Closed subspace of the $C[-1, 1]$ with inner product $(f, g) = \int \limits_{-1}^1 f(t) \bar g(t) \operatorname{d}t$.

Question

Assume linear space $H = C[-1, 1]$ with inner product $(f, g) = \int \limits_{-1}^1 f(t) \bar g(t) \operatorname{d}t$. I need to prove that subspace $H_0 = \{ f \in H \mid \int \limits_{-1}^0f(t)dt = \int \limits_0^1 f(t) dt \}$ is closed.

I know that is sufficient to show that for an arbitrary sequence $f_n$, converging to $f$, $f \in H_0$. But I can't do anything with condition $||f_n - f|| = (f -f_n, f - f_n) < \epsilon$ to deduce conditon on $H_0$.

Any hints? Thanks!

Answer 1

Consider the maps $F_1,F_2\colon C\bigl([-1,1]\bigr)\longrightarrow\mathbb C$ defined by$$F_1(f)=\int_{-1}^0f(t)\,\mathrm dt\text{ and by }F_2(f)=\int_0^1f(t)\,\mathrm dt.$$Prove that both of them are continuous. Then $F_1-F_2$ is continuous and $H_0=(F_1-F_2)^{-1}\bigl(\{0\}\bigr)$.