Convolution must be a bounded bilinear operator if it is well-defined

Question

In this answer I claimed the following.

Claim. Suppose that the convolution $f\ast g$ belongs to $L^1(\mathbb R)$ for all $f\in L^p(\mathbb R)$ and all $g\in L^1(\mathbb R)$. Then there is a constant $C>0$ independent on $f$ and $g$ such that $$\tag{1}\lVert f\ast g\rVert_1\le C\lVert f \rVert_p \lVert g \rVert_1.$$

This is an empty statement, as it is not true that $f\ast g\in L^1$ for all $f\in L^p, g\in L^1$; see this answer, for example. And indeed, the conclusion (1) is also false and it can be easily disproved by the scaling argument.

The idea of my linked answer is to prove by contradiction that $f\ast g$ may fail to be in $L^1$, using that (1) cannot hold. But then I realized that I cannot easily prove the Claim above.

Question. Can you prove the Claim?

I had carelessly thought that this Claim followed from a straightforward adaptation of the classic application of the uniform boundedness principle given, for example, in this answer. There, we prove that if $g$ is a measurable function such that $fg\in L^1$ for all $f\in L^p$, then there is a $C>0$ such that $$\left\lvert \int fg\ \right\rvert \le C\lVert f\rVert_p.$$ This follows from the uniform boundedness principle and from dominated convergence. But I don't see how to apply the same reasoning to the problem at hand.

Answer 1

First some abstract stuff: Suppose $E,F,G$ are locally convex spaces such that $F\subset G$ (with continuous embedding) and $T:E\rightarrow G$ is a continuous linear map with $T(E)\subset F$. If $E$ and $F$ are Fréchet, then the closed graph theorem implies that $T$ is automatically continuous as map $T:E\rightarrow F.$ A similar argument works for a bilinear map $B:E_1 \times E_2 \rightarrow G$ with $B(E_1,E_2)\subset F$, applying the linear result to $B(x,\cdot)$ and $B(\cdot, y)$ and noting that if $E_1,E_2$ and $F$ are Fréchet, then separate continuity in each variable implies joint continuity.

Hence, if you can show that convolution is continuous as map $L^1 \times L^p\rightarrow G$ for some locally convex space $G\supset L^1$, then the assumption $L^1\ast L^p \subset L^1$ and the abstract nonsense from above already imply continuity into $L^1$. I suppose that $G= \mathcal{D}'(\mathbb{R})$ should work but I have not worked that out.

Answer 2

Under the assumption of your question, the bilinear operator $T: L^p \times L^1 \to L^1$ by $T(f,g) = f \ast g$ is well-defined. Additionally, define $T^f:L^1 \to L^1$ and $T_g:L^p \to L^1$ for $f \in L^p$ and $g \in L^1$ by setting $T^f(g) = T(f,g) = T_g(f)$. I proceed in several steps.

Step 1: $T_g$ is bounded

This is very similar to the classic application of the UBT that you mention. Mimicking that application, set $$g_n(x):=\begin{cases} n, & \lvert g(x) \rvert \ge n\ \text{and } |x|<n, \\ g(x), & \lvert g(x)\rvert <n\ \text{and }|x| < n,\\ 0,& |x| \geq n. \end{cases}$$

By the closed graph theorem, each $T_{g_n}$ is a bounded operator. Indeed, suppose that $f_k \to f$ in $L^p$ and $T_{g_n} f_k \to h$ in $L^1$. Then notice that \begin{align*} \left | \int (f(y) - f_k(y)) g_n(x-y) dy \right | \leq \|f-f_k\|_{L^p} \|g_n\|_{L^{p'}} \leq C_n \|f-f_k\|_{L^p} \to 0 \end{align*} as $k \to \infty$. This means that $T_{g_n} f_k \to T_{g_n} f$ pointwise as $k \to \infty$ and so $h = T_{g_n} f$.

Also, we have that $|T_{g_n}f| \leq T(|f|,|g|)$ pointwise and $T(|f|,|g|) \in L^1$ by assumption. Therefore, by an application of the uniform boundedness theorem, $C_1 := \sup_n \|T_{g_n}\| < \infty$.

To conclude this step, it remains to see that $T_{g_n}f \to T_g f$ in $L^1$ as $n \to \infty$. For this, first notice that $$|f(x-\cdot) [g_n(\cdot) - g(\cdot)] | \leq 2 |f(x-\cdot) g(\cdot)|$$ and since $T(|f|,|g|) < \infty$ a.e. the right hand side is in $L^1$ for almost all $x$. Hence we can apply the dominated convergence theorem to see that $T_{g_n}f \to T_gf$ a.e. Then using the fact that $|T_{g_n}f - T_g f| \leq 2 T(|f|,|g|)$ we can apply the dominated convergence theorem again to see that $T_{g_n} f \to T_g f$ in $L^1$.

Step 2: $T^f$ is bounded

This is basically the same argument as above. Define $$f_n(x):=\begin{cases} n, & \lvert f(x) \rvert \ge n\ \text{and } |x|<n, \\ f(x), & \lvert f(x)\rvert <n\ \text{and }|x| < n,\\ 0,& |x| \geq n. \end{cases}$$ The argument then runs almost line for line the same as in step $1$ with the roles of $f$ and $g$ reversed, except that in the application of the closed graph theorem you now have $p = 1$ and $p' = \infty$ (which causes no issues at all).

Step 3: The conclusion

This is now a standard application of the UBT. Consider the set $U = \{T_g : \|g\|_{L^1} = 1\}$. Then for each $g$ with $\|g\|_{L^1} = 1$, $$\|T_g f\| = \|T^f g \| \leq \|T^f\|$$ so that by the UBT, $C_2 = \sup_{\|g\|_{L^1} = 1} \|T_g\| < \infty$. Hence for arbitrary $f \in L^p$ and $g \in L^1$ $$\|T(f,g)\|_{L^1} = \|g\|_{L^1} \|T_{\frac{g}{\|g\|_{L^1}}} f \| \leq C_2 \|g\|_{L^1} \|f\|_{L^p}$$ as desired.