Given $\displaystyle W[0,1]:\mathbb{R} \rightarrow \mathbb{C}$ defined as the space of continuously differentiable functions with inner product $\displaystyle (f,g)_{W}=\int^{1}_{0} f(t)\overline{g(t)}+f'(t)\overline{g'(t)} \ dt$.

Let $\displaystyle f \in W[0,1]$ and show:

- $\displaystyle (f,\cosh) = f(1)\sinh(1)$
- $\displaystyle M=\{f \in W[0,1]:f(1)=0\}$ is a closed subspace of $\displaystyle W[0,1]$

I am somewhat uncertain of my solution, thus asking here...

$\displaystyle (f,cosh)$

$\displaystyle = \int^{1}_{0} f(t) \overline{\cosh(t)}+f'(t)\overline{\cosh'(t)} \ dt $

$\displaystyle = [F(t)\cosh(t)]^{1}_{0} - \int^{1}_{0}F(t)\overline{\cosh'(t)} \ dt + [f(t)\overline{\cosh'(t)}]^{1}_{0} - \int^{1}_{0} f(t) \overline{\cosh''(t)} \ dt $

$\displaystyle =[f(t)\overline{\cosh'(t)}]^{1}_{0} $

$\displaystyle =f(1)\sinh(1).$

Correct?

Further...

$\displaystyle f(t)=0$ for $\displaystyle t \in [0,1]$ belongs to $\displaystyle M$ thus $\displaystyle M$ non-empty.

Assume $\displaystyle f,g \in M$ and $\displaystyle c \in \mathbb{C}$. Then $\displaystyle (f+g)(1) = f(1)+g(1) = 0$. Thus $\displaystyle (f+g) \in M$ and $\displaystyle M$ is closed under addition. Also $\displaystyle (f+cg)(1) = f(1)+(cg)(1) = f(1)+c(g)(1) = 0$. Thus $\displaystyle (f+cg) \in M$ and $\displaystyle M$ closed under scalar multiplication.

Correct?

But how do I show $\displaystyle M$ is a closed subspace? Is it enough to to state it is a continous function defined on a closed and limited interval thus attaining both its minimum and maximum value?