What is wrong with this proof that a linear, bounded, time invariant operator on $L_p$ must be a convolution?

Question

I'm trying to understand if this is true and how to prove it, "If $T$ is a bounded, time invariant operator on $L_p(\mathbb{R})$, then $T$ is a convolution operator.''

Here's an attempt at a proof that seems to work.

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Assume ( T ) is time-invariant, i.e., for any shift ( a>0 ), ( T(S_a f) = S_a (T f) ). Let $\phi_n$ be an approximation to the identity (see previous lemma), with the added condition that $\|\phi_n\|_p = 1$, e.g., $\phi_n = n^{1/p}$ on $[\pm 1/(2n)]$. Since $T$ is bounded (continuous), we see that for any $f\in L_p, \|T(f*\phi_n) - T(f)\|_p \to 0.$

Next, since $\|T \phi_n\|\leq \|T\|$, we can apply Banach-Alaoglu Theorem, and find a subsequence $n_k$ such that $T\phi_{n_k}$ converges weakly to a function $h\in L_p$. That is, for all $g\in L_q$ with $q$ conjugate exponent to $p$, we have $\langle h, g\rangle = \lim_k \langle T\phi_{n_k}, g\rangle = \lim_k \langle \phi_{n_k}, T^* g \rangle = \overline{T^*g} (0) $ if $T^*g$ is continuous at $t = 0$. Finally, \begin{align*} T(f*\phi_n)(t) &= T \int_\tau \phi_n(t-\tau)f(\tau)d\tau \\ &= \int_\tau f(\tau) T (\phi_n(t-\tau)) d\tau \text{ --- by linearity}\\ &= \int_\tau f(\tau) TS_\tau(\phi_n(t))d\tau \\ & = \int_\tau f(\tau) S_\tau T(\phi_n(t))d\tau \text{ --- by time invariance}\\ & = f*(T\phi_n) \end{align*} Letting $n\to \infty$ we have $Tf(t) = f*h$.

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So if this proof is true, then we proved there is some $h\in L_p$ with $\|h\|_p = \|T\phi_n\|_p \leq \|T\|$ and such that $\langle h, g\rangle = \bar{g}(0)$ for continuous functions $g$. I don't think such a function $h$ can exist. If so it would be the Dirac $\delta$, right? If so, what is wrong with this proof?

Is it true that $T$ bounded, linear, time invariant means it is a limit of convolutions?

Answer 1

The problem is that (as you noted by the conclusion!) your approximation of the identity is not such. The usual argument with mollifiers requires the function $\phi$ to be in $L^1$. And in $L^1$ the argument cannot possibly work because there is no Banach-Alaoglu; or there is, in a sense, but you get a measure and not necessarily a function $h$.

When $p>1$, you don't get a mollifier. Mollifiers for $p>1$ are unbounded.

Also, the first step in your argument is valid, but possibly not by the reason you state (linearity); it is a lot more than that. In general, you cannot treat the functions inside the integral as functions where you can apply your operator. Let us make a simple counterexample. Let $Tf=1_{[1,2]}\,f$. Put $h=1_{[0,1]}$, and $$\tag1 g(t)=(h*f)(t)=\int_{-\infty}^\infty h(t-\tau)\,f(\tau)\,d\tau=\int_{t-1}^tf. $$ Then $(Tg)(1)=\int_0^1f$. But $Th=1_{\{1\}}$, and so $Th*f=0$ for all $t$.

Here is another example. Let $(Tf)(t)=f(t^2)$ and put $g=h*k$. Then $$ (Tg)(t)=g(t^2)=\int_{-\infty}^\infty h(t^2-s)\,k(s)\,ds, $$ while $$ (Th)*k(t)=\int_{-\infty}^\infty h((t-s)^2)\,k(s)\,ds. $$

Having said all the above, the first step in your argument is valid; and that's because $T$ is not only linear and bounded, but translation invariant for all translations. What happens is this. You want $Tg$, where $g=h*k$. We have \begin{align} (h*k)(t)&=\int_{-\infty}^\infty h(t-s)\,k(s)\,ds =\lim_N\int_{-N}^Nh(t-s)\,k(s)\,ds\\[0.2cm] &=\lim_N\lim_M\sum_{j=1}^M h(t-s_j)k(s_j)\Delta_j\\[0.2cm] &=\lim_N\lim_M\sum_{j=1}^M h_j(t)k(s_j)\Delta_j, \end{align} where $h_j=S_{s_j}h$. Now the Riemann sum is written as a linear combination of the functions $h_j$, and the linearity and continuity of $T$ give us $$ (T(h*k))(t)=\lim_N\lim_M\sum_{j=1}^M (Th_j)(t)k(s_j)\Delta_j. $$ And because $TS_{s_j}=S_{s_j}T$, we have that $(Th_j)(t)=(Th)(t-s_j)$. Then \begin{align} (T(h*k))(t)&=\lim_N\lim_M\sum_{j=1}^M (Th)(t-s_j)k(s_j)\Delta_j\\[0.2cm] &=\int_{-\infty}^\infty (Th)(t-s)\,k(s)\,ds \end{align}

Finally, the article by Lars Hörmander: Estimates for translation invariant operators in Lp spaces, Acta Math.104 (1960), 93–140 (link here if you have access) states that translation invariant operators in $L^p(\mathbb R^n)$ are given by convolution against a distribution. Another link to the Hörmander paper