sequence of polynomials converging uniformly

Question

I have $f(x) = xsin(1/x)$ if $x>0$ and $0$ if $x=0 $. I wanna see if there exists a sequence of polynomials converging to $f$ uniformly on $I$ when (a) I=$[0,1]$ and (b) I=$(0,1]$. I think it is true for both cases but about how to justify it, I'm not sure. For case (a), $f(0) = \lim_{x\rightarrow} f(x)=0$, so $f$ is continuous and by Weirestrass approximation theorem, Bernstein polynomials converge uniformly to $f$. Is this reasoning correct?. For part (b) I'm stuck. The interval is not closed. Any help?

Answer 1

For (a), it is correct. And for (b), since $(0,1]\subset[0,1]$ and since the convergence is uniform on $[0,1]$, it is also uniform on $(0,1]$.