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 $g$. 



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> Is it true that $\Vert f\Vert\le \Vert f+\delta\_g\Vert$? 
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The norms here are in $B(\ell\_2(G))$, as convolution operators on $\ell\_2(G)$.



Norm inequality for convolution operators on groups
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](http://www.wolframalpha.com/input/?i=maximize%20%20%7Ce%5E%7Bi%20theta%7D%20%2B%20e%5E%7B%202i%20theta%7D%20-%201%20%20%2B%20e%5E%7B-%20i%20theta%7D%20%2B%20e%5E%7B%20-%202%20i%20theta%7D%7C%20%20for%20theta%20from%200%20to%202%20pi) shows that this maximum is $13/4$ attained at $z = \frac{ (\sqrt{3} + i \sqrt{5})^2 }{8}$ and thus is less than $4$.


