Chapter 8: Problem 14
Prove that every subspace of a separable normed vector space is separable.
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Chapter 8: Problem 14
Prove that every subspace of a separable normed vector space is separable.
These are the key concepts you need to understand to accurately answer the question.
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Prove that if \(V\) is a normed vector space, \(f \in V,\) and \(r>0,\) then the open ball \(B(f, r)\) centered at \(f\) with radius \(r\) is convex.
Suppose \(V\) is a real Hilbert space. The complexification of \(V\) is the complex vector space \(V_{\mathrm{C}}\) defined by \(V_{\mathrm{C}}=V \times V,\) but we write a typical element of \(V_{\mathrm{C}}\) as \(f+i g\) instead of \((f, g)\). Addition and scalar multiplication are defined on \(V_{\mathbf{C}}\) by $$ \left(f_{1}+i g_{1}\right)+\left(f_{2}+i g_{2}\right)=\left(f_{1}+f_{2}\right)+i\left(g_{1}+g_{2}\right) $$ and $$ (\alpha+\beta i)(f+i g)=(\alpha f-\beta g)+(\alpha g+\beta f) i $$ for \(f_{1}, f_{2}, f, g_{1}, g_{2}, g \in V\) and \(\alpha, \beta \in \mathbf{R}\). Show that $$ \left\langle f_{1}+i g_{1}, f_{2}+i g_{2}\right\rangle=\left\langle f_{1}, f_{2}\right\rangle+\left\langle g_{1}, g_{2}\right\rangle+\left(\left\langle g_{1}, f_{2}\right\rangle-\left\langle f_{1}, g_{2}\right\rangle\right) i $$ defines an inner product on \(V_{\mathbf{C}}\) that makes \(V_{\mathbf{C}}\) into a complex Hilbert space.
Suppose \(\left\\{f_{k}\right\\}_{k \in \Gamma}\) and \(\left\\{g_{k}\right\\}_{k \in \Gamma}\) are families in a normed vector space such that \(\sum_{k \in \Gamma} f_{k}\) and \(\sum_{k \in \Gamma} g_{k}\) converge. Prove that \(\sum_{k \in \Gamma}\left(f_{k}+g_{k}\right)\) converges and $$ \sum_{k \in \Gamma}\left(f_{k}+g_{k}\right)=\sum_{k \in \Gamma} f_{k}+\sum_{k \in \Gamma} g_{k}. $$
Give an example of a nonempty closed subset \(U\) of the Hilbert space \(\ell^{2}\) and \(a \in \ell^{2}\) such that there does not exist \(b \in U\) with \(\|a-b\|=\) distance \((a, U)\) [By 8.28, \(U\) cannot be a convex subset of \(\left.\ell^{2} .\right]\)
Prove that if \(\mu\) is a measure and \(f, g \in L^{2}(\mu),\) then $$ \|f\|^{2}\|g\|^{2}-|\langle f, g\rangle|^{2}=\frac{1}{2} \iint|f(x) g(y)-g(x) f(y)|^{2} d \mu(y) d \mu(x). $$
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