# Functions on real line which preserves dfferent modes of convergence and preserves divergence of real infinite series

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 $0.$

• Now that I think of it, I think the functions that work in 1) are exactly the ones that work in 2). I think the same basic idea will prove this. (This time use $|f(a_n)/a_n|>n^2$ maybe) I'll try to think about it tomorrow. – zhw. Jan 10 '16 at 5:53
• Awesome answer +1 ; and yes , functions that work for 2) also work for 1) ; but 1) might admit a larger class of functions ( I am not sure though ) . But I think your answer works for 3) as well , please feel free to give your further thoughts on this – user228169 Jan 10 '16 at 6:06
• For 3) we can use the result you cited. If C $\to$ AC, then obviously C $\to$ C. Thus $f$ must be linear in a neighborhood of $0.$ But clearly the only linear function giving us C $\to$ AC is the zero function. Thus the only functions that work in 3) are zero in a nbhd of $0.$ – zhw. Jan 10 '16 at 6:15
• Because if $f(x) = cx,$ then $\sum (-1)^n/n$ would go to $\sum c(-1)^n/n,$ which converges absolutely iff $c=0.$ – zhw. Jan 10 '16 at 6:25
• (4) also includes any function such that there exists $\epsilon > 0$ with $f(x) \ge \epsilon |x|$ for all $x$. – Robert Israel Jan 10 '16 at 20:23