# Show that $S^{\perp}$ is a vector subspace of $L^2$

Let $S \subset L^2(\mathbb{R})$, and let us define $S^{\perp} =\{f \in L^2(\mathbb R) | \left< f,g\right> =0, g \in S\}$. Show that $S^{\perp}$ is a vector subspace of $L^2$.

-
@alvoutila: i think your definition of $S^{\perp}$ is missing something. –  anonymous Oct 22 '10 at 8:02
what is it missing? –  laovultai Oct 22 '10 at 8:20
I think Chandru1 means you didn't define your inner product. It should be $S^{\perp} = \{f \in L^2(\mathbb{R}) : \int_\mathbb{R} f \bar{g} = 0 \forall g \in S \}$ right? Do you know what you are supposed to show and is there a specific part that you are having trouble with, or do you just not understand what you need to do? –  user1736 Oct 22 '10 at 8:56
And even more to the point: If $(V,\langle\cdot,\cdot\rangle)$ is any inner product space and $S \subset V$, then the set of vectors that are orthogonal to everything in $S$ is obviously a subspace of $V$ since the inner product is sesquilinear. –  kahen Oct 22 '10 at 9:23
@user1736: when one speaks of $L^2$, the inner product is implicit. No one sanely uses another inner product on that space... –  Mariano Suárez-Alvarez Oct 22 '10 at 12:53

I think you mean, show that $S^\bot = \lbrace f \in L^2| \int_\mathbb{R} fg dx = 0 , \forall g \in S \rbrace$ is a vector subspace of $L^2$.
Let $f, g \in S^\bot$ and $\alpha, \beta \in \mathbb{R}$, then for $h \in S$, $\int_\mathbb{R} (\alpha f + \beta g)h dx = \alpha\int_\mathbb{R}fh dx + \beta\int_\mathbb{R}gh dx = \alpha 0 + \beta 0 = 0$.
Therefore $\alpha f + \beta g \in S^\top$, and so $S^\bot$ is a vector subspace.