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I been trying to figure out if the construction of the isomorphism between $g(W)$ and $f^{*}(W^{*})$ via $$B_V(v, g(w)) = B_W(f(v),w) \quad \forall v \in V, w \in W $$ described (at least this is what I think they mean) in the wikipedia page , is meaningfull in a Banach space which is not Hilbert. One reason it wouldn't be the lack of existancee of non-degenrate bilinear forms but this is something which I can't manage to establish. Or maybe there is some other reason?

Peter Calvo
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user123124
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    I do not see any question in this question. – gerw Oct 25 '15 at 18:31
  • There is no need for the space to be Hilbert : you are adding this hypothesis from nowhere. Note: In the infinite case, an Hilbert space is not the same as a Banach space whose norm would come from a non degenerate product. – MikeTeX Oct 25 '15 at 19:14
  • The wiki article is about matrices. A continuous linear operator on a real Banach space may or may not be representable as a matrix $T$ .If it can, I don't know off-hand whether the transpose operator $T(x)$ is definable at every $x$ in the space. There is a class called $B$ algebras (and an important sub-class $C*$ algebras, which have "adjoints" which act like transposes in some ways,although bilinear forms (inner products) may not be available. This is a broad and deep subject. – DanielWainfleet Oct 25 '15 at 19:35
  • @MikeTex thats my point if the space is hilbert then evertying works out fine. But I think there might be Banach with non-degenerate biliear forms, and I wounder if there are adjonts in such spaces. – user123124 Oct 27 '15 at 08:12
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    @User2313: The concept you seem searching for goes under the name of Dual Topology for a Dual Pair. – C-star-W-star Nov 02 '15 at 17:42

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