes! Thank you. I'm glad you can access the textbook, then. – Ben Grossmann May 26 '17 at 19:51
  • I think you're right to make the connection with multiplication operators on $L^2$. This post is relevant to that end – Ben Grossmann May 26 '17 at 19:57