Let $(h_n)$ be a sequence of non-zero functions in $L_1(G)$ (where $G$ is a locally compact group) with the property $$ \left\Vert\sum_{n=1}^\infty f * h_n\right\Vert=\sum_{n=1}^\infty\Vert f*h_n\Vert $$ for all $f\in L_1(G)$. My conjecture is that there exist an angle $\theta$ and a sequence of positive reals $(r_n)$ such that $$ h_n=e^{i\theta}r_n h_1 $$ Can someone confirm or disprove it?

  • $\begingroup$ Before asking people to put work into the general case, can you tell us which special cases you have tried, or what partial results you have obtained? e.g. for T or R or Z $\endgroup$
    – Yemon Choi
    Jul 15 '14 at 12:26
  • $\begingroup$ Also, where did this problem originate? Is it from a question someone asked you; is it a step you need in order to prove some other result? $\endgroup$
    – Yemon Choi
    Jul 15 '14 at 12:28
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    $\begingroup$ @YemonChoi, I have only a vague idea why this is true. For simplicity $G$ is discrete. Then the eqaulity holds for $f=\delta_x$, $x\in G$ since they are just translations. If we take $f$ a linear combination of several deltas, then supports of $h_n$ may mess up and lhs can become smaller than rhs. And yes it is a step in my research that I can't make. $\endgroup$
    – Norbert
    Jul 15 '14 at 17:17
  • $\begingroup$ In the discrete case I think that by examining equality in the triangle inequality, $\left\Vert\sum_n f_n\right\Vert=\sum_n\Vert f_n\Vert$ implies that $f_i(x)=r_i(x)e^{i\theta(x)}$ for all $x\in G$, where $r_i$ takes values in positive reals and $\theta(x)$ doesn't depend on $i$. $\endgroup$ Jul 15 '14 at 23:44
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    $\begingroup$ @user53043 Just to clarify, my previous comment doesn't contradict the original formulation of the conjecture; what I wrote is nearly the same as $f_n(x)=r_n(x)f_1(x)$ (for some positive $r_n$), except that $f_1$ may be zero when $f_n$ is not. It does say that pointwise either the functions are zero or have the same argument. $\endgroup$ Jul 16 '14 at 16:28

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