1. Prove of closed subspace

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

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

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

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

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

Correct?

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

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

Correct?

But how do I show $M$ 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?

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

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

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

...

But how do I show $M$ is a closed subspace? Is it enough to to state it is a continuous function defined on a closed and limited interval thus attaining both its minimum and maximum value?
Hint: To show that M is closed, use part (1), which tells you that $f(1)=0\ \Longleftrightarrow\ (f,\cosh)_W = 0$.