3
$\begingroup$

Let $G$ be a discrete, finitely generated group. Let $f\in \mathbb{C} G$ be given. Consider $g\in G\setminus \operatorname{supp} f$ and let $\delta_g$ denote the Dirac delta at $g$.

Is it true that $\Vert f\Vert\le \Vert f+\delta_g\Vert$?

The norms here are in $B(\ell_2(G))$, as convolution operators on $\ell_2(G)$.

$\endgroup$
6
$\begingroup$

The operator norm of a convolution operator on $\mathbb Z$ is the supnorm of its Fourier transform.

Let $f(i) = -1$ if $i= -1 ,0,-1,-2$ and $0$ otherwise. Then the operator norm of convolution with $f$ is certainly $4$. If we add the delta function at $0$ the operator norm should be $ \max_{z \in S^1} | z + z^2 + z^{-1} + z^{-2}-1 | $. Direct calculation shows that this maximum is $13/4$ attained at $z = \frac{ (\sqrt{3} + i \sqrt{5})^2 }{8}$ and thus is less than $4$.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ Thanks! Do you know if it matters if we take $g$ sufficiently far from $\operatorname{supp} f$? $\endgroup$ – user10439561 Apr 18 '17 at 15:16
  • $\begingroup$ @user10439561 In the abelian case, it should be fine with $g$ sufficiently far away. The idea is we can adjust the character a small amount, barely affecting the character sum against $f$, to ensure that the character takes a desired value on $g$. I'm not sure about nonabelian. $\endgroup$ – Will Sawin Apr 18 '17 at 15:26
  • $\begingroup$ @user10439561 Alternately, this should work: Take a function $h$ of finite support such that $|f * h|_{\ell_2}/ |h|_{\ell_2}$ is close to the operator norm of $f$. As long as the support of $f * h$ and $\delta_g * h$ are disjoint, the $\ell^2$ norm of the sum will be the square root of the sum of the squares of the $\ell^2$ norms. This will hold as long as $g$ is sufficiently far (depending on $h$) from the support of $f$. $\endgroup$ – Will Sawin Apr 18 '17 at 16:05

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.