Consider the inner product space of continuously differentiable functions $C^1 [0,1]$ with the inner product: $$<f,g> = \int^1_0 f(x) \overline{g(x)} dx + \int^1_0 f'(x) \overline{g'(x)} dx$$
In part one of the question we are asked to show that $<f,\cosh(x)> = f(1)\sinh(1)$ for any $f \in C^1[0,1]$ . I have done this.
In part 2 we are asked to use this to show that the subspace $\{f \in C^1[0,1]: f(1) = 0\}$ is a closed subspace of $C^1[0,1]$. This is where I really need help.
I want to show the compliment of the subspace, call it say $A^c$ is open. That is $A^c = \{f \in C^[0,1]: f(1) \neq 0\}$ is open. I believe this means that for every $f \in A^c$ there exists an $\epsilon > 0$ such that $B_{\epsilon}(f) = \{g \in C^1 [0,1]: <f-g,f-g> < \epsilon\}$
But I firstly dont know if I am going about this in the right way? If so can I have some hints for the next steps? I have not tried to prove something like this before in terms of functions. Also I don't know where the relevance of part a) comes in. Are we evaluating in terms of $f(1)g(1)$ ? Thanks