The continuous function $f(t)=\sin(1/t)$ it is not uniformly approachable by polynomials?

Question

The continuous function $f: (0,1) \to \mathbb{R}$, $f(t)=\sin(\frac{1}{t})$ it is not uniformly approachable by polynomials. Why?

Any suggestions would be great!

Answer 1

Assume that $|P(x)-f(x)| < \frac{1}{2}$ for $x\in (0, 1)$. If $P(0)=\alpha$, what can you say about $P(x)$ in some neighborhood of $0$?

Answer 2

It is not difficult to show that, if $f_n:[a,b]\rightarrow \mathbb{R}$ is a sequence of continuous functions, such that $f_n\rightarrow f$ uniformly on the open interval $(a,b)$, then the convergence is uniform on $[a,b]$. This is a consequence of continuity and the Bolzano-Cauchy criteria for uniform convergence. Thus, the function $f$ must be continuous on $[a,b]$.

Recall the polynomials are continuous everywhere, in particular they are continuous on $[0,1]$. So, if you have a sequence $\{p_n\}$ of polynomials approximating $f(t)=\sin\frac{1}{t}$ uniformly on $(0,1)$ you will deduce that $\{p_n(0)\}$ converges to $f(0)$. But this is impossible, because $$\lim_{x\rightarrow 0^+}\sin\frac{1}{t}$$ doesn't exist.