1

I am following Reed and Simon's book on functional analysis, and the following is their proof of the Riesz-Fisher theorem (which states that $L^1$ is complete). I have two main questions regarding their proof, which I have also included below.

Let $f_n$ be Cauchy in $L^1$. It is enough to prove some subsequence converges so pass to a subsequence (also labeled $f_n$) with $\|f_n - f_{n+1}\|_1 \leq 2^{-n}$.

Why is it desirable to consider a subsequence instead of the original sequence? Wouldn't the rest of the proof follow without any changes if one were to use the full sequence $f_n$?

Let $$g_m(x) = \sum_{n=1}^m |f_n(x) - f_{n+1}(x)|.$$ Let $g_\infty$ be the infinite sum (which may be $\infty$). Then $g_m \nearrow g_\infty$ and $\int |g_m| \leq \sum_{n=1}^m \|f_n - f_{n+1}\| \leq 1$, so by the monotone convergence theorem, $g_\infty \in L^1$. Thus $|g_\infty(x)| < \infty$ a.e. As a result $$f_m(x) = f_1(x) - \sum_{n=1}^{m-1} \big(f_n(x) - f_{n+1}(x)\big)$$ converges pointwise a.e. to a function $f(x)$.

How does $|g_\infty(x)| < \infty$ imply $f_m \rightarrow f$ pointwise?

Moreover, $|f_m(x)| \leq |f_1(x)| + g_\infty(x) \in L^1$ so $f_n \rightarrow f$ in $L^1$ by the dominated convergence theorem.

I have understood the rest of the proof, it is just these two points that I am stuck on.

CBBAM
  • 7,149

1 Answers1

2

You have to pass to a subsequence in order to get the speed of convergence that they use. It is not necessarily true of the original sequence that $\|f_n - f_{n+1}\|_1 \le 2^{-n}$. This bound is then used to see that $\int |g_m| \le 1$ so that passing to a subsequence was necessary in this argument.

As for your second question, for $x$ such that $|g_\infty(x)|< \infty$, we have that the sum $\sum_{n} (f_n(x) - f_{n+1}(x))$ is absolutely convergent and hence convergent. In particular, this implies that $f_m(x)$ is also convergent.

Ian
  • 104,572
Rhys Steele
  • 20,326
  • Thank you for your reply. Since ${f_n}$ is assumed to be Cauchy, couldn't we take $|f_n - f_{n+1}|_1$ to be any $\epsilon$ we would like for $n \geq N$ (in this case $\epsilon = 2^{-n}$)? – CBBAM Nov 14 '22 at 20:32
  • 2
    You can adapt the proof to use a tail instead of starting at $n=1$, but that doesn't save the day if $| f_n - f_{n+1} |$ isn't summable. Which it doesn't have to be from Cauchiness, it just has to go to zero. – Ian Nov 14 '22 at 20:35
  • 1
    You can find an example of convergence in $L^1$ without convergence at any point here: https://math.stackexchange.com/questions/1415891/almost-everywhere-convergence-versus-convergence-in-measure/1415900#1415900 I also really like this example because there are "obvious" subsequences that are rapidly Cauchy in $L^1$ and consequently converge a.e. – Ian Nov 14 '22 at 20:42