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Define $$f_n(x)=\frac{1-|x|^n}{1+|x|^n}$$ Prove that the convergence is not uniform.


We have $\lim\limits_{n\rightarrow\infty}f_n(x)=\begin{cases}1\; &|x|<1\\0\; &|x|=1\\-1\; &|x|>1\end{cases}\Longrightarrow f(x)=\begin{cases}1\; &|x|<1\\0\; &|x|=1\\-1\; &|x|>1\end{cases}$

To prove $f_n$ does not converges uniformly to $f$, we use contradiction. Pick $\epsilon=\frac{1}{4}$. For any $N$, note that $f_N$ is continuous. Hence, pick $x > 1$ close to $1$ so that $|f_N (x)| <\frac{1}{4}$. But $|f(x) − f_N (x)| \geq\frac{3}{4} >\epsilon $ which is a contradiction, so the convergence is not uniform.


I am looking for different ways to do this problem. Can we do it without using $\epsilon$?

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If each $f_n$ is continuous and $f_n\to f$ uniformly, then $f$ is continuous. Since the limit function is not continuous in this case, the convergence must not be uniform.

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  • $\begingroup$ Or the $f_n$s are discontinuous. You can have a series of nowhere continuous functions which converge uniformly to a nowhere continuous function, for example. $\endgroup$ – Elliot G Mar 22 '16 at 23:43
  • $\begingroup$ The given functions are continuous. $\endgroup$ – carmichael561 Mar 22 '16 at 23:44
  • $\begingroup$ What theorem are you using? $\endgroup$ – Simple Mar 22 '16 at 23:46
  • $\begingroup$ My mistake. Might still be a useful thing to note for the OP. $\endgroup$ – Elliot G Mar 22 '16 at 23:46
  • $\begingroup$ @ Simple: The first sentence of my answer is the theorem I'm using. I don't think it has a special name. It can be proved using the triangle inequality. $\endgroup$ – carmichael561 Mar 22 '16 at 23:47

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