# Prove that $P[0,1]$ admits no norm with respect to which it is a Banach space

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.

• Why do you not think it's true? Also, see this wikipedia article. – Arthur Nov 28 '17 at 6:42
• 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? – астон вілла олоф мэллбэрг Nov 28 '17 at 6:45
• @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? – HorribleATMath Nov 28 '17 at 6:49
• @HorribleATMath Yes, that is the case – астон вілла олоф мэллбэрг Nov 28 '17 at 7:26

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.