# uniform convergence corollary

For $x \in [0,\infty)$ define $f_n(x) = \frac{1}{1+x^n}$.

(a) Find the pointwise limit function $f(x) = \displaystyle \lim_{n\to\infty} f_n(x)$.

(b) Determine if $f_n\to f$ uniformly on $[0,\infty)$.

For (a) I got a piece wise $$f(x) = \begin{cases} 1 & x < 1\\ \frac{1}{2} & x = 1\\ 0 & x > 1\end{cases}$$

Then for (b) I am not sure. There is a hint to use the following corollary:

(i) If {$f_n$} is a sequence of continuous real-valued functions on $E$ and if {$f_n$} converges uniformly to $f$ on $E$, then $f$ is continuous.

(ii) If {$f_n$} is a sequence of continuous real-valued functions on $E$ and if $\displaystyle \sum_{n=1}^{\infty} f_n$ converges uniformly on $E$ then $S(x) = \displaystyle \sum_{n=1}^{\infty} f_n(x)$ is continuous on $E$.

So it is not uniformly because $f$ is no continuous, right?

Yes, your conclusion is correct! By the corollary, if $f_n\to f$ uniformly, then $f$ would be continuous. But you correctly computed $f$, and see that it is not continuous, so the convergence cannot possibly be uniform.