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 $L^2$.
Show that $S^{\perp}$ is a vector subspace of $L^2$
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linear-algebra
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0alvoutila, you messaged me that you have many questions in your "box." What do you mean by "box" ? – 2013-01-24
1 Answers
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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.