This is the question, but I don't have a clue to solve the second part.
Prove that in $W[0, 1]$ $$<f, \cosh > = f(1) \sinh 1 $$ and deduce that
$$ \{f \in W[0; 1] | \, f(1) = 0\} $$ is a closed subspace of $W[0, 1]$.
And, the inner product is: $$\langle f,g \rangle = \int_{a}^{b} f(t)g(t)+ \acute{f}(t)\acute{g}(t) dt$$
Anyone, got any idea?
Here's a start.
If $\langle f,g \rangle = \int_{a}^{b} (f(t)g(t)+ f'(t)g'(t)) dt $ then
$\begin{array}\\ \langle f,\cosh \rangle &= \int_{a}^{b} (f(t)\cosh(t)+ f'(t)\cosh'(t)) dt\\ &= \int_{a}^{b} (f(t)\cosh(t)+ f'(t)\sinh(t)) dt\\ &= \frac12\int_{a}^{b} (f(t)(e^t+e^{-t})+ f'(t)(e^t-e^{-t})) dt \qquad\text{(don't know if you need these lines)}\\ &= \frac12\left(\int_{a}^{b} e^t(f(t)+f'(t)) dt+\int_{a}^{b} e^{-t}(f(t)-f'(t)) dt\right)\\ \end{array} $