0
$\begingroup$

I am learning convergence of sequence of functions in $\mathbb{R}$ and I would like to show the following:

Define $f_n : (0,1) \rightarrow \mathbb{R}, f_n(x)=\log nx$. Then $\{f_n(x)\}$does not convergent to any function pointwisely.

I know that uniform convergence implies pointwise convergence. But this does not help in my case. Then how do I make use of the definition to show the above:

$\{f_n(x): S\subset \mathbb{R} \rightarrow \mathbb{R}\}$ converges to $f(x)$ pointwisely iff for all $c\in S$, for all $\epsilon$, there exists $N=N(x,\epsilon)>0$ such that $|f_n(x)-f(x)|<\epsilon$ for all $n \geq N$.

Any help would be appreciated.

$\endgroup$
  • 2
    $\begingroup$ $\log(nx)=\log n+\log x$; what is $\lim_{n\to\infty}\log(nx)$? $\endgroup$ – egreg Nov 26 '14 at 8:07
  • 1
    $\begingroup$ Even when fixing a point $x\in (0,1)$, the sequence $(\log (nx))_{n=1}^\infty$ does not converges. $\endgroup$ – user99914 Nov 26 '14 at 8:08
  • $\begingroup$ For any $x$ in $(0,1)$, $nx$ goes to $\infty$ and so does $\log nx$ (the $\log$ is an unbounded function). $\endgroup$ – Yves Daoust Nov 26 '14 at 8:46
1
$\begingroup$

Pointwise convergence at $x\in(0,1)$ means that $$ \lim_{n\to\infty}f_n(x) $$ exists and is finite. But, for any $x>0$, $$ \log(nx)=\log n+\log x $$ so $$ \lim_{n\to\infty}f_n(x)=\lim_{n\to\infty}(\log n+\log x)=\ldots $$

$\endgroup$

Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.