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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.

Giuseppe Negro
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2 Answers2

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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.

Rhys Steele
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  • Great answer! I think you can avoid the closed graph theorem. Every function $g_n$ is bounded and compactly supported, thus $\lVert g_n\ast f\rVert_1 \le \lVert g_n\rVert_r \lVert f\rVert_p$ for $2=\tfrac1r+\tfrac1p$, by the inequality of Young. So, $T_{g_n}$ is a bounded operator on $L^p$ with operator norm bounded above by $\lVert g_n\rVert_r<\infty$. The same applies to $T^{f_n}$. – Giuseppe Negro Apr 29 '20 at 14:41
  • @GiuseppeNegro Doesn't that inequality require $r, p \geq 1$ and so if $p > 1$ there is no suitable $r$ satisfying the required equality for $r$ and $p$? I wanted to use Young's convolution inequality also but this seemed to be an issue. – Rhys Steele Apr 29 '20 at 14:50
  • The point is that, unlike $g$, $g_n$ is in every $L^p$ space. – Giuseppe Negro Apr 29 '20 at 17:09
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    @GiuseppeNegro Sure but $p$ corresponds to the space in which you find $f$ and not $g$. So if $p > 1$ then $\frac{1}{p} < 1$ and so we would need to have $\frac{1}{r} > 1$ in order to satisfy $\frac{1}{p} + \frac{1}{r} = 2$. As far as I know, Young's convolution inequality requires $r \geq 1$ and so we can't apply that inequality here (Obviously $g \in L^r$ for $r < 1$, that's not the thing that worries me). Am I wrong about the restriction $p,r \geq 1$ for this version of Young's? – Rhys Steele Apr 29 '20 at 17:13
  • No you are not. You are right, that's an obstruction to the application of Young, even if $g_n$ is bounded and compactly supported. Seeing also Jan's answer, I have the impression that every proof of the result of my question should use in some way the closed graph theorem. – Giuseppe Negro Apr 29 '20 at 22:46
  • I accepted the other answer because it is shorter, but it has been a hard decision, as your answer is also very good. Also I must thank you for raising that issue on the Young inequality. I had this wrong idea that convolving against a smooth and compactly supported function defined a bounded operator $L^p\to L^q$ for all $p$ and $q$. You made me realize this is not true. Thanks! – Giuseppe Negro Apr 30 '20 at 12:59
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    @GiuseppeNegro I agree with your decision to accept the other answer. There are really no non-standard ideas in this answer, even if applying those ideas takes some work. The other answer gives an elegant and more general technique which I would have preferred to use if I had been aware of it before writing this so I also learnt something new from this nice question. – Rhys Steele Apr 30 '20 at 13:25
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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.

Jan Bohr
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  • In the argument in the first paragraph, how are you seeing that if $x_n \to x$ in $E$ and $Tx_n \to y$ in the topology of F then $y$ is the limit of $Tx_n$ in the topology of $G$? In fact, if $|\cdot|_1, |\cdot|_2$ are inequivalent complete norms on a vector space $X$ and we take $E = G = (X, |\cdot|_1)$ and $F = (X, | \cdot|_2)$ then it is clear that $\operatorname{Id}: G \to G$ is continuous and $\operatorname{Id}(G) \subseteq F$ but you do not have continuity into $F$ or the norms would be equivalent by the open mapping theorem. – Rhys Steele Apr 29 '20 at 15:51
  • Oh, I forgot the important assumption that $F$ should be continuously embedded in $G$. Thanks for pointing that out! – Jan Bohr Apr 29 '20 at 16:01
  • Yes, the fact that convolution is continuous as a map of $L^1\times L^p$ into $\mathcal{D}'$ is true. Indeed, $\ast\colon L^1\times L^p\to L^p\subset \mathcal{D}'$ by the inequality of Young. – Giuseppe Negro Apr 29 '20 at 22:26
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    This is a very good answer. This explains why it is practically always the case, in analysis, that an operator that is everywhere defined must be bounded. Most operators arising in practice are continuous as maps into distributions, or similar "weak" spaces. – Giuseppe Negro Apr 29 '20 at 22:40
  • Exactly! And besides your nice application to prove a negative result, it is also very useful to establish continuity into spaces with a complicated Fréchet-topology. A nice example of this is multiplication of pseudodifferential operators $\Psi^0_K\times \Psi^{0}_L\rightarrow \Psi^{-1}$, where the subscript indicates that the principal symbol shall be contained in $K$ and $L$ respectively and we assume the sets to be compact and disjoint. – Jan Bohr Apr 30 '20 at 06:50
  • Continuity into $\Psi^0$ is clear, as multiplication is continuous, and just the fact that for $A\in \Psi^0_K$ and $B\in \Psi^0_L$ we have $AB\in \Psi^{-1}$, then implies continuity into $\Psi^{-1}$. I have been wondering whether this trick can be extended beyond bilinear maps, see here, but so far without answer. – Jan Bohr Apr 30 '20 at 06:52