Polynomials converging uniformly

Question

I want to show for each of the domains (I), if there exists a sequence of polynomials converges to $f$ where $f(x) = x \sin(1/x) $ if $x>0$ and $0$ if $x=0$.

(a)$I = (0,1]$

(b)$I = [1, \infty)$

Here's what I think: I think for both cases, there isn't any. for case (b): for any sequence of polynomials $\{P_n\}_{n=1}^\infty$ we have $\| P_n -f \|_{sup} \geq \lim_{x\rightarrow\infty}|P_n(x) - f(x)| = \infty \nrightarrow0$, so $P_n$ does not converge uniformly on R. Is this correct?

and for case (a): It's the same format but I think I have to consider $\lim_{x\rightarrow0}|P_n(x)-f(x)| $. I'm stuck here, I don't know how to show it $\nrightarrow0$.

Answer 1

For case (a), your function is continuous on $[0,1]$ and therefore you can use the Weierstrass approximation theorem in order to prove that there is such a sequence of polynomials.

Answer 2

On $[0,1]$, your $f$ is continuous. So it is a uniform limit of polynomials. The same sequence of polynomials will work on $(0,1]$.

On $[1,\infty)$, any polynomial is unbounded, while your function is bounded. So there cannot be uniform convergence.