# Pointwise & uniform boundednes

1. Jan 26, 2012

### camillio

In Baby Rudin, Theorem 7.25 states:
If $K$ is compact, $f_n \in C(K)$ for $n=1,2,3,...$ and if ${f_n}$ is pointwise bounded and equicontinuous on $K$, then
(a) ${f_n}$ is uniformly bounded on $K$

The theorem continues with point (b), which I understand.

My question is, whether point (a) needs the equicontinuity, since:
* $C(K)$ is a set of continuous bounded functions on a compact space, i.e. uniformly continuous, hence there is no function $g$ s.t. $g(x) \to \infty$.
* if the previous holds, then by the hypothesis of pointwise boundedness of $\{f_n\}$, there must exist a number $\sup_{x \in K, n=1,2,3,...} |f_n(x)| = M<\infty$, $x\in K$ arbitrary and hence the uniform boundedness should hold.

Am I wrong?

Last edited: Jan 27, 2012
2. Jan 26, 2012

### micromass

The sequence

$f_n:[0,1]\rightarrow \mathbb{R}:x\rightarrow \frac{nx}{x^2+(1-nx)^2}$

seems to be a counterexample to your claim. See example 7.21 of Rudin, I modified it a bit.

3. Jan 27, 2012

### camillio

Thanks a lot! I assume, that the reason is that $f_n$ converges pointwise for every $x\in K$, but due to the points $1/n$ in your example, it is impossible to find a global finite $M$ s.t. $|f_n(x)| \leq M$ for all $x \in K$. Right?

I really do not understand why usually things go quite well, but at some moments I get stuck with (obvious?) things :(