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$L_d = \{ \phi \in L_{[-a,a]}^2$ $\vert$ $\phi(t) = -\phi(-t)$ $\forall t \in [-a,a] \}$, prove that is a closed subspace of $L^2_{[-a,a]}$.

Now, the subspace part is always the same and I managed.

The part on closeness is where I stumble. This is what I did:

I want to prove it using closeness by sequences, so:

Let $\{\phi_n\} \subset L_d$ be a sequence, such that: $\phi_n \to \phi$ in $L^2$. I want to prove that $\phi \in L_d$.

I know that $\phi_n \to \phi$ in $L^2$ implies that there exists an extracted subsequence $\phi_{n_k}$ such that $\phi_{n_k} \to \phi'$ almost everywhere. I also know that $\phi' = \phi$ almost everywhere for uniqueness of the limit. But still, I do not know how to say that $\phi' = \phi \in L^d$. I feel like I am just missing this last step but I am not sure on how to proceed.

qcc101
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1 Answers1

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Let $ \phi^-(t) = \phi(-t). $ We want to show that $\phi+\phi^- = 0$. We have $$ \Vert \phi+\phi^- \Vert \leq \Vert \phi-\phi_n \Vert + \Vert \phi_n+\phi_n^- \Vert + \Vert \phi^--\phi_n^- \Vert. $$ The middle term is 0 since $\phi_n \in L_d$, and the other two terms converge to 0 by assumption.

MSDG
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