If $1\leq p<\infty$, it is easy to find nice necessary and sufficient equality conditions for the convolution inequality $$\lVert f*g\rVert_p\leq\lVert f\rVert_1\lVert g\rVert_p\qquad (f\in L^1(\mathbb{R}^n),\,g\in L^p(\mathbb{R}^n)),$$ for complex $f$ and $g$. But I cannot seem to determine one for $p=\infty$ nor find a reference that contains such a condition. Is there a natural equality condition in this case?

1$\begingroup$ An obvious remark is that $g=\g\_{\infty}$ on a positive measure set (a shifted version of the support of $f$). Conversely, I think for any such $g$ there will be an $f$ that gives equality. $\endgroup$– Christian RemlingJun 30, 2021 at 14:01

$\begingroup$ @ChristianRemling I don't think so  first of all, $f*g$ doesn't have to achieve its sup norm, and secondly I think you're assuming $f$ is positive. If $f = 1_{[0,1]}  1_{[1,0]}$ for instance, then $f * g = 0$ for any constant function $g$. $\endgroup$– Nik WeaverJun 30, 2021 at 14:27

$\begingroup$ Your first statement, that $g$ must satisfy this condition, is false in two ways. One is that if $f$ is not positive then $g = \g\_\infty$ won't help you because you need to counteract the modulus of $f$. The other is that if $f \geq 0$ then $g$ needn't achieve its sup norm, only come arbitrarily close. E.g. $g(t) = 0$ on $(\infty, 1)$ and $g(t) = 1  \frac{1}{t}$ on $[1,\infty)$ will satisfy $\f*g\_\infty = \f\_1\g\_\infty$ for any $f \geq 0$. $\endgroup$– Nik WeaverJun 30, 2021 at 17:52
2 Answers
Let $\theta_f = \frac{f}{f}$ (with any convention for $\frac{0}{0}$, it doesn't matter), let $A_\epsilon = \{t \in \mathbb{R}: f(t) \geq \epsilon\}$, and let $g_s$ be the shifted function $g_s(t) = g(s  t)$.
The condition is: For every $\epsilon > 0$ there exists $s \in \mathbb{R}$ and a scalar $a$ of modulus 1 such that $g_s$ is uniformly within $\epsilon$ of $a\cdot\g\_\infty\cdot\theta_f$ on a subset of $A_\epsilon$ of measure at least $\mu(A_\epsilon)  \epsilon$.
It's easy to see that if $f$ and $g$ satisfy this condition then there will be points $s \in \mathbb{R}$ at which $f*g(s)$ gets arbitrarily close to $\f\_1\g\_\infty$. Conversely, if there are such points then $g$ must satisfy the stated condition: if $g_s$ is too far from $\g\_\infty\cdot \theta_f$ on $A_\epsilon$ then there will be cancellations which keep us away from $\f\_1\g\_\infty$. But it would take a bit of work to write this out.

$\begingroup$ Note that if $f \geq 0$ then $\theta_f$ becomes constantly $1$, which simplifies the condition somewhat. $\endgroup$ Jun 30, 2021 at 14:34

$\begingroup$ This condition may be sufficient, but it is not necessary. Take $g(t)=0$ on $(\infty, 1)$, $g(t)=1\frac{1}{t}$ on $[1, \infty)$ and choose $f\geq 0$ so that $\{t>a: f(t)\geq 1\}$ has positive measure for every $a$. $\endgroup$ Jul 1, 2021 at 12:19

$\begingroup$ Yeah, good point. I have to add an epsilon in there. $\endgroup$ Jul 1, 2021 at 15:19

$\begingroup$ The condition still fails if $f$ is multiplied by $1$ in the above example. Perhaps some additional rotation of $g_s$ should be allowed? $\endgroup$ Jul 2, 2021 at 14:32

$\begingroup$ How would multiplying $f$ by $1$ change either $\f*g\_\infty$ or $\f\_1\g\_\infty$? $\endgroup$ Jul 2, 2021 at 14:35
Here is a fairly simple condition.
It uses the following notion: A family of functions $f_t\in L^1$ depending on a parameter $t$ in a measure space $X$ is said to tend to $f$ somewhere if the essential infimum of $\lVert f_tf\rVert_1$ over $X$ is $0$.
Put $g_s(t)=g(ts)$. The condition is that $afg_s\to\lvert f\rvert\lVert g\rVert_\infty$ somewhere, for some constants $a=a(s)$ with $a=1$.
Clearly, this condition implies equality. For the converse, choose $a$ so that $\lvert\int fg_s\rvert=\int\Re(afg_s)$. Then equality implies that $\Re(afg_s)\to\lvert f\rvert\lVert g\rVert_\infty$ somewhere, and the condition follows via the inequality $\lVert\Im f\rVert_1^2\leq2\lVert f\rVert_1\lVert\lvert f\rvert\Re f\rVert_1$. ($\Re$ and $\Im$ denote real and imaginary parts.)