Dual space of product of Hilbert spaces What can we say about the dual space of the product of Hilbert spaces? Suppose $V = A \times B$ and the inner product of two vectors in this space is the obvious one (add the two separate inner products). Then is $V^* = A^*\times B^*$ and is $\langle f, u \rangle = \langle f_1, u_1 \rangle + \langle f_2, u_2\rangle$?
 A: Consider $\phi \in V^*$. It means that $\phi: A \times B \to \mathbb{F}$ is a continuous linear functional. As $A \times B \simeq A \oplus B$, we have inclusions $i_A: A \to A \times B, i_B: B \to A \times B$, and composing them with $\phi$ give us two continuous linear functionals $\phi \circ i_A: A \to \mathbb{F}, \phi \circ i_B: B \to \mathbb{F}$. Conversely, given two continuous linear functionals $f: A \to \mathbb{F}, g: B \to \mathbb{F}$, by universal property of a direct sum, these two give us a map $f \oplus g: A \times B \to \mathbb{F}$. Thus every $\phi \in V^*$ gives us a pair of functionals $f \in A^*, g \in B^*$, and every such pair gives us a functional in $V^*$. It's not hard to check that this actually gives us a canonical isomorphism $(A \times B)^* \simeq A^* \times B^*$
In down-to-earth terms, for $u = (u_a, u_b) \in A \times B$, we have $u = (u_a, 0) + (0, u_b)$ (in the earlier terminology, $u = i_A(u_a) + i_B(u_b)$). As $\phi$ is linear, $\phi(u) = \phi((u_a, 0)) + \phi((0, u_b))$. If we define $f: A \to \mathbb{F}$ as $f(a) = \phi((a, 0))$ (in earlier terminology $f = \phi \circ i_A$), and similarly $g: B \to \mathbb{F}$, we see that $\phi(u) = f(u_a) + g(u_b)$. Thus mapping $\phi \mapsto (f, g) = (\phi \circ i_A, \phi \circ i_B)$ gives us map $(A \times B)^* \to A^* \times B^*$. You can easily check that it's linear isomorphism.
