Show that $x \sin (\frac {1 } {x } ) $ is uniformly continuous on $(0,1) $ I want to prove that $f(x)=x \sin (\frac {1 } {x } ) $ is uniformly continuous on $0<x<1$.
If we consider the same function with the extra condition that $f $ is defined to equal zero at $x=0 $. then this new function would be continuous on $[0,1 ] $ and thus uniformly continuous.
Now my function isn't defined outside $(0,1)$ is it possible to claim that $f $ is uniformly continuous on $(0,1) $ from this?
Thanks in advance!

The function $$\bar{f} (x) =\begin{cases} 0 \mbox{ if } x=0 \\ x\sin\frac{1}{x} \mbox{ if } 0<x\leq 1\end{cases}$$ is continuous and hence uniformly continuous since $[0,1]$ is compact set. Now since $f=\bar{f}|_{(0,1)}$ the function $f$ is uniformly continuous.
