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Let $X,Y$ be Banach spaces over $\mathbb{R}$ and $B:X\times Y\rightarrow \mathbb{R}$ be a continuous bilinear function such that $X\rightarrow L(Y,\mathbb{R}):x\mapsto B(x,-)$ and $Y\rightarrow L(X,\mathbb{R}):y\mapsto B(-,y)$ are injective.

Let $T:X\rightarrow X$ be a continuous linear transformation. Then, does there exist a continuous linear transformation $T^*:Y\rightarrow Y$ such that $B(Tx,y)=B(x,T^*y)$ for all $(x,y)\in X\times Y$?

Rubertos
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  • I think the answer is yes for Hilbert spaces (using Riesz's representation theorem) and no for general Banach spaces, but I don't have a reference. – Daniel Robert-Nicoud May 25 '17 at 11:28

2 Answers2

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In general, such a $T^{\ast}$ doesn't exist. For an example, let $Y$ a non-reflexive Banach space, $X$ its dual, and $B$ the evaluation map. Let $\eta \in Y^{\ast\ast}\setminus Y$ and $x_1 \in X \setminus \{0\}$. Define $T(x) = \eta(x)\cdot x_1$. Then the Banach space adjoint (or transpose) of $T$ is ${}^tT \colon X^{\ast} \to X^{\ast}$ given by ${}^tT(\xi) = \xi \circ T \colon x \mapsto \eta(x)\cdot \xi(x_1)$, i.e. ${}^tT(\xi) = \xi(x_1)\cdot \eta$. If $T^{\ast}$ existed, it would be the restriction of ${}^tT$ to $Y$, but for $y \notin \ker x_1$, we have ${}^tT(y) = x_1(y)\cdot \eta \notin Y$.

Generally, as vector spaces, we can identify $Y$ with a subspace of $X^{\ast}$, and have $B$ the restriction of the evaluation map to $X\times Y$. Then the existence of $T^{\ast} \colon Y \to Y$ satisfying $B(Tx,y) = B(x,T^{\ast}y)$ for all $x,y$ means that ${}^tT(Y) \subset Y$, where ${}^tT$ is the transpose of $T$. That is the case if and only if $T$ is continuous with respect to $\sigma(X,Y)$. In that case, the closed graph theorem yields the continuity of $T^{\ast} = {}^tT\lvert_Y$ with respect to the chosen norm on $Y$ (which generally isn't equivalent to the norm induced by $X^{\ast}$).

Daniel Fischer
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  • Apologies for me being a bit late to the party, but could you please explain (or give a reference for) why $^tT(Y)\subset Y$ is equivalent to the continuity of $T$ w.r.t. $\sigma(X,Y)$? – skewfield Aug 11 '22 at 18:46
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Just came across this old post. This is an excellent question and I came across it in my own research!

The answer is yes. The generalization is known as Singer's representation theorem, which was originally published in Russian in 1957.

I. Singer, "Linear functionals on the space of continuous mappings of a compact Hausdorff space into a Banach space", Rev. Math. Pures Appl. 2 (1957), pp. 301--315.

Just like the Riesz representation theorem, the continuity (more generally, the Hausdorff condition) and a boundedness assumption ensure the existence of the representation. An inf-sup condition ensures the uniqueness of the representation.

See e.g. the following papers in English for its proof and its further generalizations.