# Prove that there exists a sequence of polynomials $(p_n)$ that converges uniformly to $f$ on $[a,b]$

I've been stuck on the following problem for a some time but have it started somewhat and not sure whether it is correct or how to proceed.

Let $f$ be a continuous function on $[a,b]$. Prove that there exists a sequence of polynomials $(p_n)$ that converges uniformly to $f$ on $[a,b]$

I know that since $f$ is a continuous function on $[a,b]$, then applying the Weierstrass Approximation Theorem there exists a sequence $(p_n)$ of polynomials such that $p_n \to f$ uniformly on $[a,b]$.

Then,

Considering a sequence of polynomials $p_n$ which converges uniformly to $f$.

If we let $q_n(x)=p_n(x)+f(a)-p_n(a)$.

Then $q_n(a)=f(a)$, $q_n$ must be a polynomial and

$$|q_n(x)-f(x)| \leq |p_n(x)-f(x)|+|f(a)-p_n(a)|$$ which proves that $q_n$ converges uniformly to $f$.

So is this enough to prove that there exists a sequence of polynomials $(p_n)$ that converges uniformly to $f$ on $[a,b]$?

• "applying the Weierstrass Approximation theorem, there exists a sequence $(p_n)$ of polynomials such that $p_n\to f$ uniformly on $[a,b].$" Is this conclusion not exactly what you're trying to prove? – spaceisdarkgreen Mar 24 '17 at 3:24
• @spaceisdarkgreen ah yes I got a little confused, would you be able to help me out? – jh123 Mar 24 '17 at 3:25
• Not really as it's unclear what help is needed. You are correct that the Weierstrass approximation theorem says that there is a sequence of polynomials that converges uniformly to $f$. And that's what you're trying to prove. The proof is done. – spaceisdarkgreen Mar 24 '17 at 3:31
• @spaceisdarkgreen so then all I need to say is, since $f$ is a continuous function on $[a,b]$, then applying the Weierstrass Approximation Theorem there exists a sequence $(p_n)$ of polynomials such that $p_n \to f$ uniformly on $[a,b]$. ? seems a little short – jh123 Mar 24 '17 at 3:33
• Whether you've 'played by the rules' for whatever exercise this is is less clear. Perhaps you were given a different formulation of the Weierstrass approximation theorem and you need to show explicitly that it implies the existence of that sequence. Perhaps the exercise is to prove the Weierstrass approximation. – spaceisdarkgreen Mar 24 '17 at 3:33

## 1 Answer

Let $\epsilon>0.$ Once we pick our polynomials $p_n$, to prove uniform convergence we must show there is an $N$ such that $|f(x)-p_n(x)|<\epsilon$ for all $n>N$ and all $x\in[0,1].$

Consider an arbitrary integer $n$. By the Weierstrass approximation theorem (as you stated it in the comments), we can choose a polynomial $p_n$ that satisfies $|p_n(x)-f(x)| < 1/n$ for all $x\in[0,1].$ Choose $p_n$ this way for all $n.$

Now, pick an integer $N >1/\epsilon.$ Then for any $n>N,$ we have $|p_n(x)-f(x)|< \frac{1}{n}<1/N < \epsilon$ for all $x\in [0,1].$

• thanks very much, so this completes the full details of the proof? and we do not need to choose a particular pn? – jh123 Mar 24 '17 at 4:01
• @jh123 You choose them to get closer and closer to $f$ as $n$ increases (close in the sense that $|f-p_n|$ is small for all $x$). You can choose polynomials arbitrarily close by your statement of the Weierstrass approximation theorem. There aren't explicit polynomials to pick (if that's what you mean), just those whose existence the theorem guarantees. (Actually, there are explicit polynomials one can pick en.wikipedia.org/wiki/Bernstein_polynomial , but I doubt you need a constructive proof here.) – spaceisdarkgreen Mar 24 '17 at 4:12