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Consider the following linear space \begin{equation}P([a,b],\mathbb{K}) = \{p:[a,b] \rightarrow\mathbb{K}: p \text{ is a polynomial} \} \end{equation}

and for $p \in P([a,b],\mathbb{K})$ the following norm is defined:

\begin{equation} ||p|| = \max\{|a_{k}| : k=0,\dots,\deg(p)\} \end{equation} When \begin{equation}\sum^{\deg(P)}_{k=0} a_{k}x^{k} \end{equation} Also the common sup norm is defined $||p||_{\infty} = \sup_{x \in [a,b]}|p(x)|$.

For $[a,b] =[0,1]$ I need to show that these two norms are not equivalent. I've tried to find a function such that $N||p|| \leq ||p||_{\infty} \leq M||p||$ cannot hold but so far I didn't succeed. I was thinking of a sequence of polynomials such that it converges to $0$ for the regular sup norm and $1/n$ for the $||.||$ norm. Can anyone help me out?

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    Is $\mathbb{K}$ either reals or complex, or is it allowed to be something else? – 2017-02-17
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    Yes $\mathbb{K}$ is either real or complex – 2017-02-17

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Consider the sequence $\{p_n\}$ in $P([0,1],\mathbb K)$ defined by $$p_n(x)=\sum_{k=0}^nx^k.$$ Then $\|p_n\|=1$ while $\|p_n\|_\infty=n+1.$

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    "Let's assume the scalar field is $\mathbb{R}$" isn't that a bit too convenient assumption to have in this case? I'm not sure similar example would work in a general field $\mathbb{K}$ – 2017-02-17
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    @user160738, what do you mean by a more general field? In Banach space theory there are either real or complex scalars. – 2017-02-17
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    @user160738 by the stated definition of $P([a,b],\mathbb K)$, the scalar field has to at least contain $\mathbb R$. Can you think of a meaningful field to use other than $\mathbb{R}$ or $\mathbb C$? – 2017-02-17
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    I am not allowed to choose my own domain, $[a,b] =[0,1]$ is given. – 2017-02-17
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    @johnd1992 Sorry I didn't pay close enough attention to that first sentence in the last paragraph. Editted. – 2017-02-17