# When is $\lVert f*g\rVert_\infty=\lVert f\rVert_1\lVert g\rVert_\infty$?

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?

• 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. Jun 30, 2021 at 14:01
• @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 $g$. Jun 30, 2021 at 14:27
• 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 $f \geq 0$. Jun 30, 2021 at 17:52

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.

• Note that if $f \geq 0$ then $\theta_f$ becomes constantly $1$, which simplifies the condition somewhat. Jun 30, 2021 at 14:34
• 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 $a$. Jul 1, 2021 at 12:19
• Yeah, good point. I have to add an epsilon in there. Jul 1, 2021 at 15:19
• The condition still fails if $f$ is multiplied by $-1$ in the above example. Perhaps some additional rotation of $g_s$ should be allowed? Jul 2, 2021 at 14:32
• How would multiplying $f$ by $-1$ change either $\|f*g\|_\infty$ or $\|f\|_1\|g\|_\infty$? 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 $$\lVert f_t-f\rVert_1$$ over $$X$$ is $$0$$.

Put $$g_s(t)=g(t-s)$$. 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.)