# Absolutely but not uniformly convergent

Let $$f_n(x) = \begin{cases} 0, & \text{if }x<\frac{1}{n+1}, \\ \sin^2\frac{\pi}{x}, & \text{if }\frac{1}{n+1}\le x\le\frac{1}{n}, \\ 0, & \text{if }x>\frac{1}{n}. \end{cases}$$Show that $\{f_n\}$ converges to a continuous function, but not uniformly. Use the series $\sum f_n$ to show that absolute convergence, even for all $x$, does not imply uniform convergence.

My proof: It's easy check that limit function $f(x)$ of $\{f_n\}$ is null function, i.e. $f(x)\equiv 0$. But $\{f_n\}$ does not uniformly converges since $$M_n=\sup \limits_{\mathbb{R}}|f_n(x)-f(x)|=\sup \limits_{[\frac{1}{n+1}, \frac{1}{n}]}\sin ^2\left(\frac{\pi}{x}\right)=\frac{1}{2}-\frac{1}{2}\inf \limits_{[\frac{1}{n+1}, \frac{1}{n}]}\cos \left(\frac{2\pi}{x}\right)=1$$ which does tends to zero as $n\to \infty$.

Also $\sum \limits_{n}f_n(x)$ absolutely converges for any $x\in \mathbb{R}$ but does not converges unifromly since $$M_n=\sup \limits_{\mathbb{R}}\left|\sum \limits_{k=n+1}^{\infty}f_k(x)\right|=\sup \limits_{\mathbb{R}}(f_{n+1}(x)+f_{n+2}(x)+\dots)\ge \sup \limits_{\mathbb{R}}f_{n+1}(x)=$$$$=\sup \limits_{[\frac{1}{n+2}, \frac{1}{n+1}]}\sin ^2\left(\frac{\pi}{x}\right)=1.$$

What did you think about my proof? Are my reasoning correct?

• I don't see mistakes in you proof. Personally, the only thing I would have done differently is $$\sup\left\{\sin^2\frac{\pi}{x}\,:\,\frac{1}{n+1}\le x\le\frac1n\right\}=\\=\sup\left\{ \sin^2(\pi t)\,:\, n\le t\le n+1\right\}=1$$ because I find it easier. But it's unimportant. – user228113 Dec 8 '15 at 19:31
• @G.Sassatelli, Thank you very much! – ZFR Dec 8 '15 at 20:04