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$\newcommand\norm[1]{\left\lVert#1\right\rVert}$

This is a question from Royden's book:

Let $X$ be the linear space of all polynomials defined on $\mathbb{R}$. For $p \in X$, define $\norm{p}$ to be the sums of the absolute values of the coefficients of $p$. Show that this is a norm on $X$. For each $n$ define $\psi_n: X \to \mathbb{R}$ by $\psi_n(p)=p^{(n)}(0)$. Use the properties of the sequence $\psi_n$ in $L(X,\mathbb{R})$ to show that $X$ is not a Banach Space.

To show that $\norm{p}$ is a norm is straight forward. But how can I show that $X$ is not a Banach space by using this definition? Should I use Banach-Saks-steinhaus Theorem?

starry1990
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    In fact, since $X$ has countable dimension, you can use the Baire category theorem to show that no matter what norm you choose, $X$ will not be a Banach space. – Nate Eldredge Nov 22 '15 at 17:13
  • Thanks !! I know this idea, I just don't want to use it here since it requires me to use properties of $\psi_n$ – starry1990 Nov 22 '15 at 17:17

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Choose the sequence {$p_n$};$\space n\ge 1$, of polynomials defined by $$p_n(x)=\frac{x^n}{n^2}+ \frac{x^{n-1}}{(n-1)^2}+\cdot\cdot\cdot+\frac{x^2}{2^2}+x+1$$

You can easily verify it is a Cauchy sequence but its limit is not a polynomial because of infinitely many terms, hence the sequence is not convergent and $X$ is not Banach.

Piquito
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  • "its limit" doesn't really make sense does it? The intuition is there, but you really want to show that for any $p\in X,|p-p_n| \not \to 0.$ – zhw. Nov 22 '15 at 18:34
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    @zhw: If supposing you don't speak neither French nor Spanish then I would wanted see you write in both languages a simple sentence.....This is my case for English. I hope you understand me. – Piquito Nov 23 '15 at 11:15
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You want to use Banach-Steinhaus, and it's a good idea. You have a collection of (obviously linear) functions $\phi_n$. Show that they are continuous and that they are pointwise convergent. Are they uniformly bounded ?

Tlön Uqbar Orbis Tertius
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