# Can $\sin(1/x)$ be approximated pointwise by polynomials over $(0,\infty)$

Can the function $$f(x)=\sin(1/x)$$ on $$(0,\infty)$$ be approximated by a sequence of polynomials pointwise on the domain?I am sure that uniform approximation is not possible because $$\lim_{x\to 0+}\sin(1/x)$$ does not exist.But is there a possibility of pointwise approximation by a polynomial sequence? [Note: I am an undergraduate student and the only thing that I can use is Weierstrass polynomial approximation and any other independent idea,but I know nothing of approximation theory,so I am expecting some elementary answer.]

• $f(x)=\sin(1/x)$ is continuous and bounded on $\Bbb{R}^*$ thus $f_n(x)=\int_{-\infty}^\infty f(y)ne^{-\pi n^2(x-y)^2}dy$ is analytic and $f_n\to f$ uniformly on every closed interval where $f$ is continuous and for $K_n$ growing fast enough the sequence of Taylor approximations $f_{n,K_n}$ of $f_n$ satisfies your requirements. Note $f_{n,K_n}$ is continuous and uniformly bounded on $[-A,A]$ for all $A$. – reuns Dec 13 '19 at 3:55

Short answer: Yes, because any continuous function on a compact interval can be approximated arbitrarily sharply by a polynomial. So, at step $$n$$ of your approximating sequence, consider the compact subset $$[1/n, n]$$ of $$[0, \infty)$$, and find (by Weierstrass approximation) a polynomial $$P_n(x)$$ which is at distance $$\leq 1/n$$ from your $$\sin(1/x)$$ function, uniformly on $$[1/n, n]$$. Then for any fixed point $$x$$ of $$(0, \infty)$$, $$P_n(x)$$ will converge to $$\sin(1/x)$$, since for $$n$$ large enough you will always have $$x \in [1/n, n]$$, and therefore $$\left|\sin(1/x) - P_n(x)\right| \leq 1/n \stackrel{n \to \infty}\to 0$$.
Actually there is nothing specific to $$\sin(1/x)$$ in this argument: it works the same for any function that is continuous on any open interval of $$\mathbf{R}$$.
I feel like you can do $$\sup_{x\in[1/n,n]}|f(x)-p_{n}(x)|<1/n$$ by applying Weierstrass on each $$[1/n,n]$$.
Let $$I_n = [{1 \over n},n]$$, this is compact and we can choose a polynomial $$p_n$$ such that $$\sup_{x \in I_n}|p_n(x)-f(x) | < {1 \over n}$$.
Then for any fixed $$x>0$$ we have $$p_n(x) \to f(x)$$.