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Let $h:[0,1]\to \mathbb R$ be a continuous and strictly monotone function. Let $f:[0,1] \to \mathbb R$. Prove that there exists a sequence of polynomials $p_n :\mathbb R \to \mathbb R$ such that $p_n(h(x)) \to f(x)$.

If I prove the case $h:[0,1]\to [0,1]$ with $h$ homeomorphism, then I'm done. I think that this case is easier, because $f(x)=f(h(y))$ for a unique $y\in [0,1]$. I think that I have to consider a sequence of polynomials $p_n(h(y))=p_n(x) \to f(x)=f(h(y))$ uniformly. But that is not helpful. Please give me some help.

Gerry Myerson
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Miguel
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1 Answers1

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Suppose that $h([0,1])=[a,b]$, then $h:[0,1]\to[a,b]$ is a homeomorphism. The function $f\circ h^{-1}$ is continuous on $[a,b]$. So there is a sequence of polynomials $(P_n)_n$ that converges uniformly on $[a,b]$ to $f\circ h^{-1}$. This is equivalent to the fact that $(P_n\circ h)_n$ converges uniformly to $f$ on $[0,1]$.$\qquad\square$

Omran Kouba
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  • It has been quite a while since you answered this, but would you mind explaining why is $f\circ h^{-1}$ continuous? – Fhoenix Mar 17 '19 at 16:21