Is a point evaluation a continuous linear functional on $H^1(0, 1)$?

Question

Like the question title suggests, is the mapping $u \mapsto u(0)$ from $H^1(0, 1)$ into $\mathbb{R}$ necessarily a continuous linear functional on $H^1(0, 1)$?

Answer 1

As $H^1(I)$ is embedded into $C([0,1])$, the mapping is well defined.

The linearity is easy to verify.

As the said embedding is continuous, there exists a constant $c_0$ such that $$|u(0)|\leq \sup_{x\in[0,1]} |u(x)|\leq c_0\|u\|_{H^1},\quad\forall \ u\in H^1(0,1)$$ and thus the mapping is also continuous.

So, the answer is yes.

Remark. The proof that $H^1(I)$ is embedded into $C([0,1])$ is similar to the proof of Fact 1 here. And the proof that the embedding is continuous is similar to the proof of Fact 2 in the same link.

Answer 2

Explicitly, you can observe that $$u(0) = \int_0^1 \left(u(x) - \int_0^x u'(t)\,dt\right)\,dx.$$ Then it's easy to see that the absolute value of the right side is bounded by $\|u\|_{L^2} + \|u'\|_{L^2} \le 2 \|u\|_{H^1}$.

Formally speaking, the identity given above holds for $u \in C^1([0,1])$, which is dense in $H^1$. So although strictly speaking $u(0)$ is not defined for general $u \in H^1((0,1))$, the right side of the identity above defines a continuous linear functional which extends the map $u \mapsto u(0)$. Therefore this functional may as well be taken as the definition of $u(0)$.