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!

Yes it's fine; being uniformly continuous on a set implies it is on any subset. I believe being able to extend a continuous function on $(a,b)$ to a continuous function on $[a,b]$ is actually equivalent to being uniformly continuous on $(a,b)$.
