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Could you help me with this problem?

What functions $f: \mathbb{R} \rightarrow \mathbb{R}$ satisfy the following implication?

$\sum _{n=1} ^{\infty} x_n < \infty \ \Rightarrow \sum _{n=1} ^{\infty} f(x_n) < \infty $

The necessary condition for convergence of series $\sum _{n=1} ^{\infty} a_n $ is that $\lim a_n =0$. So I think that $f$ should be decreasing.

Series is convergent $\iff$ $\forall \epsilon>0 \ \ \exists N\in \mathbb{N} \ : \ \forall q \ge p \ge k \ : \ |\sum _{n=p} ^q a_n|< \epsilon$. But I don't know how to use it. Could you help me?

Thank you.

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marked as duplicate by mrf, Alex Becker, saz, Alexander Gruber, lhf Feb 26 '13 at 11:37

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

    
Do you need the $ x_{n} $’s to be non-negative? –  Haskell Curry Feb 26 '13 at 10:05
    
No, I don't think so. But if you could help me with nonnegative $x_n$'s, that would also be very helpful. –  Hagrid Feb 26 '13 at 10:08
    
Certainly $f$ need not decrease, for example let $f$ be the identity. –  Alex Becker Feb 26 '13 at 10:13
2  
You certainly want $f(0)=0$. –  UwF Feb 26 '13 at 10:18
1  
I just don't remember, but I read, in some posting in this community, that a function $f : \Bbb{R} \to \Bbb{R}$ satisfies the condition $$ \sum f(x_n) \text{ converges whenever } \sum x_n \text{ converges for } \{ x_n \} \subset \Bbb{R} $$ if and only if $f$ is linear (possibly in some neighborhood of $0$?). But I'm not sure if OP intended to include the case $\sum x_n = -\infty$ in the condition $\sum x_n < \infty$, which would clearly alter the answer. –  sos440 Feb 26 '13 at 10:58

1 Answer 1

up vote 3 down vote accepted

For a real-valued function $f : \Bbb{R} \to \Bbb{R}$, we consider the following property:

Whenever $\sum_{n=1}^{\infty} x_{n}$ converges for $\{x_{n}\}_{n=1}^{\infty} \subset \Bbb{R}$, the series $\sum_{n=1}^{\infty} f(x_{n})$ also converges.

We say that $f$ has property $L$ if $f$ satisfies the condition specified above.

If $f$ is linear on some neighborhood of $0$, then it obviously has property $L$. Conversely, assume that $f$ has property $L$. We prove that $f(x)$ is linear near $0$.

Before the proof, we remark that $f(0) = 0$ and $f(x_n) \to 0$ whenever $x_n \to 0$.

Step 1. We must have $f(-x) = -f(x)$ near $x = 0$

Suppose not. Then there exists a monotone decreasing sequence $p_k \downarrow 0$ such that $f(-p_k) \neq -f(p_k)$. Then the difference of these two values $d_k = \left|f(p_k) + f(-p_k)\right|$ is positive. Now it is clear that we can choose a sequence of natural numbers $N_k$ such that $N_{k+1} d_{k+1} > 3N_k d_k $. Now consider the following sequence:

$$(x_n) = ( \underbrace{p_1, -p_1, \cdots, p_1, -p_1}_{2N_1\text{-terms}}, \underbrace{p_2, -p_2, \cdots, p_2, -p_2}_{2N_2\text{-terms}}, \underbrace{p_3, -p_3, \cdots, p_3, -p_3}_{2N_3\text{-terms}}, \cdots ).$$

It is clear that $\sum_{n=1}^{\infty} x_n = 0$. But we also have

\begin{align*} \left| \sum_{n=1}^{2(N_{1}+\cdots+N_{m})} f(x_n) \right| &= \left| \sum_{k=1}^{m} N_{k}\left(f(p_k) + f(-p_k)\right) \right| \\ &\geq N_m d_m - (N_{m-1}d_{m-1} + \cdots + N_1 d_1) \\ &\geq N_m d_m - \left(\frac{N_m d_m}{3} + \cdots + \frac{N_m d_m}{3^{m-1}} \right) \\ &\geq \frac{1}{2}N_m d_m \geq \frac{3^{m-1}}{2}N_1 d_1, \end{align*}

which clearly diverges as $m \to \infty$. Therefore $f(-x) = -f(x)$ near $x = 0$.

Step 2. $f$ is linear near $x = 0$.

Suppose not. Then there exists a sequence of real numbers $(p_k)$ such that $\left| p_k \right| \downarrow 0$ and

$$ \frac{f(p_k)}{p_k} \neq \frac{f(p_{k+1})}{p_{k+1}}. $$

Note that we may assume, by choosing an appropriate subsequence, that $0 < p_k < 1$, that $\left| f(p_k) \right| < 1$, and that $\sum p_k < \infty$.

Now we want to construct a sequence $(x_n)$ such that $\sum_{n=1}^{\infty} x_n $ converges while $\sum_{n=1}^{\infty} f(x_n)$ diverges. To this end, we need the following lemma:

Lemma. Given two positive numbers $p > q > 0$, suppose that there exists two positive integers $a$ and $b$ such that $\left|ap - bq\right| \leq 2p$. Let $$ (r_{1}, \cdots, r_{a+b} ) = ( \underbrace{p, \cdots, p}_{a\text{-temrs}}, \underbrace{-q, \cdots, -q}_{b\text{-temrs}} )$$ Then there exists a rearrangement $(r'_k)$ of $(r_k)$ such that $$ \left| \sum_{k=1}^{m} r'_k \right| \leq 2p $$ for all $1 \leq m \leq a+b$. That is, if $ap - bq$ is comparable to $p$, then we can rearrange the order of summation so that every partial sum is also comparable to $p$.

We give a sketch of proof at the end of this answer.

Now for each $k$, we find that $(p_k, p_{k+1})$ and $(f(p_k), f(p_{k+1}))$ are not parallel, forming a basis of $\Bbb{R}^2$. Thus we can find two positive real numbers $a'_k$ and $b'_k$ such that

$$ a'_k p_k - b'_k p_{k+1} = 0 \quad \text{and} \quad \left| a'_k f(p_k) - b'_k f(p_{k+1}) \right| = 3^{k}. $$

Now let $a_k = \langle a'_k \rangle$ and $b_k = \langle b'_k \rangle$ be a nearest integer of $a'_k$ and $b'_k$, respectively. Then we have

$$ \left| a_k p_k - b_k p_{k+1} \right| \leq 2p_{k} \quad \text{and} \quad \left| \left| a_k f(p_k) - b_k f(p_{k+1}) \right| - 3^{k} \right| \leq 2. $$

Put $N_k = a_k + b_k$ for notational simplicity. Let $(r'_{k,1}, \cdots, r'_{k,N_k})$ be a rearrangement of the sequence $(p_{k}, \cdots, p_{k}, -p_{k+1}, \cdots, -p_{k+1})$ obtained by Lemma, and let

$$ (x_n) = ( \underbrace{r'_{1,1}, \cdots, r'_{1, N_1}}_{k=1}, \underbrace{r'_{2,1}, \cdots, r'_{2, N_2}}_{k=2}, \underbrace{r'_{3,1}, \cdots, r'_{3, N_3}}_{k=3}, \cdots ) $$

be the sequence obtained by concatenating those rearrangements in increasing order of $k$. Then for $N_1 + \cdots + N_{k-1} < m < l$, we have

$$ \left| \sum_{n=m}^{l} x_{n} \right| \leq 2\left( p_{k} + p_{k+1} + \cdots \right). $$

This shows that partial sums of $(x_n)$ is Cauchy, hence $\sum_{n=1}^{\infty} x_{n}$ converges. On the other hand, we have

\begin{align*} \left| \sum_{n=1}^{N_1+\cdots+N_k} f(x_{n}) \right| &\geq \left| a_{k}f(p_{k}) - b_{k}f(p_{k+1}) \right| \\ &\qquad - \left( \left| a_{1}f(p_{1}) - b_{1}f(p_{2}) \right| + \cdots + \left| a_{k-1}f(p_{k-1}) - b_{k-1}f(p_{k}) \right| \right) \\ &\geq 3^{k} - 2 - \left( (3 + 2) + \cdots + (3^{k-1} + 2) \right) \\ &\geq \frac{3^{k}}{2} - 2k. \end{align*}

This shows that $\sum_{n=1}^{\infty} f(x_{n})$ does not converges, a contradiction! Therefore $f$ is linear near $x = 0$.

Sketch of proof. For each permutation $\sigma$ on $\{1, \cdots, a+b\}$, we associate a piecewise-linear function $p_{\sigma}(x)$ on $[0, a+b]$ as follows:

  • $p_{\sigma}(m) = \sum_{k \leq m} r_{\sigma(k)}$ for $m = 0, \cdots, a+b$.
  • The graph of $y = p_{\sigma}(x)$ is the polygonal path joining $(0, 0), (1, p_{\sigma}(1)), \cdots, (a+b, p_{\sigma}(a+b))$.

Now, starting from the graph $y = p_{\mathrm{id}}(x)$, we fold down each local maximum of the graph of $y = p_{\mathrm{\sigma}}(x)$ as follows:

If $y = p_{\mathrm{\sigma}}(x)$ has local maximum with value $\geq 2p$ at $k$ on $[k-1, k+1] \subset [0, a+b]$, then we have $r_{\sigma(k)} = p$ and $r_{\sigma(k+1)} = -q$. Now interchange these two values so that $x = k$ reduces to a local minimum. Repeat this process until every local extremum lies within the band $[-2p, 2p]$. With the resulting permutation $\sigma$, $r'_k = r_{\sigma(k)}$ is a rearrangement with the desired property.

The following animation is a simulation with $(p, q) = (1, 1/\sqrt{2})$ and $(a, b) = (10, 14)$.

Simulation with $(p, q) = (1, 2^{1/2})$ and $(a, b) = (10, 14)$.

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1  
Mountain erosion sequence! lol –  Squirtle Apr 9 at 15:37

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