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I'm looking at two different functional analysis exercises, and I feel that I've hit the same road block with them. I was wondering if there is a kind of "systematic" way of fixing these solutions (or if I need to abandon hope and try a new approach).

The first exercise asks to show that $\ell^2$ is complete. My approach is

Take $\{f_n\}$ Cauchy in $\ell^2$ and observe that $\{ f_n(m) \}$ is Cauchy in $\mathbb{C}$ for each natural number $m$. I define $f \in \ell^2$ by $f(m) := \lim_{n \to \infty} f_n(m)$, but when I try to show that $f_n \rightarrow f$, I end up looking at the sum $$\sum_{j=1}^\infty |f_n(j)-f(j)|^2,$$ and trying show that it tends to $0$. I can control each individual summand, but since I have to deal with infinitely many, I'm not sure how to proceed.

The second exercise asks to show that weak convergence implies strong convergence in $\ell^1$. My approach is

Take $\{f_n\}$ weakly convergent in $\ell^1$. Then $$|| f_n - f || = \sum_{j=1}^{\infty} |f_n(j)-f(j)| = \sum_{j=1}^{\infty} |\pi_j(f_n-f)|,$$ where $\pi_j$ is the projection to the $j$-th coordinate of $f$. Again, I can control each of these summands (by weak convergence and continuity of these projections), but I don't know how to deal with infinitely many. (For this particular exercise, the hint is to use the fact that $\ell^1$ is dual to $\ell^{\infty}$, but I'm still trying to use this).

To reiterate: I'd like to know if there's some general technique for dealing with these situations. If so, what do I do, and how should I think about it? If not, are there any possible ad hoc fixes, or should I pursue a different course of action?

Thanks in advance!

Bachmaninoff
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  • "I define $f\in\ell^2$ by $f(m) := \lim_{n \to \infty} f_n(m)$" - how do you know that this $f$ is in fact in $\ell^2$? – Math1000 Mar 03 '16 at 06:35
  • For the equivalence of weak and strong convergence in $\ell^1$ see here: http://math.stackexchange.com/questions/42609/strong-and-weak-convergence-in-ell1 – Math1000 Mar 03 '16 at 06:39

2 Answers2

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For the first question, since $\{f_n\}$ is a Cauchy sequence, for each $\varepsilon>0$ there is an $N$ such that if $n,m\geq N$ then $$ \sum_{j=1}^{\infty}|f_n(j)-f_m(j)|^2<\varepsilon^2 $$

Let $f(j)=\lim_{n\to\infty}f_n(j)$ as in your question. Then for each fixed $J$, we have $$ \sum_{j=1}^{J}|f_n(j)-f_m(j)|^2\leq \sum_{j=1}^{\infty}|f_n(j)-f_m(j)|^2<\varepsilon^2 $$ if $m,n\geq N$, hence taking $m\to\infty$ in the first sum yields $$ \sum_{j=1}^{J}|f_n(j)-f(j)|^2<\varepsilon^2$$

Then taking $J\to\infty$ shows that $$ \sum_{j=1}^{\infty}|f_n(j)-f(j)|^2\leq \varepsilon^2$$ hence $f_n\to f$ in $\ell^2$.

carmichael561
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First, a Cauchy sequence is always bounded. This is because there exists $N$ such that $\|f_n-f_k\| < 1$ whenever $n,k \ge N$. Then $$ \|f_k\| \le M=\max\{\|f_1\|,\|f_2\|,\cdots,\|f_N\|+1\},\;\;\; k=1,2,3,\cdots. $$ Another standard thing is to note that the point evaluations are continuous: $$ |f_j(n)-f_{k}(n)| \le \|f_j-f_k\|, $$ which means that $f(n)=\lim_jf_j(n)$ exists for all $n$ because the field $\mathbb{C}$ or $\mathbb{R}$ is complete. So, for any fixed upper limit $N$, $$ \sum_{n=1}^{N}|f_j(n)|^2 \le \|f_j\|^2 \le M^2. $$ The above holds for all $j$ and for all $N$. Hence, the following holds for all $N$: $$ \sum_{n=1}^{N}|f(n)|^2 \le \lim_{j}\sum_{n=1}^{N}|f_j(n)|^2 \le M^2. $$ Therefore $f\in \ell^2$. For every $\epsilon > 0$ there exists $J$ such that $\|f_j-f_k\| < \epsilon/2$ for all $j,k \ge J$. Therefore, for all $N$, $$ \sum_{n=1}^{N}|f_j(n)-f_k(n)|^2 < \frac{\epsilon^2}{4},\;\;\; j,k \ge J \\ \sum_{n=1}^{N}|f(n)-f_k(n)|^2 \le \frac{\epsilon^2}{4},\;\;\; k \ge J. $$ Because this holds for all $N$, $$ \sum_{n=1}^{\infty}|f(n)-f_k(n)|^2 \le \frac{\epsilon^2}{4} < \epsilon^2,\;\;\; k \ge J. $$ Because $\epsilon > 0$ was arbitrary, by the definition of the vector limit, $$ \lim_{j} f_j = f. $$

Disintegrating By Parts
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