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.

  • 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

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 $$ \lim_{n\to\infty}f_n(x)=\lim_{n\to\infty}(\log n+\log x)=\ldots $$


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