I'm reading Fourier Analysis by Stein and Shakarchi and it seems to me that there's a mistake in the proof of Lemma 2.3 in Chapter 3. I wonder if there's something I'm missing or if someone could suggest an alternative proof. Any insight would be appreciated!
Lemma 2.3 states: "Suppose that the Abel means $A_r = \sum_{n = 1}^\infty r^n c_n$ of the series $\sum_{n=1}^\infty c_n$ are bounded as $r$ tends to 1 (with $r < 1$). If $c_n = O
(1/n)$, then the partial sums $S_N = \sum_{n = 1}^N c_n$ are bounded."
The proof is as follows.
Let $r = 1 – 1/N$ and choose $M$ so that $n|c_n| \leq M$. We estimate the difference
$S_N – A_r = \sum_{n = 1}^N (c_n – r^n c_n) – \sum_{n = N+1}^\infty r^n c_n$
as follows:
$|S_N – A_r| \leq \sum_{n = 1}^N |c_n| (1-r^n) + \sum_{n= N+1}^\infty r^n |c_n| \leq M \sum_{n=1}^N (1-r) + \frac{M}{N} \sum_{n = N+1}^\infty r^n \leq MN(1-r) + \frac{M}{N} \frac{1}{1-r} = 2M$,
where we have used the simple observation that
$1-r^n = (1-r)(1+r+\cdots + r^{n-1}) \leq n(1-r)$.
So we see that if $M$ satisfies both $|A_r| \leq M$ and $n|c_n| \leq M$, then $|S_N| \leq 3M$.
The mistake I notice here is in the line where the authors claim that $MN(1-r) + \frac{M}{N} \frac{1}{1-r} = 2M$. By my calculations, using their assumption that $r = 1 – 1/N$, we instead have the following:
$MN(1-r) + \frac{M}{N} \frac{1}{1-r} = M(N-1) + \frac{M}{N-1} = M(\frac{(N-1)^2 + 1}{N-1}) = M \frac{N^2 – 2N + 2}{N-1} = M \frac{(N-1)(N-2)}{N-1} = M(N-2)$. This clearly isn't bounded.
Thank you again! Any insight would be appreciated!
Best Answer
Note that $1-r=1-(1-1/N)=1/N$. Then:
$$MN(1-r)+\frac{M}{N}\frac{1}{1-r}=MN(1/N)+\frac{M}{N}\frac{1}{1/N}=M+\frac{M}{N}\cdot N=2M$$