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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?

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    $\begingroup$ 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. $\endgroup$ Commented Jun 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 Weaver
    Commented Jun 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 Weaver
    Commented Jun 30, 2021 at 17:52

2 Answers 2

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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.

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  • $\begingroup$ Note that if $f \geq 0$ then $\theta_f$ becomes constantly $1$, which simplifies the condition somewhat. $\endgroup$
    – Nik Weaver
    Commented 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 $a$. $\endgroup$
    – apanpapan3
    Commented Jul 1, 2021 at 12:19
  • $\begingroup$ Yeah, good point. I have to add an epsilon in there. $\endgroup$
    – Nik Weaver
    Commented 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? $\endgroup$
    – apanpapan3
    Commented Jul 2, 2021 at 14:32
  • $\begingroup$ How would multiplying $f$ by $-1$ change either $\|f*g\|_\infty$ or $\|f\|_1\|g\|_\infty$? $\endgroup$
    – Nik Weaver
    Commented Jul 2, 2021 at 14:35
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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.)

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