Why is convolution with Fejer kernel enough to prove this result? Exercise 2.9 in Katznelson's Introduction to Harmonic Analysis reads as follows. 

Show that for $f \in L^{1}(\mathbb{T})$ the norm of the operator $ F: g \mapsto f * g$ on $L^{1}(\mathbb{T})$ is $||f||_{L^{1}}$. 

I can see easily enough how $||f * g)||_{1} \leq ||f||_{1}||g||_{1} \ \forall \ g \in L^{1}(\mathbb{T})$ and thus $||F||_{\text{operator}}$ is bounded by using Fubini's theorem, but was struggling to prove the actual equality. 
Then my professor told me that the solution was just to recall that for the Fejer summability kernel $K_{n}$,
$$ f * K_{n} \to f \text{ as } n \to \infty$$
and so $||f * K_{n}||_{L^1} \to ||f||_{L^1}$, therefore $||F||_{\text{operator}} = ||f||_{L^1}$.
My question is, why is it enough to just show this? I have looked at my definitions for Fejer's kernel and summability kernels, and can't make the logical step from trying to show "$\sup_{||g||_{L^{1}} \leq 1} ||f * g||_{L^{1}} = ||f||_{1}$" being equivalent to showing "$||f * K_{n}||_{L^1} \to ||f||_{L^1}$", and so think I must be missing something very obvious. 

Note that $\|K_n\|_{L^1}=1$ (using Katznelson's normalisation). Let the
operator norm of $g\mapsto f*g$ be $A$. We know $A\le \|f\|_{L^1}$.
Then $\|f\ast K_n\|_{L_1}\le A\|K_n\|_{L^1}=A$. But $\|f\ast K_n\|_{L^1}\to
\|f\|_{L^1}$, so in the limit, $\|f\|_{L^1}\le A$.
