Functions preserving convergent series

Find all functions $f:\mathbb{R} \rightarrow \mathbb{R}$ that preserve convergent series. (That is the series $\sum \limits_{n=1}^{\infty} f(a_n)$ converges whenever the series $\sum \limits_{n=1}^{\infty} a_n$ converges)

Solution

To see what actually is going on here let us consider some examples. The series $\sum \limits_{n=1}^{\infty} \frac{(-1)^{n}}{\sqrt{ n}}$ converges as an alternating one, but the series $\sum \limits_{n=1}^{\infty} \frac{1}{n}$ diverges since it is the harmonic series. Hence $x^2 , \; x \in \mathbb{R}$ cannot preserve convergent series. Similarly $x^3$ cannot also preserve convergent series as well neither can $x^k, \; k \geq 4$. Hence no mononym can preserve convergent series. Therefore , neither a polynomial can preserve a convergent series. So, we are left with only one option. That the functions that are most likely to preserve convergent series are of the form:

$$f(x)=ax, \; x \in (-\delta, \delta)$$

In fact these are all the functions that indeed preserve convergent series and this is exactly what are proving here. 

Proof

Let $f:\mathbb{R} \rightarrow \mathbb{R}$ be a function. Then the following statements are equivalent. 

1. $f$ preserves convergent series , that is the series $\sum \limits_{n=1}^{\infty} f(a_n)$ converges whenever the series $\sum \limits_{n=1}^{\infty} a_n$ converges.
2.  $f$ is linear in a neighbourhood of zero , that is it is of the form $f(x)=ax, \; x \in (-\delta, \delta)$ 

The direction $(2) \implies (1)$ is obvious. We only need to prove the direction $(1) \implies (2)$.

Claim 1: $f$ is odd in a neighbourhood of zero. 

  
Proof:

Suppose on the contrary that this does not hold. Then there exists a sequence $x_n$ such that $\lim x_n =0$ that for each  $ n \in \mathbb{N}$ holds $y_n =f\left ( x_n \right )+f\left ( -x_n \right ) \neq 0$. For all $n \in \mathbb{N}$ we choose a positive integer $k_n$ such that $\displaystyle \left | y_n \right |>\frac{1}{k_n}$. We consider the sequence $a_m$ that is defined piecewise as follows:

$$\underbrace {{x_1}, - {x_1},{x_1}, - {x_1}, \ldots ,{x_1}, - {x_1}}_{2{k_1}{\rm{ }}},\underbrace {{x_2}, - {x_2},{x_2}, - {x_2}, \ldots ,{x_2}, - {x_2}}_{2{k_2}{\rm{ }}}, \ldots$$

$$\ldots ,\underbrace {{x_n}, - {x_n},{x_n}, - {x_n}, \ldots ,{x_n}, - {x_n}}_{2{k_n}{\rm{ }}}, \ldots$$

that is its $n$-th part is consisted of $2k_n$ terms that are alternating equal to  $x_n$ and $-x_n$ respectively. Since $\lim x_n =0$ this means that $\displaystyle \sum_{n=1}^{\infty}a_n =0$. However from Cauchy's criterion the series $\displaystyle \sum_{n=1}^{\infty} f(a_n)$ does not converge because the sum of the terms of the $n$-th part is equal to $k_n |y_n|>1$. Hence claim 1 is complete. 

Claim 2: There exists a $\delta_1$ such that for all $x, y \in (-\delta_1, \; \delta_1)$ holds: $$f(x+y)=f(x)+f(y)$$ That is $f$ is linear in a neighbourhood of zero.

Proof:

Like above we choose $k_n$ such that for all $n \in \mathbb{N}$ to hold $\displaystyle \left | z_n \right |>\frac{1}{k_n}$. We consider the sequence $a_m$ that is defined piecewise as follows:

Its $n$-th part is consisted of  $3k_n$ that are $x_n +y_n, -x_n, -y_n$ repeated $k_n$ for'es. Since $ \lim x_n =\lim y_n =0$ we get $\displaystyle \sum _{n = 1}^\infty a_n = 0 $ but the series $\displaystyle \sum _{n = 1}^\infty f(a_n)$ does not converge from Cauchy's criterion because the sum of the terms of its $n$-th part is equal to  $k_n |z_n|>1$. This proves claim 2.

Claim 3: There exist $\delta_2>0$ and a constant $C>0$ such that forall $x \in (-\delta_2, \delta_2)$ to hold: $|f(x)|\leq Cx$

Proof:

Suppose that the claim is false. Then there exists a sequence $x_n$ such that forall $n \in \mathbb{N}$ to hold $|x_n|<\dfrac{1}{2^n}$ kai $|f(x_n)|\geq 2^n |x|$. We note that $x_n \neq 0$ for all  $n \in \mathbb{N}$ (since from claim (1) holds $f(0)=0$) .

Hence, for every $ n \in \mathbb{N}$ we choose a positive integer $k_n$ such that it holds:

$$\frac{1}{2^{n+1}}\leq k_n \left | x_n \right |\leq \frac{1}{2^n}$$

We consider the sequence $a_m$ that is defined piecewise again as follows:
its $n$-th part is consisted of $2k_n$ terms that are of the form:

$$\underbrace {{x_n},{x_n}, \ldots ,{x_n}}_{{k_n}},\underbrace { - {x_n}, - {x_n}, \ldots , - {x_n}}_{{k_n}}$$

Again we have that $\displaystyle \sum_{n=1}^{\infty} a_n =0$ but the series $\displaystyle \sum_{n=1}^{\infty} f(a_n)$ does not converge from Cauchy's criterion since the absolute value of the sum of the first $k_n$ terms is:

$$k_n \left | f(x_n) \right |>k_n 2^n \left | x_n \right |\geq 2^n \cdot \frac{1}{2^{n+1}}=\frac{1}{2}$$
This proves claim 3.

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Now to complete the exercise we set   $\delta=\frac{1}{2}\min \left \{ \delta_1, \delta_2 \right \}$ and we note that forall $x, y \in (-\delta,\delta)$ holds: $|x-y|< \min \left \{ \delta_1, \delta_2 \right \}$. Therefore from Claims $(1)$ and $(2)$ follows that:


$$\left| {f\left( x \right) - f\left( y \right)} \right| = \left| {f\left( {x - y} \right)} \right| \le C\left| {x - y} \right|$$

Hence $f$ is continuous, additive on $(-\delta, \delta)$ and linear. This completes the exercise.

The exercise can also be found in mathimatikoi.org and mathematica.gr .