For $f\in L^1(\mathbb T)$ and $p\in [1,\infty]$, consider the bounded linear operator $T_f:L^p(\mathbb T)\to L^p(\mathbb T)$ given by $T_f(g)=f\ast g$. If $f\ge 0$, then prove that $\Vert T_f\Vert_{p-p}=\Vert f\Vert_1$.
Here, $\mathbb{T}$ is the unit circle, and we have defined $$ (f * g)(x) = \frac{1}{2\pi}\int_{-\pi}^\pi f(x - t) g(t)\:dt.$$
I can show that $\Vert f*g\Vert_p \leq \Vert f\Vert_1\Vert g\Vert_p$, and $g=1$ proves the equality.
But, is the result still true for complex valued $f$?
The conclusion is not true for complex valued $f.$ Let $p=2.$ We have $$\widehat{f*g}(n)=\widehat{f}(n) \widehat{g}(n)$$ where $\widehat{h}(n)$ denotes the $n$th Fourier coefficient of $h\in L^2.$ By Parseval identity the operator $T_f$ is unitarily equivalent to the operator $M_f$ acting on $u\in\ell^2(\mathbb{Z})$ by $$(M_fu)(n)=\widehat{f}(n)u(n)$$ The norm of $M_f$ is equal to $\sup_n|\widehat{f}(n)|\le \|f\|_1.$ The inequality may be strict, like for example when $f(t)=e^{-it}+e^{it}.$ Indeed $\|M_f\|=1$ but $$\|f\|_1={1\over 2\pi}\int\limits_{-\pi}^{\pi}2|\cos t|\,dt={4\over \pi}$$