Support is defined as the closure of $\{x:f(x)\neq 0\}$. Now consider $f(x)=\frac{1}{x}\chi_{(0,1)}$. So the support is $[0,1]$ which is compact. However, any continuous function with compact support is uniformly continuous. $f(x)$ is continuous on $(0,1)$ but it is not uniformly continuous.
I am confused. Could anyone explain where I made a mistake.
On
The function $f(x)=\frac{1}{x}\chi_{(0,1)}$ is not continuous on $\mathbb{R}$ nor on $[0,1]$. It has jump discontinuities at $x=0$ and $x=1$.
It is continuous on $(0,1)$ of course, but its support is all of $(0,1)$ which is not compact. Because, as a subspace of $(0,1)$, the closure of $(0,1)$ is still $(0,1).$ So it does not have compact support on this domain.
But if you further restrict to any compact set within $(0,1)$, you will see that the function is actually uniformly continuous. Eg it is uniformly continuous on the interval $[1/4,3/4]$.
It is not continuous. As surprising as it may seem.
What is the value of $f(0)$?
$f(0)$ is well defined. $f(0) =0$ as $0$ is outside of $(0,1)$.
But what is $ \displaystyle \lim_{x \to 0^+} f(x) $ ?