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Consider the convolution operator on $L^1(\Bbb R)$, $f→f∗g$, where $g$ is some $L1$ function. I need to show that the norm of this operator equals to $||g||_1$.
I have seen a question here which gives an answer to the case when $g$ is positive here:
Limit of convolution

I assume that the way of showing equality won't change for a negative function, but i have troubles with approach for a 'mixed' function, for which $Im(g) \in R$. Would be very glad for any help.

Антон Рябченко
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    The answer in the link you gave assumes no positivity in $g$, unless I miss it. The OP raises this as a criteria but I don't believe the answer posed by Martin-Blas assumes this a priori. –  May 14 '20 at 11:08
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    @808GroundState As i understood it, the proof uses positiveness of g only once in approximation ||g||1 with an integral from −n to n. – Антон Рябченко May 14 '20 at 11:41
  • Let $f_\varepsilon$ be an approximation of identity, and let $\epsilon\rightarrow 0$, you can get the result. – Antimonius Sep 04 '23 at 10:29

1 Answers1

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Let $f=\chi_{[0,1]}$.

$$\int_{R^d}[\int_{R^d}f(x-y)g(y)dy]dx = \int_{R^d}f(x-y)dx\int_{R^d}g(y)dy = \int_{R^d}g(y)dy$$ which is not neccessarily equal with $||g||_{L^1(R^d)}$
Actually what you can say is that $||*|| \leq ||g||_{L^1(R^d)}$

Khaled Alekasir
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