This is an excerpt from Chapter 4 of Fourier Analysis by Stein and Shakarchi.
Define $$f_\alpha (x)=f(x)=\sum_{n=0}^{\infty} 2^{-n\alpha}e^{i2^n x}.$$ Define the delayed means by $$\vartriangle_N (g)=2\sigma_{2N}(g)-\sigma_N (g),$$ where $\sigma_N(g)$ is the $N$th Cesaro mean of the Fourier series. We can get the formula for $\sigma_N$ by multiplying $a_n e^{inx}$ by $1-|n|/N$ for $|n|\le N$ and $0$ for $|n|>N$.
Also, $\vartriangle_N$ arises by multiplying $a_n e^{inx}$ by $1$ if $|n|\le N$, by $2(1-|n|/(2N))$ for $N\le |n| \le 2N$, and $0$ for $|n|>2N$.
Then, in particular for our function $f=f_\alpha$ $$S_N (f)=\vartriangle_{N'}(f) \qquad (6)$$ where $N'$ is the largest integer of the form $2^k$ with $N'\le N$. This is clear by examining Figure 5 and the definition of $f$.
My questions is, how do we clearly see the identity $(6)$ from Figure 5 and the definition of $f$? Also, how can this identity be proven algebraically from the formulas for $\vartriangle_N (g)$ and $S_N (g)$? This seems to be easy to show from the author's comment, but I don't understand why.
If N is of the form $2^k$ then $2N=2^{k+1}$. From the definition of $f$, its Fourier coefficients are zero between $N+1$ and $2N$. So its delayed sum has identical Fourier coefficients as f for $n\leq N$ (as a delayed sum does for any function) but the rest of the terms in the delayed sum are zero. The idea of the figure is that because each nonzero frequency is twice as far out as the last one, when you do the delayed sum, the function never "sees" the slanted part (the averaging), it only sees the horizontal part which is the same as the partial sum.