# Real Analysis, Folland problem 5.5.54 Hilbert spaces

problem 5.5.54 - For any nonempty set $A$, $l^{2}(A)$ is complete.

Attempted proof (inspired by the answer below): Let $x\in A$, and $\{T_n(x)\}_{n\in\mathbb{N}}$ be a Cauchy sequence in $l^{2}(A)$. This means for $\epsilon = \frac{1}{k}$ we can find an $n_k(\epsilon)\in\mathbb{N}$ for $k\in\mathbb{N}$ such that $$|T_n(x) - T_m(x)|\leq \lVert T_n - T_m\lVert \leq \frac{1}{k} \ \ \forall \ n,m \geq n_k(\epsilon)$$

Take $S\subset A$, where $S$ is finite, for $x\in S$ it follows $$\sum_{x\in S}|T_{n_k}(x) - T(x)|^2 = \lim_{m\rightarrow \infty}\sum_{x\in S}|T_{n_k}(x) - T_m(x)|^2 \leq \left(\limsup\lVert T_{n_k} - T_m\rVert\right)^{2} \leq \frac{1}{k^2}$$ Thus, $$\lVert T_{n_k} - T\rVert \leq \lim\sup \lVert T_{n_k} - T_m\rVert \leq \frac{1}{k}$$ And, $T = T_{n_1} - (T_{n_1} - T)\in l^{2}(A)$

I am not exactly sure this is correct, any suggestions is greatly appreciated.

• Where is the confusion? $l^2$ is a normed space, the definition of Cauchy sequence is straight forward. Where do you stuck at proving that every Cauchy sequence converges? Btw. it is the same proof for every $L^p$ space, $1\le p < \infty$. – user251257 Mar 26 '16 at 19:26
• I see, so if I show the $T_n$ $n\in\mathbb{N}$ is a Cauchy sequence in $l^2(A)$ I am done? – Wolfy Mar 26 '16 at 19:30
• No, $T_n$ is a Cauchy sequence in $l^2$. Then, you have to show that there is a $T\in l^2$ such that $T_n\to T$ in $l^2$. Hint: To obtain $T$, use the fact that $T_n$ is Cauchy and show it has one convergent subsequence (Cauchy sequence has one limit point at most). – user251257 Mar 26 '16 at 19:33
• Ok sorry I should of re-edited my comment I meant that if I let $T_n$ $n\in\mathbb{N}$ be a Cauchy sequence in $l^2$ and I show that $T_n\rightarrow T\in l^2$ then I am done? – Wolfy Mar 26 '16 at 19:35
• Yes. You need to construct $T$ explicitly. – user251257 Mar 26 '16 at 19:36

The case $l^2(A)$:

1. For every $x\in A$ we have $|T_l(x) - T_m(x)| \le \|T_l - T_m\|$. Thus, $(T_n(x))_{n\in \mathbb N}$ is Cauchy and has a limit $T(x)\in\mathbb R$.

2. Now, for every $k\in\mathbb N$ let $n_k\in\mathbb N$ such that for every $l,m\ge n_k$ it follows $\|T_l - T_m \| \le 1/k$. For every finite subset $S\subset A$ we have $$\sum_{x\in S} |T_{n_k}(x) - T(x)|^2 = \lim_{m\to\infty}\sum_{x\in S} |T_{n_k}(x) - T_m(x)|^2 \le \left( \limsup_{m\to\infty} \|T_{n_k} - T_m\| \right)^2 \le \frac1{k^2}.$$ Thus $$\| T_{n_k} - T \| \le \limsup_{m\to\infty} \|T_{n_k} - T_m\| \le \frac1{k} \to 0.$$

3. In particular, we have $T_{n_1} - T\in l^2(A)$ and $T= T_{n_1} - (T_{n_1} - T) \in l^2(A)$.

Further, as a Cauchy sequence has at most a cluster point, the cluster point $T$ is also the limit of the Cauchy sequence $(T_n)_{n\in\mathbb N}$.

The proof of Riesz-Fischer theorem on arbitrary measure space:

1. For every $k\in\mathbb N$ there is some $n_k\in\mathbb N$ such that for every $l,m\ge n_k$ it follows $\| T_l - T_m \| < 2^{-k}$ .

2. It is sufficient to show that the subsequence $T_{n_k}$ converges in $l^2$. Wlog let that subsequence be the complete sequence, that is $$\| T_{n+k} - T_n \| < 2^{-n}$$ for every $n,k\in\mathbb N$.

3. Let $S_n = \sum_{k=1}^n |T_{k+1} - T_k|$ (pointwise absolute value). Using Monotone Convergence/Fatou's lemma to obtain $$\int_A \lim_{n\to\infty} S_n^2 = \lim_{n\to\infty} \int_A S_n^2 = \left(\lim_{n\to\infty} \|S_n\| \right)^2 \le \left( \sum_{n=1}^\infty \| T_{n+1} - T_n \| \right)^2 \le 1.$$

4. Thus, for almost every $x\in A$ the limit $\lim_{n\to\infty} S_n(x)$ exists and $(T_n(x))_{n\in\mathbb N}$ is a Cauchy sequence. Thus, there exists some $T:A\to\mathbb R$ such $T_n\to T$ pointwise a.e.

5. Using Fatou's lemma, we obtain $$\int_A |T - T_n|^2 \le \liminf_{k\to\infty} \int_A |T_k - T_n|^2 = \left( \liminf_{k\to\infty} \|T_k - T_n\| \right)^2 \le (2^{-n})^2 \to 0$$ for $n\to\infty$.

Notes:

• The theorem is called / due to Riesz-Fischer

• If you replace the 2's with $p \in [1,\infty)$, you obtain that $L^p$ is complete.

• I am trying to incorporate these points in my proof. But I having some trouble. Do you think my set-up to start is ok? – Wolfy Mar 26 '16 at 21:26
• @Wolfgang: I add the version for counting measure, without convergence theorems. It is a little bit easier. – user251257 Mar 26 '16 at 21:27
• I see that now, for 1 though I am confused a tad after when you say "For every $x\in A$ we have...". Why is that the case? Can't one just start with assuming we have a Cauchy sequence like I did in the above? – Wolfy Mar 26 '16 at 21:29
• @Wolfgang: I did and I use the fact that $T_n$ is Cauchy at many places. Like I said before, you have to explicitly construct a limit, here $T$. – user251257 Mar 26 '16 at 21:31
• Ok, I will attempt that and re-post my proof shortly – Wolfy Mar 26 '16 at 21:32