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 $\langle f, u \rangle = \langle f_1, u_1 \rangle + \langle f_2, u_2\rangle$?


1 Answer 1


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.

  • $\begingroup$ Thanks, what do you mean by $A \otimes B$? Direct sum? But we may not be able to add elements of $A$ and $B$ together; they can be different spaces completely. But I agree every functional in $V^*$ corresponds to two elements, one in $A^*$ and one in $B^*.$ But I still can't see why it must be true that $\langle f, u \rangle_{V^*,V} = \langle f_1, u_1 \rangle_{A^*,A} + \langle f_2, u_2\rangle_{B^*,B}$? $\endgroup$
    – soup
    Apr 20, 2013 at 17:24
  • 1
    $\begingroup$ By $A \oplus B$ I mean the coproduct in the category of Banach spaces. You may have encountered it under the name "external direct sum". For finite sums of Banach spaces, the notions of direct product $A \times B$, and (external) direct sum $A \oplus B$ coincide, but in other categories (for instance in the category of groups), the direct product of groups is something different than the coproduct (direct sum) of groups. $\endgroup$
    – xyzzyz
    Apr 20, 2013 at 18:44
  • $\begingroup$ @xyzzyz How to show that the isomorphism is isometric? $\endgroup$
    – aere
    May 22, 2013 at 13:08
  • $\begingroup$ @aere: how do you define the norm in $A \times B$ and $A \oplus B$? $\endgroup$
    – xyzzyz
    May 24, 2013 at 11:05
  • $\begingroup$ @xyzzyz I would say $\lVert (a,b)\rVert^2 := \lVert a \rVert^2_{A} + \lVert b \rVert^2_B$. $\endgroup$
    – aere
    May 24, 2013 at 12:26

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.