From this question The set of functions which map convergent series to convergent series , it is known that the set of functions on real line which maps convergent series to convergent series is well-studied and completely characterized . My question is ; has any one of the following situations been studied ?
1) Functions $f:\mathbb R \to \mathbb R$ which maps every absolutely convergent series $\sum_{n=1}^\infty a_n$ to a convergent series $\sum_{n=1}^\infty f(a_n) $
2) Functions $f:\mathbb R \to \mathbb R$ which maps every absolutely convergent series $\sum_{n=1}^\infty a_n$ to an absolutely convergent series $\sum_{n=1}^\infty f(a_n) $
3)Functions $f:\mathbb R \to \mathbb R$ which maps every convergent series $\sum_{n=1}^\infty a_n$ to an absolutely convergent series $\sum_{n=1}^\infty f(a_n) $
4)Functions $f:\mathbb R \to \mathbb R$ which maps every divergent series $\sum_{n=1}^\infty a_n$ to a divergent series $\sum_{n=1}^\infty f(a_n) $
I have included all these situations in one question because of their similar motivation . A necessary condition for all the functions in 1),2),3) is that $f(0)=0$ and $f$ should be continuous at $0$ . Functions satisfying $|f(x)|\le k|x|$ in a neighbourhood of $0$ satisfy conditions 1) and 2) but I can't figure out whether these characterize all such functions . For 3) , I have no-idea . For 4) , I have only figured that $f(x) \ne 0$ for $x\ne 0$ . Any help , reference , link regarding any of these will be highly appreciated . Thanks in advance
The result in 1) holds iff $|f(x)/x|$ is bounded in some deleted neighborhood of $0.$
Proof: As you said, if $|f(x)/x|$ is bounded, then $\sum f(a_n)$ is absolutely convergent whenever $\sum a_n$ is absolutely convergent. That is more than enough for this direction and easy to prove.
Now suppose $\sum f(a_n)$ is convergent whenever $\sum a_n$ is absolutely convergent. Assume, to reach a contradiction, that $|f(x)/x|$ fails to be bounded. Then there is sequence $a_n \to 0$ such that $|f(a_n)/a_n|> n^2.$
Now there is a subsequence $n_k$ such that $|a_{n_k}|<1/k^6.$ For large $k$ we can say the following: There exists $m_k \in \mathbb N$ such that
$$\tag 1 1/(k+1)^2 \le m_k|a_{n_k}| \le 1/k^2.$$
The reason is that the length of $[1/(k+1)^2, 1/k^2]$ is about $1/k^3.$ So we start with $|a_{n_k}|< 1/k^6$ and move up from there in increments of $|a_{n_k}|.$ We have to land in the above interval at some point because the increments are smaller than the length of the interval.
We now design a series in blocks. The $k$th block is $a_{n_k} + a_{n_k} + \cdots + a_{n_k},$ where there are exactly $m_k$ terms. This series is absolutely convergent. In fact $\sum|a_n| =\sum_{k=1}^{\infty} m_k|a_{n_k}|.$ By $(1),$ this sum is finite.
Claim: $\sum_{n=1}^{\infty} f(a_n)$ diverges. The claim gives us the desired contradiction. To prove the claim, let $B_k$ be the $k$th block of indices. I'll show
$$\tag 2 |\sum_{n\in B_k} f(a_n)| > 1/2$$
for large $k.$ This proves the desired divergence. Why? It shows the sequence of partial sums of $\sum_{n=1}^{\infty} f(a_n)$ is not Cauchy.
Now the left side of $(2)$ equals
$$|m_kf(a_{n_k})| \ge m_k|n_k^2a_{n_k}|.$$
Because $n_k \ge k,$ $(1)$ shows the above is at least
$$k^2m_k|a_{n_k}| \ge k^2/(k+1)^2,$$
which is $> 1/2$ for large $k.$ That gives $(2)$ and proves the claim.
On to 2). Claim: $f$ takes AC to AC iff $|f(x)/x|$ is bounded in some deleted neighborhood of $0.$ In other words, from the solution to $(1),$ $f$ takes AC to AC iff $f$ takes AC to C. The proof is easy: Clearly if $|f(x)/x|$ is bounded, then $f$ takes AC to AC. Suppose $f$ takes AC to AC. Because AC $\subset$ C, we see $f$ works in 1), hence $|f(x)/x|$ is bounded.
3). The only functions that work here are identically $0$ in a neighborhood of $0.$ Proof: If $f$ takes C to AC, then $f$ takes C to C. From the result you cite in the first line of your question, $f(x) = cx$ in a neighborhood of $0.$ Consider the series $\sum (-1)^n/n.$ Applying $f$ to this gives a series whose terms for large $n$ are $c(-1)^n/n.$ That series doesn't converge absolutely unless $c=0$ and we're done.
4). Haven't really thought about it. Note that any $f(x) = cx, c\ne 0,$ will take D to D. It may be that any $f$ that works in 4) must be equal to one of these in a neighborhood of $0.$