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While studying Weierstrass approximation Theorem, I realized continuity on a compact set is necessary to solve the Theorem. Then, here is my question.

Does there exist a sequence $(p_n)_{n\ge 1}$ of polynomial such that it converges uniformly to $f$ on (0,1), where f is bounded continuous function from $(0,1)$ to $\Bbb R$ ?

I think it is not true because when it comes to $f(x)= sin\frac{1}{x}$, any sequence of polynomial wouldn't converge, but I can't prove it.

Here is Further question

Then, If sequence of polynomial converge uniformly to $f$ on some open interval, $f$ can be continuously extended to the end point..?

I want you to give me a few hints to solve my first question, and give me an counterexample of the other question if it is not true. Thanks.

J. W. Tanner
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1 Answers1

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If $p_n \to f$ uniformly on $(a,b)$ where $p_n$'s ar polynomials and $f$ is continuous then, for any $\epsilon >0$ we can find $n$ such that $|p_n(x)-f(x)| <\epsilon$ for all $x$. If you combine this with the fact that $p_n$ is uniformly continuous on $[a,b]$ you can check that $f$ is uniformly continuous on $(a,b)$. This implies that $f$ extends to a continuous function on $[a,b]$.

Your example $f(x)=\sin (\frac 1 x)$ works fine because this function is not uniformly continuous. Consider the points $x_n=\frac 1 {n\pi}$ and $y_n=\frac 1 {(n+\frac 1 2) \pi} $ to show that $f$ is not uniformly continuous.

Kavi Rama Murthy
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