Example of $L_1$ and $L_p$ function whose convolution is not in $L_1$

Question

Given functions $f \in L_p(\mathbb{R})$ and $g \in L_1(\mathbb{R})$, one can show that the convolution $f*g$ is well defined and Young's convolution inequality tells us that $\|f*g\|_{p} \leq \|f\|_{p}\|g\|_{1}$, thus showing that $f*g \in L^{p}$. What I want to know is whether it's true that $f*g \in L_1(\mathbb{R})$ for all such $f,g$. On the outset, it seems unlikely that it's true, but I'm having a difficult time finding a counter example. I have tried using some properties of Fourier tranforms and their relations with convolutions to show the existence of $f$ and $g$ with no success. So my question is, is it true that $f*g \in L_{1}$, if it is not, then how does one go about coming with a counterexample? More generally is $f*g \in L_{r}(\mathbb{R})$ for any $r \neq p$?

Answer 1

Giuseppe Negro answer is valid, he is showing you that if the inequality was true, then the constant in the inequality could not be uniform on all functions $f$ and $g$ (which is a contradiction by the Banach–Steinhaus theorem).

If you really want an example, take $g∈ C^\infty_c$ nonnegative (and so in $L^1$) and $f(x) = \frac{1}{(1+|x|)^a}$ with $a\in\left(\frac{1}{p},1\right)$ (you can take for example $a=\frac{1+p}{2\,p}$).

Then $f∈ L^p$, $f\notin L^1$ and the behavior of $f*g$ is the same as the behavior of $f$ when $x\to \infty$, i.e. you can find constants such that $$ \frac{C_1}{(1+|x|)^a}≤ f*g(x) ≤ \frac{C_2}{(1+|x|)^a} $$ and so $f*g∈L^p$ but $f*g∉ L^1$.

Answer 2

No, a quick way of seeing that this cannot work is the scaling argument.

If $f\ast g\in L^1$ for all $f\in L^p, g\in L^1$, by functional analysis arguments$^{[1]}$ you should have an inequality $$\tag{1} \lVert f\ast g\rVert_1\le C \lVert f\rVert_p \lVert g\rVert_1,$$ for some $C>0$. But this cannot be true, because the scaling here is wrong. Indeed, if you let $$ f_\lambda(x):=\lambda^{\frac1p}f(\lambda x),\qquad g_\lambda(x):=\lambda g(\lambda x), $$ then $\lVert f_\lambda\rVert_p=\lVert f\rVert_p$ and $\lVert g_\lambda \rVert_1=\lVert g\rVert_1$, but $$ f_\lambda \ast g_\lambda (x)=\lambda^\frac1p f\ast g(\lambda x), $$ and so $$ \lVert f_\lambda \ast g_\lambda \rVert_1 =\lambda^{\frac1p-1}\lVert f\ast g\rVert_1.$$ This shows that (1) cannot hold for all functions; the left-hand side can be made arbitrarily big by taking $\lambda$ close to $0$, while the right-hand side is invariant.

This is the mathematician version of a typical physicist's tool: dimensional analysis. You can't compare meters and radians, say. But (1) is exactly that; in the left-hand side you have a dimensioned quantity, that is, one that has a certain degree of homogeneity, while in the right-hand side you have a dimensionless quantity. It makes no sense to compare them.

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[1] I cheated here, I don't really know which "functional analysis arguments" should be applied. But I asked a follow-up question on this.

It is generally true that if a linear or bilinear operator is defined everywhere, then it is bounded; the precise version of this fuzzy principle is the closed graph theorem. Thus, the fact that (1) does not hold tell us that something is terribly wrong; this has been my train of thoughts.