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I've the following exercise:

Prove that every continuous function $f:[a,b]\rightarrow \mathbb R$ is the uniform limit of a sequence of even polynomials if and only if $(a,b)$ doesn't contain the origin.

Proof:

Suppose $(a,b)$ doesn't contain the origin. And for simplicity consider $[0,1]$ instead of $[a,b]$.

It's equivalent to work with $[0,1]$ because of this proof by @zhw at From [a,b] to [$0$,$1$]:

Suppose we have the result for $[0,1].$ If now $f\in C([0,b]),$ define $g:[0,1]\to [0,b]$ by setting $g(x) = bx.$ Then $f\circ g\in C([0,1]).$ So there exists a sequence $P_n$ of even polynomials such that $P_n\to f\circ g$ uniformly on $[0,1].$ This implies $P_n\circ g^{-1}\to f$ uniformly on $[0,b].$ Since each $P_n\circ g^{-1}$ is an even polynomial, we have the result for $[0,b]$.

If now $0<a<b$ and $f\in C([a,b]),$ we can extend $f$ to a continuous function $F$ on $[0,b],$ simply by setting $F(x) = f(a)$ for $x\in [0,a).$ By the above, there is a sequence $P_n$ of even polynomials such that $ P_n \to F$ uniformly on $[0,b]$. This of course implies $ P_n \to F$ uniformly on $[a,b]$. Since $F=f$ on $[a,b]$, we have the desired result for $[a,b]$.

With a similar argument we obtain the result for $a<b\le 0$.

So we have this $\Rightarrow$ side of the proof. Now for this $\Leftarrow$ side, I think it's easier if we prove it by contradiction.

So we need to start the proof from this (below) and to find a contradiction:

Exists an interval $[a,b]$ s.t. every continuous function $f:[a,b]\rightarrow\mathbb R$ is the uniform limit of a sequence of even polynomials and $(a,b)$ contains the origin.

I've read @zhw proof many times but I can't find a contradiction considering now that (a,b) contains the origin.

Can someone tell me where is the possible contradiction?

Help will be appreciated.

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  • $\begingroup$ See my A to Q 2292801 $\endgroup$ – DanielWainfleet May 24 '17 at 3:47
  • $\begingroup$ but I have to follow this idea, I just need to find one contradiction @DanielWainfleet $\endgroup$ – user441848 May 24 '17 at 3:52
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    $\begingroup$ Consider the function $f(x)=x$ on $[a,b].$ $\endgroup$ – zhw. May 25 '17 at 6:34
  • $\begingroup$ A not even function $f$. right $\endgroup$ – user441848 May 25 '17 at 19:06
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Taking an smaller interval, we can suppose wlog that $[a,b] = [-\delta,\delta]$, $\delta>0$. But if every continuous function in $[-\delta,\delta]$ is uniform (even pointwise) limit of even functions (polynomials in this case), then every continuous function is even and this is obviosly false.

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