I just got out of an introductory real analysis exam. I had this question which I tried to solve for more than $1.5$ hours and yet I accomplished nothing.

Let $f:\Bbb R \to\Bbb R$ be a differentiable function with $f(0)=f'(0)=0$.

Prove that $$ \sum_{n=1}^\infty \left|f\left(\frac 1 n \right)\right|^r $$ Converges for all $r>1$.

I tried to use the ratio test, root test, Lagrange's theorem, and a few more things I know but I couldn't crack this.

Could someone explain how to solve this? (We only studied basic things about series)

  • $\begingroup$ This looks like it's false. If you set $f(x)=x^2$ and $r=1/2$, then the sum is $\sum_{n=1}^\infty \sqrt{1/n^2}=\sum_{n=1}^\infty 1/n=\infty$. $\endgroup$ – user281392 Mar 4 '16 at 22:51
  • $\begingroup$ @user281392 sorry I miscopied!! It's $r>1$. $\endgroup$ – YoTengoUnLCD Mar 4 '16 at 22:52
  • $\begingroup$ That doesn't converge(it's $1/2^r \cdot \infty$). $\endgroup$ – YoTengoUnLCD Mar 4 '16 at 22:57
  • $\begingroup$ Can you assume $f^\prime$ continuous? $\endgroup$ – sinbadh Mar 4 '16 at 23:01
  • $\begingroup$ @sinbadh No.${}$ $\endgroup$ – YoTengoUnLCD Mar 4 '16 at 23:03

Since $f$ is differentiable at $0$, the expression $$ {f(\frac1n)-f(0)\over\frac1n} $$ tends to $f'(0)=0$ as $n\to\infty$. But $f(0)=0$, so this means $ nf(\frac1n) $ tends to zero. In particular for all large $n$ we have $$ \textstyle\left|nf(\frac1n)\right|\le1. $$ Rearranging, this implies $$ \left|f({\textstyle\frac1n})\right|^r\le\frac1{n^r} $$ for all large $n$, which is enough to show $\sum|f(\frac1n)|^r$ converges.

  • $\begingroup$ Thanks for your answer. We're using the definition $$f'(0)=\lim_{x\to 0} \frac {f(x)-f(0)}{x}$$ Could you explain how to get the equivalence between that and $\lim_{n\to \infty}\frac{f(\frac 1 n)- f(0)}{\frac 1 n}$? $\endgroup$ – YoTengoUnLCD Mar 5 '16 at 18:33
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    $\begingroup$ @YoTengoUnLCD : Please see here $\endgroup$ – user99914 Mar 6 '16 at 5:11

For another approach: Since $1/n\to 0$ and $f(0)=0$, by definition of derivative, $$\left|\frac{f(\frac{1}{n})}{\frac{1}{n}}\right|\to f^\prime(0)$$.

That is $|nf(1/n)|\to |f^\prime(0)|$. Therefore, for fixed $r$, $|nf(1/n)|^r\to |f^\prime(0)|^r$. Now, let $a_n=|f(1/n)|^r$. Then $$\lim_{n\to\infty}n^ra_n=\lim_{n\to\infty}n^r|f(1/n)|^r=\lim_{n\to\infty}|nf(1/n)|^r=|f^\prime(0)|^r.$$

By the Pringsheim test, we conclude $\sum_{n=1}^\infty|f(\frac{1}{n})|^r$ is a real number.

Please, note that we don't use $f^\prime(0)=0$. Indeed, we only use $f$ differentiable in 0.

  • $\begingroup$ Very nice! +1.${}$ $\endgroup$ – YoTengoUnLCD Mar 30 '16 at 16:36

By the MVT, $f(\frac{1}{n})=\frac{1}{n}\frac{f(\frac{1}{n})-f(0)}{\frac{1}{n}}=\frac{1}{n}f^\prime(x_n)$ with $0<x_n<\frac{1}{n}$. Therefore $x_n\to 0$. Then $|f(\frac{1}{n})|=\frac{1}{n}|f^\prime(x_n)|$. Thus $f^\prime(x_n)\to f^\prime(0)=0$ since $nf(\frac{1}{n})=f^\prime(x_n)$ and $f$ is continuous. It implies that $\{f^\prime(x_n)\}$ is bounded. Say $|f^\prime(x_n)|\le M$.

Therefore $\sum_{n=1}^\infty|f(\frac{1}{n})|^r\le \sum_{n=1}^\infty|\frac{1}{n}f^\prime(x_n)|^r\le\sum_{n=1}^\infty\frac{1}{n^r}M^r=M^r\sum_{n=1}^\infty\frac{1}{n^r}<\infty$

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    $\begingroup$ I think you're assuming that $f'$ is continuous, but this is not in the hypotheses. $\endgroup$ – YoTengoUnLCD Mar 4 '16 at 23:12
  • $\begingroup$ No. I'm using $f^\prime(x_n)=nf(\frac{1}{n})$ $\endgroup$ – sinbadh Mar 4 '16 at 23:13
  • $\begingroup$ Something like $$f(x) = \begin{cases} x^2 \sin(1/x^2) & \text{if } x\neq 0 \\ 0 & \text{if } x= 0 \end{cases}$$ shows that $|f'(x_n)| \le M$ is not possible. $\endgroup$ – user99914 Mar 4 '16 at 23:26
  • $\begingroup$ Why the downvote? Basically it is the same solution above and he has at least 3 upvotes $\endgroup$ – sinbadh Mar 4 '16 at 23:33
  • $\begingroup$ I did not downvote it! I don't understand how you get that $f'(x_n)\to f'(0)$, could you elaborate a bit on that? $\endgroup$ – YoTengoUnLCD Mar 5 '16 at 0:03

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