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The space $P[0,1]$ is the set of polynomial functions with domain $[0,1]$ and range $\mathbb{R}$.

I'm having trouble understand this question. Am I suppose to prove that $P[0,1]$ any any norm is not a Banach space? I don't think the statement is actually true in that case.

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  • $\begingroup$ Why do you not think it's true? Also, see this wikipedia article. $\endgroup$ – Arthur Nov 28 '17 at 6:42
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    $\begingroup$ Prove that any Hamel basis of a Banach space must be uncountable, maybe using Baire Category theorem. Can you find a countable Hamel basis for the polynomials which is countable? $\endgroup$ – астон вілла олоф мэллбэрг Nov 28 '17 at 6:45
  • $\begingroup$ @астонвіллаолофмэллбэрг Thank you! To confirm, if I can prove that statement, then I can use the countable basis $\{1,x,x^2,...\}$ of the polynomials, together with the contrapositive of the statement to show $P[0,1]$ cannot be a Banach space right? $\endgroup$ – HorribleATMath Nov 28 '17 at 6:49
  • $\begingroup$ @HorribleATMath Yes, that is the case $\endgroup$ – астон вілла олоф мэллбэрг Nov 28 '17 at 7:26
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You need the following result:

if $X$ is a Banach space, if $Y$ is a subspce of $X$ qand if $Y$ has an interior point, then $Y=X$.

Now suppose that there is a norm on $X=P[0,1]$ which makes $X$ to a Banach space. For $n \in \mathbb N_0$ let $Y_n $ be the supspace of $X$ of all plynomials with degree $ \le n$. Then $ \dim Y_n =n+1 < \infty$, hence $Y_n$ is closed and

$X= \bigcup_{n \ge 0}Y_n$

By Baire, there is $m$ such that $Y_m$ contains an interior point.

Consequence: $X=Y_m$, a contradiction.

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