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Let $\mathcal{S}(\mathbb{R})$ be the Schwartz-space, I define on it the two operators

$$\tau_h(f)(x) := f(x-h)$$ $$Df(x) := f'(x)$$

Now let $T : \mathcal{S}(\mathbb{R}) \to \mathcal{S}(\mathbb{R})$ be an operator that commutes with $D$, is it true that $T$ commutes with $\tau_h$ for all $h$?

The heuristic idea behind this is the following

$$\tau_h(f)(x) = \sum_{k=0}^{\infty}{D^kf(x)\frac{h^k}{k!}}$$

So that If I apply $T$ both sides and commuting $T$ with $D^k$ I get

$$T\tau_hf = \sum_{k=0}^{\infty}{TD^kf\frac{h^k}{k!}} = \bigg[\sum_{k=0}^{\infty}{\frac{h^k}{k!}D^k}\bigg]Tf = \tau_hTf$$

Of course the problem is that is not true in general that $\sum_{k=0}^{\infty}{D^kf(x)\frac{h^k}{k!}} = f(x+h)$.

Actually the converse implication is really easy, just let $h \to 0$ in the equality

$$T(\frac{1}{h}(f - \tau_hf)) = \frac{1}{h}(Tf - \tau_hTf)$$

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  • $\begingroup$ We're in the schwartz space, so my first thought is to take the Fourier transform of both identities and see what there is to see. $\endgroup$
    – Charles Hudgins
    May 26 at 0:50
  • $\begingroup$ I get that it's equivalent to prove that if $\hat{T}$ is an operator on $\mathbb{S}_{\xi}(\mathbb{R})$ that commutes with $\xi \,\cdot$ (multiplication by $\xi$) then $\hat{T}$ commutes with $e^{ih\xi}\,\cdot$. It's easy to prove the equality $\hat{T}(\sum_{k=0}^{n}{\frac{(ih\xi)^k}{k!}\hat{f}}) = \sum_{k=0}^{n}{ \frac{(ih\xi)^k}{k!}\hat{T}\hat{f} } $, now I would like to let $n \to \infty$, the RHS converges pointwise to $e^{ih\xi}\hat{T}\hat{f}(\xi)$ but I'm unable to say anything about the LHS $\endgroup$
    – Paul
    May 26 at 9:42

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