I am struggling with this seemingly easy problem.
Let $E = \mathbb{R}[X]$ equipped with the following norm: $||P||_\infty := \sup_{x \in [0, 1]} | P(x) |$ and define $\delta_a : E \rightarrow \mathbb{R}$ the linear transformation defined by: $\forall P \in E, \delta_a(P) = P(a)$. This transformation gives nothing but the value of $P$ at point $a$. For which $a \in \mathbb{R}$ is $\delta_a$ continuous? For such $a$, compute the norm of $\delta_a$.
From the definition given in my lessons, the transformation is continuous if there exists a $c > 0$ such that $$ \forall P \in E, ||\delta_a(P)||_{\mathbb{R}} \leq c ||P||_\infty \iff |P(a)| \leq c \sup_{x \in [0, 1]} |P(x)| $$
I have no idea how to find such a constant. The case $a \in [0, 1]$ obviously works, but since the norm we're provided with is only defined on this interval, what can we say about other values of $a$ ?
When $a \in [0, 1]$, I don't know how to work out a more explicit expression for the norm of the operator, other than the definition of the operator norm:
$$|||\delta_a||| = \sup_{P \neq 0} \frac{|\delta_a(P)|}{||P||_\infty} = \sup_{P \neq 0} \frac{|P(a)|}{\sup_{x \in [0, 1]}|P(x)|} $$
Any help or guidance is welcome…
According to tfp , it remains to consider the case when $a\in [-1,0)$.
$\textbf{Proposition}$. $\delta_{-1}$ is not continuous.
$\textbf{Proof}$. For every integer $p>0$, let $y_p=x^{2p}-x^{2p-1}$.
$y_p'=x^{2p-2}(2px-2p+1)$.
$\min_{[0,1]}(y_p)<0$ is obtained for $x_p=1-\dfrac{1}{2p}$ and is equal to $m_p=(1-\dfrac{1}{2p})^{2p}-(1-\dfrac{1}{2p})^{2p-1}$. When $p\rightarrow +\infty$, $m_p\sim \dfrac{-1}{2pe}$.
Finally, when $p$ is large,
$\dfrac{|y_p(-1)|}{|m_p|}\sim 4pe$ and we are done.
EDIT. In fact, it is easier than I thought.
For every $a<0$, $\delta_a$ is not continuous.
It suffices to consider $y_p=(2x-1)^p$.
indeed $||y_p||=1$ and $y_p(a)=(2a-1)^p$ with $|2a-1|>1$.