Closed Range Convolution Operator

Question

Does there exists a nontrivial $f \in L^1(\mathbb{R})\cap L^\infty(\mathbb{R})$ such that $f\ast L^\infty(\mathbb{R})$ is a closed subspace of $L^\infty(\mathbb{R})$?

I couldn't find any good reference on closed range convolution operators.

Answer 1

I am assuming that $f$ is a nonnegative function in $L^1(\mathbb R)$. Then the range of the operator $Tg:=f\ast g$ is closed only in the trivial case $f=0$. Here $T\colon L^\infty(\mathbb R)\to L^\infty(\mathbb R)$.

Proof. Letting $g_n(x):=\mathbf{1}_{\left[-\tfrac1n, \tfrac1n\right]}(x)$, we have
$$ \lVert g_n \rVert_{\infty}=1,\quad \lVert Tg_n\rVert_\infty \to 0,\quad Tg_n\ne 0;$$ the last of the three being a consequence of $f\ge 0$, while the second is a consequence of $f\in L^1(\mathbb R)$.

Now, the existence of such a sequence $g_n$ implies that the range of $T$ is not closed. Indeed, $T$ is injective (here I use again that $f\ge 0$). If it had a closed range, its inverse would be bounded. And this is precisely what the sequence $g_n$ above disproves.