I found this example online:

Let $\{f_{n}\}$ be the sequence of functions on $(0,\infty)$ defined by $f_{n}(x)=\frac{nx}{1+n^{2}x^{2}}$ .This function converges pointwise to zero. Indeed, $(1+n^{2}x^{2})\sim n^{2}x^{2}$ as $n$ gets larger. So, $\underset{n\rightarrow\infty}{\lim}f_{n}(x)=\underset{n\rightarrow\infty}{\lim}f_{n}(x)=\frac{nx}{n^{2}x^{2}}.=\frac{1}{x}\underset{n\rightarrow\infty}{\lim}\frac{1}{n}=0.$

But for any $\epsilon<1/2$ , we have

$|f_{n}(\frac{1}{n})-f(\frac{1}{n})|=\frac{1}{2}-0>\epsilon$ .

Hence$ \{f_{n}\}$ is not uniformly convergent.

  1. By the definition of pointwise convergence: $f_n(x)\rightarrow f(x)$ for all $x\in X$. But this does not seem to happen when $x=\frac{1}{n}$. So how can ${f_n}$ converge pointwise to zero for all $x\in (0,\infty)$ if $f_{n}(\frac{1}{n})=\frac{1}{2}$?

  2. Is it correct to think like this: If $ \{f_{n}\}$ was uniformly convergent, then if someone gave me an $\epsilon<1/2$, I should be able to find and $n\geq N$ so that $|f_{n}(x)-f(x)|$ became smaller than $1/2$ for any $x \in (0,\infty)$, but this doesn't hold for any $x$ because we saw that $|f_{n}(\frac{1}{n})-f(\frac{1}{n})|=\frac{1}{2}-0>\epsilon$, so ${f_n(x)}$ cannot be uniformly convergent.

  • $\begingroup$ pointwise means you fix an $x$. $\endgroup$ – Yimin Mar 13 '13 at 18:32
  • $\begingroup$ @Yimin What do you man by "fix an x"? What happens if i fix $x=1/n$? $\endgroup$ – john.abraham Mar 13 '13 at 18:33
  • $\begingroup$ $1/n$ is changing... $\endgroup$ – Yimin Mar 13 '13 at 18:33
  • $\begingroup$ @Yimin Ohh, of course. Thanks! :) $\endgroup$ – john.abraham Mar 13 '13 at 18:34

Pointwise convergence $f_n(x)\underset{n\to\infty}{\to}{f(x)}$ is convergence by $n$ at every (but fixed) argument $x$.


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