4
$\begingroup$

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

$\endgroup$
2
+25
$\begingroup$

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 $0.$

$\endgroup$
  • $\begingroup$ 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. $\endgroup$ – zhw. Jan 10 '16 at 5:53
  • $\begingroup$ 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 $\endgroup$ – user228169 Jan 10 '16 at 6:06
  • $\begingroup$ 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 $0.$ $\endgroup$ – zhw. Jan 10 '16 at 6:15
  • 1
    $\begingroup$ Because if $f(x) = cx,$ then $\sum (-1)^n/n$ would go to $\sum c(-1)^n/n,$ which converges absolutely iff $c=0.$ $\endgroup$ – zhw. Jan 10 '16 at 6:25
  • 2
    $\begingroup$ (4) also includes any function such that there exists $\epsilon > 0$ with $f(x) \ge \epsilon |x|$ for all $x$. $\endgroup$ – Robert Israel Jan 10 '16 at 20:23

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy