Proof that $\|S_N\|_p < \infty $ is equivalent to $\|S_N f - f\|_p \to 0$ as $N \to \infty$ I am having difficulties with the proof of proposition 1.9 in the book "Classical and multilinear harmonic analysis, Vol. 1" by C. Muscalu and W. Schlag.

The following statements are equivalent for any $1 \leq p \leq \infty$:
$(i)$ for every $f \in L^p(\mathbb{T})$ (or $f \in C(\mathbb{T})$ if $p = \infty$) one has $$ \|S_Nf - f\|_p \to 0 \text{  as } N \to \infty; $$
  $(ii)$ $\sup_N \|S_N\|_{p \to p} < \infty.$

The proof for the fact that $(i)$ implies $(ii)$ is as follows.

Suppose that $\sup_N \|S_N\|_{p \to p} = \infty$. For every positive integer $\ell$, one can therefore find a large integer $N_\ell$ such that 
  $$ \|S_{N_\ell}f_\ell\|_p > 2^\ell $$ 
  where $f_\ell$ is a trigonometric polynomial with $\|f_\ell\|_p = 1$. 
  Now let 
  $$ f(x) = \sum_{\ell = 1}^\infty \frac{1}{\ell^2}e(M_\ell x)f_\ell(x), $$ 
  with integers $\{M_\ell \}$ to be specified. Notice that $\|f\|_p < \infty$. Now choose a sequence $\{M_\ell\}$ tending to infinity so rapidly that the Fourier support of 
  $$ e(M_\ell x)f_\ell(x) $$ 
  is to the right of the Fourier support of 
  $$ g_\ell(x) := \sum_{j=1}^{\ell-1} \frac{1}{j^2}e(M_j x)f_j(x) $$ 
  for every $j \geq 2$. We also demand that $M_\ell - N_\ell \to \infty$ as $\ell \to \infty$ and that $S_{M_\ell - N_\ell-1}f = g_\ell$ and $S_{M_\ell+N_\ell}f = g_{\ell+1}$. Then
  $$ \|(S_{M_\ell+N_\ell} - S_{M_\ell-N_\ell-1})f\|_p = \frac{1}{\ell^2} \|S_{N_\ell}f_\ell\|_p > \frac{2^\ell}{\ell^2}.$$

The part I am having trouble with is the definition of the numbers $\{M_\ell\}$. I get that we can impose that they grow fast enough so that
$$ S_{M_\ell-N_\ell-1} f = g_\ell $$
since $g_\ell$ does not depend on $M_\ell$, but how is it that we can have 
$$ S_{M_\ell+N_\ell}f = g_{\ell+1} $$
when the term $e(M_\ell x)f_\ell(x)$ appears in $g_{\ell+1}$? I seems to me that the Fourier support of this term is the Fourier support of $f_\ell$ shifted to the right by a factor of $M_\ell$. Then, don't we need $f_\ell(k) = 0$ for $k > N_\ell$ so that all the terms of $g_{\ell+1}$ appear in $S_{M_\ell+N_\ell}f$?
Also, can we deduce from this that
$$ g_{\ell+1} = S_{M_\ell+N_\ell}g_{\ell+1} $$
which seems needed in order to obtain the last equality?
 A: Let us forget about what they write after the definition of $g_\ell$. I will try to reconstruct what is meant.
Let $M'_\ell$ denote the size of the Fourier support of $f_\ell$. So, $\widehat{f_\ell}(k)=0$ if $|k|>M'_\ell$. 
We have $M'_\ell>N_\ell$ by choice of $f_\ell$.
We want $M_\ell$ to be chosen large enough so that the intervals $[M_\ell - M'_\ell, M_\ell + M'_\ell]$ (the Fourier support of $e(M_\ell x)f_\ell(x)$) are disjoint. 
The idea is now to recover $S_{N_\ell} f_\ell$ by an appropriate clipping in frequency. The operator $S_{A}-S_{B-1}$ (where $A\ge B$) cuts out the frequency annulus $[-A,-B]\cup [B,A]$. So we have that
$$(S_{M_\ell+N_\ell} - S_{M_\ell-N_\ell-1})f$$
selects precisely the components of $f$ with frequency in the interval $[M_\ell-N_\ell, M_\ell + N_\ell]\subset [M_\ell-M'_\ell, M_\ell+M'_\ell]$, so it equals 
$$1/\ell^2 e(M_\ell x) S_{N_\ell} f_\ell.$$
Now we proved the last line in the proof you quoted. So in the book one should just drop the definition of $g_\ell$ and these two identities regarding $g_\ell$ entirely. We don't need them and they are flawed as we discussed above in the comments.
From here it is easy to finish the proof. Suppose $M_\ell$ is large enough such that $M_\ell - N_\ell\to\infty$ as $\ell\to\infty$. Then $(S_{M_\ell+N_\ell} - S_{M_\ell-N_\ell-1})f$ is a Cauchy sequence in $\ell$ by assumption. So we arrive at a contradiction as $\ell\to\infty$.
