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Let $f\in C[0,1]$ be real-valued. Prove that there is monotone increasing sequence of polynomials $\{p_n(x)\}^\infty_{n=1}$ converging uniformly on $[0,1]$ to $f(x)$.

Yea, it should be done by stone-Weierstrass theorem, there I will get simply a sequence of polynomials that converges to $f$. Is there any way we can define this polynomial to make monotone increasing polynomials?

thanks for helping

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  • $\begingroup$ You should atone for your "stone" or you will go it alone. $\endgroup$ – marty cohen Jun 24 '14 at 21:11
  • $\begingroup$ what do you mean "go it alone". $\endgroup$ – Toeplitz Jun 24 '14 at 21:15
  • $\begingroup$ Don't make a rhyme if you can't do the time. $\endgroup$ – marty cohen Jun 24 '14 at 21:22
  • $\begingroup$ It's just a sillygism. $\endgroup$ – marty cohen Jun 24 '14 at 21:22
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    $\begingroup$ You want, I think, for each $x$ the sequence $(p_n(x))$ to be monotone. Approximate $g_n=f-1/n$ by a polynomial to within $\epsilon_n$ for appropriate $\epsilon_n$. $\endgroup$ – David Mitra Jun 24 '14 at 21:24
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This is just a technicality, the essential result is that the polynomials are dense in $C[0,1]$.

Take any strictly decreasing sequence $\alpha_n \downarrow 0$ that converges to zero.

Let $\beta_n ={1 \over 2} (\alpha_n+\alpha_{n-1})$, $\delta_n = {1 \over 2} (\alpha_{n-1}-\alpha_n)$. Note $\beta_n>0, \delta_n>0$ and $\beta_n \to 0$, $\delta_n \to 0$, and $\beta_n+\delta_n=\alpha_{n-1}$, $\beta_n-\delta_n = \alpha_n$

Choose $p_n$ to be a polynomial such that $\|p_n-(f-\beta_n)\| < \delta_n$. In particular, we have $f(t)-\alpha_{n-1} < p_n(t) < f(t) - \alpha_n$ for all $t \in [0,1]$.

It follows that for any fixed $t$ that $p_{n+1}(t) > p_n(t)$, and since $\|p_n-f\| \le \|p_n-(f-\beta_n)\| + \beta_n$, we see that $p_n \to f$.

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