Let $E = \mathbb{R} [X]$ equipped with the norm $||p|| = \int_0^1 (|p(t)| + |p'(t)|) \ d t $.
Check if the functional $\psi : E \ni p \rightarrow p(0) \in \mathbb{R}$ is continuous, and if it is, find its norm.
$p,q \in E, \ ||p-q||= \int_0^1 (|p(t)-q(t)| + |p'(t) - q'(t)|) \ dt = $
$\int_0^1 (|p(t)-q(t)|) dt + \int_0^1 (|p'(t) - q'(t)|) \ dt = $
$\int_0^1 (|p(t)-q(t)|) dt +|p(1)-q(1)| - |p(0) + q(0)|$
And $||\psi (p) - \psi (q)|| = |p(0) - q(0)|$
Could you tell me if it is correct and what I can do next?
Thank you!
$\psi$ is linear, so we need only check that $\psi$ is bounded.
$|p(0)| = |\int_0^1 p(0) dx| = |\int_0^1 (p(0)-p(x) + p(x)) dx| \le \int_0^1 (|p(0)-p(x)| + |p(x)| ) dx$.
Note that $p(0)-p(x) = \int_0^x -p'(t) dt$ (this is the key part of this solution), hence $|p(0)-p(x)| \le \int_0^1 |p'(t)| dt$.
Then $|p(0)| \le \int_0^1 (\int_0^1 |p'(t)| dt ) + |p(x)| ) dx = \|p\|$. Hence $\|\psi\| \le 1$.
If we let $p(x) = 1$, we have $\phi(p) = 1$ and $\|p\| = 1$, hence $\| \psi \| = 1$.