# Norm inequality for convolution operators on groups

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 $\ell_2(G)$.

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 $4$.
• Thanks! Do you know if it matters if we take $g$ sufficiently far from $\operatorname{supp} f$? – user10439561 Apr 18 '17 at 15:16
• @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. – Will Sawin Apr 18 '17 at 15:26
• @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 $f$. – Will Sawin Apr 18 '17 at 16:05