Any function in $C[a,b]$ can be uniformly approximated by increasingly sequence of polynomials.

how to prove this?

My attempt:

By Weierstrass Approximation Theorem, for each $f\in C[a,b]\text{ and }\epsilon>0$,there is polynomial sequence $\{p_n\}$ such that exist $N>0$, for all $n>N$, $|f-p_n|\leq \epsilon$.

So for each $\epsilon>0$, for each $k$, define $\epsilon_k=\epsilon/2^k$, then there is polynomial $p_k$ such that $|f-p_k|\leq \epsilon/2^k$, then define $p_{k+1}=p_k+\epsilon_k$, then $p_k$ is increasing, and $|p_k-f|\leq \sum_{i=1}^{k-1}\epsilon/2^i<\sum_{i=1}^{\infty}\epsilon/2^i=\epsilon,k=1,2,3,...$

Is this proof correct?

  • $\begingroup$ Isn't that just the statement of the Weierstrass approximation theorem $\endgroup$ – Someguy May 12 '16 at 18:40
  • $\begingroup$ @user1952009 Original Weierstrass theorem doesn't have monotone $\endgroup$ – DuFong May 12 '16 at 18:42
  • $\begingroup$ you interpret increasingly as the sequence of polynomials is monotone ? i.e. $\max_{x \in [a,b]} |f(x)-P_n(x)| \to 0$ and $P_{n+1}(x) \ge P_n(x)$ ? $\endgroup$ – reuns May 12 '16 at 18:42
  • $\begingroup$ Yes. Though it seems trivial. $\endgroup$ – DuFong May 12 '16 at 18:54
  • $\begingroup$ and I understand nothing with what you are trying to do in your proof. you start from a sequnece $p_n$ such that $\|f-p_n\|_\infty \to 0$. then what ? (your $p_k$ doesn't seem depending on $p_n$) $\endgroup$ – reuns May 12 '16 at 18:57

No, what you wrote is not correct. It's very confused; the notation $p_k$ seems to mean more than one thing. Or something; you start with a sequence $p_k$ and then you define $p_{k+1}$ in terms of $p_k$...

Start with polynomials $p_k$ such that $|f-p_k|<2^{-k-3}$. Then $$f-2^{-k-3}<p_k<f+2^{-k-3}.$$So if you let $q_k=p_k-2^{-k}$ you have $$q_k<f-2^{-k-3}-2^{-k}<f-2^{-k-2}-2^{-k-1}<p_{k+1}-2^{-k-1}=q_{k+1}.$$

  • $\begingroup$ I mean Pick up one polynomial, then we can construct polynomial sequence recurringly $\endgroup$ – DuFong May 12 '16 at 19:57
  • $\begingroup$ @Benjamin Not by letting $p_{k+1}=p_k+\epsilon_k$ you can't. There's no way that's going to converge to $f$. $\endgroup$ – David C. Ullrich May 12 '16 at 20:46

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