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Let $f$ be a function defined on $R$. for any absoulutely convergent series $\sum_{n=1}^{\infty}a_{n}$,the series $\sum_{n=1}^{\infty}f(a_{n})$ converges,Prove that $$ f(x)=O(x)\qquad (x\to 0) $$.

I have tried to prove the contradiction side,assume that there exist a series which makes $$ \lim_{n\to\infty}\left|\frac{f(a_{n})}{a_{n}}\right|=+\infty $$ Then I stuck here,Can anybody help me? Thank you very much!

pxchg1200
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  • Suppose that f(x) = c + g(x) s.t. g(x)->0 as x->0 and c>0. What happens to the sum? – Paul Jun 06 '14 at 05:35
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    @Paul $\lim f=c\ne0$ is hardly the only way $f=O(x)$ can fail. – anon Jun 07 '14 at 04:17
  • True. As another example, if f(x) = sqrt(x) and a[n]=1/n^2, then f is not O(x) as x->0 and can be seen to transform the convergent series to a divergent one. – Paul Jun 07 '14 at 04:35

1 Answers1

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If $f(x)$ is not $O(x)$ as $x\to0$, we can find a sequence $x_k\ne0$ tending monotonically to $0$ such that all the $f(x_k)$ have the same sign and $$ \lim_{k\to\infty}\left|\frac{f(x_k)}{x_k}\right|=\infty\tag{1} $$ For $n\gt0$, find $k_n$ so that for $k\ge k_n$, we have $|x_k|\le1/n^2$ and $$ \left|\frac{f(x_k)}{x_k}\right|\ge n\tag{2} $$ Let $j_n=\left\lfloor\dfrac{1/n^2}{|x_{k_n}|}\right\rfloor$, then $$ \frac1{2n^2}\lt\overbrace{|x_{k_n}+x_{k_n}+\dots+x_{k_n}|}^{j_n\text{ terms}}\le\frac1{n^2}\tag{3} $$ Summing the right inequality of $(3)$ gives $$ \left|\sum_{n=1}^\infty\ \overbrace{x_{k_n}+x_{k_n}+\dots+x_{k_n}}^{j_n\text{ terms}}\right| \le\sum_{n=1}^\infty\frac1{n^2}\tag{4} $$ which converges. However, $(2)$ and the left inequality of $(3)$ say $$ \left|\sum_{n=1}^\infty\ \overbrace{f(x_{k_n})+f(x_{k_n})+\dots+f(x_{k_n})}^{j_n\text{ terms}}\right| \ge\sum_{n=1}^\infty\frac1{2n}\tag{5} $$ which diverges.

Therefore, $f(x)=O(x)$.

A Note On Inequality $(3)$

The first of the equivalent inequalities in $(6)$ follows from the definition of the floor function: $$ 0\le\frac{1/n^2}{|x_{k_n}|}-j_n\lt1\tag{6a} $$ $$ j_n|x_{k_n}|\color{#C00000}{\le}1/n^2\color{#00A000}{\lt}(j_n+1)|x_{k_n}|\tag{6b} $$ $$ \frac{j_n}{j_n+1}1/n^2\color{#00A000}{\lt} j_n|x_{k_n}|\color{#C00000}{\le}1/n^2\tag{6c} $$ and since $j_n\ge1$, $\frac{j_n}{j_n+1}\ge\frac12$, we get $$ \frac1{2n^2}\lt j_n|x_{k_n}|\le\frac1{n^2}\tag{7} $$ which is inequality $(3)$.

robjohn
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  • I think choose $|x_{k}|\leq \frac{1}{2n^2}$ is suitable,because $j_{n}|x_{k_{n}}|\geq \frac{1}{n^2}-x_{k_{n}}\geq \frac{1}{2n^2}$ – pxchg1200 Jun 08 '14 at 00:20