Prove that the space of square integrable vector valued functions is separable

Question

I'm trying to solve problem 12 of chapter 2 from Volume I of the book series Methods of modern mathematical physics. The problem goes as follow:

We say that a vector valued function $f$ from a measure space $(X,\mu)$ to a separable Hilbert space $\mathcal{H}$ is measurable, if $\langle \vec{y},f(x) \rangle_{\mathcal{H}}$ is measurable for each $\vec{y}\in\mathcal{H}$.

a) Show that if $f$ and $g$ are measurable vector valued functions, then $\| f(x) \|^2_{\mathcal{H}}$ and $\langle f(x),g(x) \rangle_{\mathcal{H}}$ are measurable.

b) Let $\{ \phi_k \}_{k=1}^{\infty}$ be a basis for $\mathcal{H}$. Prove that if $g\in L_{2}(X,\mu; \mathcal{H})$, then $$\sum_{k=1}^{N} \langle \phi_k,g(x) \rangle_{\mathcal{H}}\phi_k\rightarrow g,$$ and if $f\in L_{2}(X,\mu; \mathcal{H})$, then $$(f,g)=\sum_{k=1}^{\infty}\int_{X}\langle f(x),\phi_k \rangle_{\mathcal{H}} \langle \phi_k,g(x) \rangle_{\mathcal{H}}d\mu(x).$$

c) Assume that $L_2(X,\mu)$ is separable and prove that $L_{2}(X,\mu; \mathcal{H})$ is separable.

I have proved part a) and part b) but I'm having problems with part c).

By part b), I can see that $\sum_{k=1}^{N}\langle \phi_k,g(x) \rangle_{\mathcal{H}}\phi_k$ can approximate $g$. However, I don't know how use this to deduce or construct a countable orthonormal basis of $L_{2}(X,\mu; \mathcal{H})$. Also, I don't know if it would be easier to find an arbitrary dense countable subset of $L_{2}(X,\mu; \mathcal{H})$, not necessarily a basis. If $L_2(X,\mu)$, how can I use this to find a dense countable subset of $L_{2}(X,\mu; \mathcal{H})$? I just need a suggestion or hint please.

Correction: Updated chapter number. I apologize for the mistake.

Answer 1

The form $(f,g)$ is the inner product in $L^2(X,\mu;\mathcal{H}).$ Let $\{\psi_l\}_{l=1}^\infty$ be an orthonormal basis of $L^2(X,\mu).$ Then $\psi_l(x)\phi_k\in L^2(X,\mu;\mathcal{H}).$ The collection $\psi_l(x)\phi_k$ is linearly dense in $L^2(X,\mu;\mathcal{H}).$ Indeed, assume $(f,\psi_l(x)\phi_k)=0$ for all $k,l. $ Then $$0= \int\limits_X\langle f(x),\phi_k\rangle_\mathcal{H}\,\overline{\psi_l(x)}\,d\mu(x)$$ Hence $$\langle f(x),\phi_k\rangle_\mathcal{H} =0,\ k\in \mathbb{N}$$ Thus $f(x)=0,$ $\mu$ a.e.

Answer 2

Let $g_k : x \mapsto \langle \phi_k , g(x)\rangle_\mathcal{H}$. The $g_k$s belong to $L^2(X,\mu)$, and you've shown in $b)$ that $\left(x \mapsto \sum_{k = 1}^N g_k(x) \phi_k\right)_N$ converges to $g$ in $L^2(X, \mu; \mathcal{H})$.
Now consider an orthonormal basis $(\psi_l)_{l \geq 1}$ of $L^2(X,\mu)$, so that $g_k = \sum_{l = 1}^\infty \langle g_k, \psi_l\rangle_{L^2(X,\mu)} \psi_l$ in $L^2(X,\mu)$.
These two convergence results combined with the density of $\mathbb{Q}(i) := \mathbb{Q} + i\mathbb{Q}$ in $\mathbb{C}$ then yield that the countable set $$E := \left\{x \mapsto \sum_{k = 1}^N \sum_{l = 1}^M \alpha_{k,l}\psi_l(x) \phi_k\mid \alpha_{k,l} \in \mathbb{Q}(i), N \geq 1, M \geq 1\right\}$$ will be dense in $L^2(X, \mu; \mathcal{H})$, and $L^2(X, \mu; \mathcal{H})$ is separable.
You can also show that $(\psi_l \phi_k)_{k,l}$ is an orthonormal family of $L^2(X, \mu; \mathcal{H})$ with a somewhat quick calculation using the second part of $b)$.

Answer 3

Here is yet another approach. Let $\{ \phi_{k} : k\in\mathbb{N} \}$ be an orthonormal basis for $\mathcal{H}$. You can use part (b) to immediately conclude that \begin{equation} (\| \langle f(\cdot ), \phi_{k} \rangle_{\mathcal{H}} \|_{L_{2}(X)} )_{k\in\mathbb{N}} \in \ell_{2} \; \text{and} \; \|f\|_{L_{2}(X; \mathcal{H})}^{2} = \sum_{k\in\mathbb{N}} \|\langle f(\cdot ), \phi_{k} \rangle_{\mathcal{H}} \|_{L_{2}(X)}^{2} \tag{$*$} \end{equation} for every $f\in L_{2}(X, \mu ; \mathcal{H})$. Define $\Phi \colon L_{2}(X, \mu ; \mathcal{H}) \to \oplus_{k\in\mathbb{N}} L_{2}(X, \mu )$ by \begin{equation} \Phi f := (\langle f(\cdot ), \phi_{k} \rangle_{\mathcal{H}} )_{k\in\mathbb{N}}. \end{equation} By ($*$) the map $\Phi$ is a well-defined isometry. Hence $L_{2}(X, \mu ; \mathcal{H})$ is isometrically embedded into the Hilbert space $\oplus_{k\in\mathbb{N}} L_{2}(X, \mu )$, the latter of which is separable as it is a countable (Hilbert space) direct sum of separable Hilbert spaces. Therefore $L_{2}(X, \mu ; \mathcal{H})$ is separable as desired. (Note. With some more work it can be shown that $\Phi$ is a unitary operator.)

Answer 4

This follows easily from a general result on the tensor product of Hilbert spaces (for a proof see here):

Proposition: Let $\mathcal{H}_{1}, \mathcal{H}_{2}$ be Hilbert spaces and $M_{i} \subset \mathcal{H}_{i}$ with $\overline{ \operatorname{span} M_{i}} = \mathcal{H}_{i}$ for $i \in \{1,2\}$. Then $\overline{ \operatorname{span} \{ \varphi \otimes \psi : \varphi \in M_{1} , \psi \in M_{2} \} } = \mathcal{H}_{1} \hat{ \otimes} \mathcal{H}_{2} $.

Here $\mathcal{H}_{1} \hat{ \otimes} \mathcal{H}_{2} $ denotes the Hilbert space tensor product.

$L^2 (X,\mu, \mathcal{H}) = L^2(X,\mu) \hat \otimes \mathcal{H}$. Let $M_1$ be any countable dense subset of $L^2(X,\mu)$ and $M_2$ any countable dense subset of $\mathcal{H}$. Clearly we can approximate any element in $$\operatorname{span} \{ \varphi \otimes \psi : \varphi \in M_{1} ,\psi \in M_{2} \}$$ by an element of $$\operatorname{span}_{\mathbb{Q}+i\mathbb{Q}} \{ \varphi \otimes \psi : \varphi \in M_{1} , \psi \in M_{2} \} $$ (linear combinations with scalars that have only rational real and imaginary parts). This latter set is again countable, since $\mathbb{Q}$ and $M_1, M_2$ are. By the above proposition we conclude that $$ \overline{ \operatorname{span}_{\mathbb{Q}+i\mathbb{Q}} \{ \varphi \otimes \psi : \varphi \in M_{1} , \psi \in M_{2} \}} = L^2(X,\mu,\mathcal{H})$$ and so $L^2(X,\mu,\mathcal{H})$ has a dense countable subset. Actually the argument works for the tensor product of any two separable Hilbert spaces.