How would I show a functional is linear and bounded?

Question

Using the following results, for any $f \in H^1(a,b)$, $f$ is continuous on $[a,b]$, and therefore, $$ \int_a^b f(x) dx = f(\zeta)$$ for some $\zeta \in (a,b)$. In addition $$f(c) = f(\zeta) + \int_\zeta^c f'(x)dx.$$ How would I go about showing the functional defined as $\ell \in (H^1 (a,b))'$ by $$ \ell(v) = v(c), v \in H^1 (a,b)$$ for some $c \in [a,b]$, is linear and bounded on $H^1(a,b)$?

Answer 1

You have all the ingredients ready:

To show that $\ell$ is bounded, we need $|\ell(f)| \le C||f||_{H^1}$ for some $C$. Now there is $\zeta$ (depending on $f$) so that

$$\ell(f) = f(c) = f(\zeta) + \int_\zeta^c f'(x)dx = \int_a^b f(x) dx + \int_\zeta^c f'(x)dx$$

Then

$$|\ell (f)| \le \int_a^b |f(x) |dx + \int_\zeta^c |f'(x)| dx \le \cdots$$