Challenge Aren't you tired of counterexamples already?

[andrewkirk]

2. In any Banach space $X$ and given any series $\sum a_n$ in $X$, then the series converges absolutely (that is: $\sum \|a_n\|$ converges) if and only if the series converges unconditionally (that is: for any bijection $\pi:\mathbb{N}\rightarrow \mathbb{N}$ holds that $\sum_n a_{\pi(n)}$ converges to the same number).

I think I can prove the forward direction.

Let $a_n$ be a sequence whose series is absolutely convergent to $A$.
Let $\sigma:\mathbb N\to\mathbb N$ be a bijection. We claim that $\sum_{n=1}^\infty a_{\sigma(n)}=A$.

Attempted Proof:
Define the following :

$A_n=\sum_{k=1}^n a_n$
$S_n=\sum_{k=1}^n\|a_n\|$
$S=\lim_{n\to\infty}s_n$. We know this limit exists because the series absolutely converges.
$A'_n=\sum_{k=1}^n a_{\sigma(n)}$

Given $\epsilon>0$ choose $N\in\mathbb N$ such that for any $n\geq N$ both $\|S-S_n\|$ and $\|A-A_n\|$ are less than $\frac\epsilon9$.

Let $M=\max\{\sigma^{-1}(k)\ |\ k\leq N\}$.

We claim that $n\geq M\Rightarrow \|A'_n-A\|<\epsilon$, which is sufficient to prove the forward direction.

Let $n$ be any integer that is $\geq M$, and let $B$ be the set of integers in $\{1,...,n\}$ that are not in $\sigma^{-1}\left(\{1,...,N\}\right)$.

Here goes:

\begin{align*}
\|A'_n-A\|&=\|(A_N+\sum_{k\in B}a_{\sigma(k)})-A\|\\
&\leq \|A_N-A\|+\|\sum_{k\in B}a_{\sigma(k)}\|\\
&\leq \frac\epsilon9+\sum_{k\in B}\|a_{\sigma(k)}\|\\
&\leq \frac\epsilon9+\sum_{k>N}\|a_k\|\\
&= \frac\epsilon9+\|S-S_N\|\\
&\leq\frac\epsilon9+\frac\epsilon9<\epsilon
\end{align*}

That seemed quite easy (although it might be because there's a mistake in there). If so then I'm guessing that the reverse direction will be the harder one. I'll need to look at that later.

Edit: corrected typo in first line of aligned equations, pointed out by Samy (see post below)

 

[Samy_A]

Here goes:

\begin{align*}
\|A'_n-A\|&=\|(A_n+\sum_{k\in B}a_{\sigma(k)})-A\|\\
&\leq \|A_n-A\|+\|\sum_{k\in B}a_{\sigma(k)}\|\\
&\leq \frac\epsilon9+\sum_{k\in B}\|a_{\sigma(k)}\|\\
&\leq \frac\epsilon9+\sum_{k>N}\|a_k\|\\
&= \frac\epsilon9+\|S-S_N\|\\
&\leq\frac\epsilon9+\frac\epsilon9<\epsilon
\end{align*}.

Is there a typo in the first line? $A_n$ should be $A_N$, I think.

That seemed quite easy (although it might be because there's a mistake in there). If so then I'm guessing that the reverse direction will be the harder one. I'll need to look at that later.

This was the difficult part.

 

[andrewkirk]

This was the difficult part.

Ah, silly me, I didn't read the instructions in the OP closely enough. I was looking to prove the opposite direction, but all I needed was a counterexample to disprove that direction. I couldn't see that anybody had done that yet above (maybe I missed it). So:

Consider the Hilbert space $l^2$. Let $b_1,b_2,...$ be such that $b_k$ is the sequence that is all zeroes except for a 1 in the $k$th place. It is easy to show that this is an orthonormal basis for the space. Then the sequence $a_k=\frac{b_k}k$ is not absolutely convergent, as
\begin{align}\sum_{k=1}^\infty \|a_k\|=
\sum_{k=1}^\infty \frac1k\end{align}
which does not converge.
But, given $\epsilon>0$ we can choose $N$ such that if $n\geq N\Rightarrow \sum_{k=n}^\infty\frac1{k^2}<\frac{\epsilon^2}9$. So, for $m,n\geq N$ we have (defining $S_k\equiv\sum_{k=1}^n s_n$):
\begin{align}
\|S_m-S_n\|&=\left\|\sum_{j=n+1}^m a_j\right\|\\
&=\sqrt{\left\langle \sum_{j=n+1}^m a_j,\sum_{j=n+1}^m a_j\right\rangle}\\
&=\sqrt{\sum_{j,k=n+1}^m \langle a_j,a_k\rangle}\\
&=\sqrt{\sum_{j,k=n+1}^m \frac{\delta_{jk}}{jk}}\\
&=\sqrt{\sum_{k=n}^\infty \frac1{n^2}}<\frac\epsilon3<\epsilon\\
\end{align}
So the sequence $S_n$ is Cauchy and, since the space is complete, must converge, to some point $p$. Choose $R$ such that $n\geq R\Rightarrow \|S_n-p\|<\frac\epsilon9$.

Now let $\sigma:\mathbb N\to\mathbb N$ be any rearrangement and set

$M=\max\{\sigma^{-1}(k)\ |\ k\leq N\}$

$S'_k=\sum_{j=1}^n a_{\sigma(j)}$

$H\equiv\max(R,M,N)$

$n\geq H$

$B(n)=\sigma(\{1,...,n\})-\{1,...,H\}$

Then
\begin{align*}
\|S'_n-p\|&\leq \|S'_n-S_n\|+\|S_n-p\|\\
&< \|S'_n-S_H\|+\|S_H-S_n\|+\frac\epsilon9\\
&= \left\|\sum_{k\in B(n)}a_k\right\|+\frac\epsilon9+\frac\epsilon9\\
&= \sqrt{\left\langle\sum_{k\in B(n)}a_k,\sum_{k\in B(n)}a_k\right\rangle}+\frac{2\epsilon}9\\
&= \sqrt{\sum_{\substack{j,k\in B(n)}}\left\langle a_j,a_k\right\rangle}+\frac{2\epsilon}9\\
&= \sqrt{\sum_{\substack{j,k\in B(n)}}\frac{\delta_{jk}}{jk}}+\frac{2\epsilon}9\\
&= \sqrt{\sum_{\substack{k\in B(n)}}\frac1{k^2}}+\frac{2\epsilon}9\\
&\leq \sqrt{\sum_{\substack{k\in B(n)}}\frac1{k^2}}+\frac{2\epsilon}9\\
&\leq \sqrt{\sum_{\substack{k=N+1}}^\infty\frac1{k^2}}+\frac{2\epsilon}9\\
&<\frac\epsilon3+\frac{2\epsilon}9=\frac{5\epsilon}9<\epsilon\\
\end{align*}

So the re-arranged series $S'_n$ converges to $p$ and hence the series is unconditionally convergent. Hence the sequence $a_n$ is a counter-example to the reverse direction.

 

[micromass]

Hint for (7): every pointswise limit of continuous functions has only countably many discontinuity points. Use Baire's theorem.

 

[ProfuselyQuarky]

Hint for (7): every pointswise limit of continuous functions has only countably many discontinuity points. Use Baire's theorem.

Can you post the counterexample since it seems like no one is going to do this one? I've been waiting to see it.

 

[micromass]

I will give credit to Samy_A for question $7$ since he came up with an answer to the question before I revised it. His answer still answers one direction too.

So, here's the result for $7$. On one side, continuous functions are Borel measurable. Limits of Borel measurable functions are Borel measurable. This the limit of a continuous limit is definitely Borel measurable.

On the other hand, not all Borel measurable functions are the limit of a sequence of continuous functions. One counterexample here is the indicator function of $\mathbb{Q}$. The idea is that this function has "many" many discontinuity points, indeed: every point of $\mathbb{R}$ is a discontinuity point. I will prove now however that if $f = \lim_n f_n$ is a limit of continuous functions then $f$ has a discontinuity set that is of first category.

Some terminology:
A closed set is called nowhere dense if its interior is empty. A set is called of first category if it is contained in the countable union of nowhere dense sets. A set is called of second category if it is not of first category.
For any function $f:U\rightarrow \mathbb{R}$ define $\text{osc}_g(x_0) = \lim_{\delta\rightarrow 0} \text{sup}_{|x-x_0|<\delta} |g(x) - g(x_0)|$. This is called the oscillation of $g$ at $x_0$. Note that $g$ is continuous at $x_0$ if and only if $\text{osc}_g(x_0) = 0$.

Lemma: If $U\subseteq \mathbb{R}$ is open and $\varepsilon>0$, then
$$\overline{\{x\in U~\vert~\text{osc}_g(x)>2\varepsilon\}}\subseteq \{x\in U~\vert~\text{osc}_g(x)\geq \varepsilon/2\}$$

Proof: We need to see that the limit points of the set on the left are in the set on the right. Thus suppose that $\text{osc}_g(x_n)\geq 2\varepsilon$ for all $n$ and that $x_n\rightarrow x_0$. For each $n$, choose $x_{n,m}$ such that $\lim_m x_{n,m} = x_n$ and let $|g(x_{n,m}) - g(x_n)|\geq \varepsilon$ for all $m$. Because convergence of $x_{n,m}$ to $x_n$, we may choose, for each $n$, an integer $m_n$ such that $|x_{n,m_n} - x_n| < |x_0-x_n|$ and then $\lim_n x_{n,m_n} = x_0$ by the triangle inequality. From $|g(x_{n,m_n}) - g(x_n)|\geq \varepsilon$, the triangle inequality forces
$$|g(x_{n,m_n}) - g(x_0)|\geq \varepsilon/2~\text{or}~|g(x_n) - g(x_0)|\geq \varepsilon/2.$$
Defining $y_n$ to be $x_{n,m_n}$ or $x_n$ according to the first or second inequality, we have $y_n\rightarrow x_0$ and $|g(y_n) - g(x_0)|>\varepsilon/2$. This proves the lemma.

In the following, we will use the Baire category theorem which states that $\mathbb{R}$ is of second category. Furthermore, all nonempty open subsets of $\mathbb{R}$ are of second category.

Theorem: If $f_n:\mathbb{R}\rightarrow \mathbb{R}$ are continuous functions converging pointswise to a function $f:\mathbb{R}\rightarrow \mathbb{R}$, then the set of discontinuous is of the first category.

Proof: In view of the lemma and the fact that $\mathbb{R}$ is not of the first category, it is enough to prove for each $\varepsilon>0$, that $\{x\in \mathbb{R}~\vert~\text{osc}_f(x)\geq \varepsilon\}$ does not contain a nonempty open subset of $\mathbb{R}$. Assuming the contrary, suppose that it contains the nonempty open set $V$. Define
$$A_{mn} = \{x\in V~\vert~|f_m(x) - f_n(x)|\leq \varepsilon/4\}.$$
This is closed in $V$. Then $A_m = \bigcap_{n\geq m} A_{mn}$ is closed in $V$. If $x\in V$, then the fact that $\{f_n(x)\}$ is Cauchy implies that there is some $m$ such that $x\in A_{mn}$ for $n\geq m$. Hence $\bigcup_m A_m = V$. The Baire theorem implies that some $A_m$ has nonempty interior. Fix that $m$ and let $W$ be its nonempty interior. Since
$$A_m\subseteq \{x\in V~\vert~|f_m(x) - f(x)|\leq \varepsilon/4\},$$
every point of $W$ has $|f_m(x) - f(x)|\leq \varepsilon/4$ and $\text{osc}_f(x)\geq \varepsilon$. Let $x_0\in W$ and choose $x_n\rightarrow x_0$ wth $|f(x_n) - f(x_0)|\leq 3\varepsilon/4$. From $|f_m(x_n) - f(x_n)|\leq \varepsilon/4$ and $|f_m(x_0) - f(x_0)|\leq \varepsilon/4$, we obtain $|f_m(x_n) - f_m(x_0)|\geq \varepsilon/4$. Since $x_n$ converges to $x_0$, this inequality contradicts the continuity of $f_m$ at $x_0$. This proves the theorem.

Finally, the characteristic function has entire $\mathbb{R}$ as discontinuity set, but this is not of first category by the Baire category theorem.

For more information, see https://en.wikipedia.org/wiki/Baire_function and Knapp's "Basic real analysis" from which I adapted this proof.