I was about to post yet another question about a comment in my notes. But I think my key point in misunderstanding jumps in proofs is how the $L_{p}$ spaces are nested in eachother.
The example that always makes me question this is the function $f(x) = \frac{1}{\sqrt{x}}$ which is not bounded on $[0,1]$ yet is integrable.
The following remark gives rise to my question:
If $\{\phi_{n}\}$ is a summability kernel and $f\in L_{p}(\mathbb{T})$, then we immediately have $f*\phi_{n}\in L_{p}(\mathbb{T})$.
When I begin to compute $\int_{-\pi}^{\pi}|f(t-s)\phi_{n}(s)|^pds$ I get this
\begin{eqnarray*} \int_{-\pi}^{\pi}|f(t-s)\phi_{n}(s)|^pds &=& \int_{-\pi}^{\pi}|f(t-s)|^{p}\cdot|\phi_{n}(s)|^pds \end{eqnarray*}
if I had $\phi_{n}\in L_{\infty}(\mathbb{T})$, then I could immediately get $ \begin{eqnarray*} \int_{-\pi}^{\pi}|f(t-s)|^{p}\cdot|\phi_{n}(s)|^{p}ds &\leq& \int_{-\pi}^{\pi}|f(t-s)|^{p}\cdot||\phi_{n}||_{\infty}^{p}ds\\ &\leq& ||\phi_{n}||_{\infty}^{p}\int_{-\pi}^{\pi}|f(t-s)|^{p}ds\\ &<& \infty \end{eqnarray*}$
This would make the answer yes. But the example I mentioned above makes me think this might not be true; and thus the answer to my question would be non-trivial.