closed subspace in $W[0,1]$

Question

Let $W[0,1]$ be the space of all continuously differentiable functions on $[0,1]$ with values in $\mathbb{C}$, with the following inner product

$(f,g)_W =\int_0^1 {f(t)\overline{g(t)}+f'(t)\overline{g'(t)}}dt$.

We need to prove that $(f,\cosh)_W=f(1)\sinh1$ and then use this fact to prove that the subspace $\{f \in W[0,1]: f(1)=0 \}$ is $\textbf{closed}$.

I was able to prove the identity, but I couldn't find how can I apply it to prove the subspace is closed.

I know it is closed if it contains all its limits points. But couldn't start the proof.

Answer 1

Hint: rewriting the identity you proved, we have

$$ f(1) = \left(f,\frac{\cosh}{\sinh(1)}\right)_W, $$ and in general for any $u$ we know that $(-,u)_W$ is a linear operator, continuous with respect to the topology induced by $(-,-)_W$ (in fact, Riesz's representation theorem tells us all continuous functionals are of this form).

The subspace of interest is $\ker(ev_1)$; hence its closedness is equivalent to the continuity of $ev_1$, as the latter is a functional on $W[0,1]$. But this stems from the fact that $ev_1$ is of the form $(-,u)_W$ with $u = \frac{\cosh}{\sinh(1)}$.