Prove or give a counterexample to the following converse of theorem: A continuous function on a compact set K(subset R) is uniformly continuous.

Question

Thanks a lot for people who offered help in this question

Answer 1

The converse is not true in general.

First note that $X$ is compact if and only if every $f\in C(X)$ is bounded. One direction is trivial. To see the other, note that once $C(X)$ is bounded then $\|f\|=sup_X|f(x)|$ becomes a norm on $C(X)$ and it is easy that $C(X)$ is a unital abelian $C^*$-algebra under this norm. By Gelfand's theorem, $C(X)$ is $*$-isomorphic to some $C(X')$, where $X'$ is compact Hausdorff. Then one can show $X$ itself is compact since it is homeomorphic to $X'$.

Now it suffices to show 1) every $f\in C(X)$ is uniformly continuos is not equivalent to 2) every $f\in C(X)$ is bounded. But this is easy, one just take $X=\mathbb{N}$ then 1) is true but 2) is not.

Answer 2

In general, to determine the converse, you need to formulate the expression as "$X$ implies $Y$." The converse is "$Y$ implies $X$." Then to provide a counterexample to the coverse, you need to show an example where $Y$ is true but $X$ is not.

In this case $X$ is: "$K\subset\mathbb R$ is compact."

$Y$ is: "$\forall f:K\to\mathbb R$, $f$ continuous implies $f$ is uniformly continuous."

So a counterexample would be a set $K$ for which $Y$ is true, but not $X$.