# Orthogonal complement to continuous function

Question: Suppose $$C[−1, 1]$$ is the vector space of continuous real-valued functions on the interval $$[−1, 1]$$ with inner product given by $$\langle f, g\rangle = \int_{a}^b f(x)g(x)dx$$

Let $$U = {f ∈ C[−1, 1] : f(0) = 0}$$ be the subspace of $$C[−1, 1]$$. Which of the following statement(s) is(are) correct?Justify your answer.

(a) $$C[−1, 1] = U ⊕ U^\bot$$

(b) $$U^\bot = \{0\}$$

(c) $$U^\bot$$ is a proper and non-trivial subspace of $$C[−1, 1]$$

Difficulty: I am sure that option (b) is correct but not able to write a proof of it. Also about option (a) is valid for finite-dimensional subspace but it is not am I correct about it.

• Welcome to MSE. In order to get $\{x\}$, you should type \{x\}. Sep 29 '20 at 7:38

Hint: If $$g \in U^{\perp}$$ then $$\int fg=0$$ for all $$f \in U$$. Let $$f_n(x)=g(x)$$ for $$|x| >\frac1 n$$, $$f_n(0)=0$$ and $$f_n$$ have a straight line graph in $$[-\frac 1 n ,0]$$ as well as $$[0, \frac 1 n]$$. Then $$\int g f_n=0$$ and letting $$n \to \infty$$ yields $$\int g^{2}=0$$. Hence $$g=0$$. This proved b).
It is obvious from (b) that (a) is false. The constant function $$1$$ is in LHS but not in RHS. (c) is also answered by (b).