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In $\mathbb C[X]$, we consider the norm $\left\lVert P \right\rVert = \sup \left|a_i\right|$ for $P(X) = \sum_{i=1}^na_ix^i$. For all $x_0$ we consider the linear form $\phi : \mathbb C[X] \to \mathbb C, P \mapsto P(x_0)$. Determine all the $x_0$ such that $\phi$ is continuous and calculate its norm.

Here's what I did : $\phi$ is continuous in $x_0$ if and only if it is $k$-Lipschitz in $x_0$, which is equivalent to say that there exists an $M$ such that

$$ \left\lvert P(x_0) \right\rvert \le M \left\lVert P \right\rVert, \forall P \in \mathbb C [X]$$

We have, $$\left| P(x_0) \right| = \left| \sum_{i=1}^n a_i x^i\right| \le \sum_{i=1}^n|a_i||x_0^n| \le M_n\sup|a_i| = M_n\left\lVert P \right\rVert$$, where $M_n$ is a constant that depends on the degree of $P$.

By contradiction I could show that $\phi$ is continuous only for $x_0 = 0$ but I'm pretty sure I did something wrong, because I'm also supposed to find the norm of $\phi$ which is not defined for $x_0 = 0$. Any thoughts?

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I don't understand why your polynomials do not include a constant term $a_0$. I included it below.

Let's first recognize that bounding $P(x_0)$ by $M_n\|P\|$ where "$M_n$ is a constant that depends on the degree of $P$" is not specific enough for conclusion either way. Just saying "depends on the degree" does not mean it cannot be bounded independently of the degree. For example, $\frac{2n}{n+1}$ depends on $n$, but we can bound it by $2$ just fine.

Let's also recognize that $|x_0|^i\le |x_0|^n$ is not always true.

Here's my version:
$$|P(x_0)|\le \sum_{i=0}^n|a_i||x_0|^i \le \sup|a_i| \sum_{i=0}^n |x_0|^i = \|P\| \sum_{i=0}^n |x_0|^i $$ Do you see the geometric series here? If $|x_0|<1$, the sum can be bounded independently of $n$: $$\sum_{i=0}^n |x_0|^i \le \sum_{i=0}^\infty |x_0|^i =\frac{1}{1-|x_0|}$$

There's your $M$. The number $\frac{1}{1-|x_0|}$ is indeed the norm, which you can demonstrate with appropriate choices of polynomials $P$. Hint: Consider $P(x)=\sum_{i=0}^n x^i$ if $x_0>0$ and $P(x)=\sum_{i=0}^n (-1)^i x^i$ if $x_0<0$.

If $|x_0|\ge 1$, the functional is unbounded. Use the polynomials in the hint.

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