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By the Weierstrass approximation theorem for $f\in C[a,b]$ there exists a sequence ($Q_n$) of polynomials such that $Q_n(x) \rightrightarrows f(x)$ on $[a,b]$ or equivalently for each $\varepsilon >0$ there exists a polynomial $Q$ such that $|Q(x)-f(x)|<\varepsilon$.

Let $P_n$ be athe space of all polynomials (one real variable) of degree $\leq n$, and $E_n(f)=dist(f,P_n)$.

How to show using the Weierstrass theorem that for continuous $f: [a,b]\rightarrow \mathbb R$ : $$lim_{n\rightarrow \infty} E_n(f)=0.$$

The Weierstrass theorem says nothing about degrees of approximating $f$ polynomials.

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It's obvious that $\bigl(E_{n}(f)\bigr)$ is non-increasing in $n$, and Weierstrass approximation says $E_{n}(f)$ can be made arbitrarily small by taking $n$ sufficiently large.

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  • $\begingroup$ Weierstrass approximation says that for every $\varepsilon > 0$, there exists a polynomial $Q$ such that $\|f - Q\|_{\infty} < \varepsilon$. If $N = \deg(Q)$, then $E_{N}(f) < \varepsilon$. Consequently, $E_{n}(f) < \varepsilon$ for all $n \geq N$. (Added in edit: Since the question this comment addresses was removed, I'm guessing the issue is resolved? :) $\endgroup$ – Andrew D. Hwang May 24 '14 at 18:20
  • $\begingroup$ Many thanks. Now I understand. $\endgroup$ – Alex May 24 '14 at 18:22
  • $\begingroup$ You're very welcome. :) $\endgroup$ – Andrew D. Hwang May 24 '14 at 18:39

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