if we have a closed subspace of $L^p$ called $X \cong l^2$ where the topologies of $L^p$ and $L^2$ coincide (we assume $p>2$). Then we can regard $X$ as a subspace of $L^2$, which means that he is a separable Hilbert space and we can find an orthonormal basis $(\phi_n)_n$ such that $P:L^2 \rightarrow L^2, f \mapsto \sum_{i=1}^{\infty} \langle f, \phi_i \rangle \phi_i$ is a projection on $X$; so far so good. Now the question is raised, whether this is also a continuous projection on $L^p$? This reference says yes: reference; page 95 last line of the proof of corollary 9.8
But I have troubles to understand this reasoning. If we take $P: L^p \rightarrow L^p$( is this, what the author wants to do in his reasoning?), then $\langle f, \phi_i\rangle$ is not necessarily defined, cause $f$ does not have to be in $L^2$ and therefore I don't understand why the argumentation with $L^2$ is possible. Also $||f||_2$ is not defined. I understand that you can say something about $Pf$ in both norms, cause this is an element of $X$ and there both norms are equivalent. But how can they talk about $f$ in both norms? Essentially, it would completely answer my question if you could explain to me what happens in this last line in the given reference!
If anything is unclear, please do not hesitate to comment.
This is $L^p([0,1])$, right? Then $\|f\|_2\le\|f\|_p$ by Jensen's inequality (for the concave function $t\mapsto t^{2/p}$).