1. Apr 3, 2008

jostpuur

If X and Y are norm spaces, T:X->Y is a continuous linear mapping, and T(X) is closed in Y, is T*(Y*) always closed in X*? Here X* and Y* are dual spaces, and T*:Y*->X* is the adjoint of T.

2. Apr 4, 2008

Peeter

This question is interesting to me, not because I can answer it, which I can't, but because it's by far the most abstract definition I've seen of the adjoint.

The second most, from "Geometric Algebra for Physicists" was the following (roughly). For a linear transformation F : V -> V, in a product space that can produce a scalar component, the adjoint:

$$\bar{F}$$

is defined implicitly by:

$$\langle A F(B) \rangle = \langle \bar{F}(A) B \rangle$$

where $$\langle \cdots \rangle$$ selects the scalar component of the product.

From that definition, for a real vector space and the scalar component defined by an inner product, I can see how this is consistent with the matrix transpose, if one introduces a basis and reciprocal frame vectors, and expand the linear transformation in terms of components for this basis.

However, it appears to me that your adjoint definition is formulated in a way that allows for it to apply to infinite dimensional spaces. If that's the case, to calculate the adjoint from this definition couldn't use the basis method above (at least I'm not sure how one would calculate the reciprocal frame vectors).

From your definition how one would calculate an adjoint for a given linear mapping. Would you mind giving an example of a mapping T and it's adjoint that fits your definition?

3. Apr 4, 2008

jostpuur

I have in fact encountered two somewhat different definition for the adjoint. This is the other one. When X and Y are norm spaces, we set

$$X^* := \{x^*:X\to\mathbb{C}\;|\; x^*\;\textrm{is continuous and linear}\}$$

and similarly Y*. When T:X->Y is continuous linear mapping, we set

$$(T^* y^*)x = y^*(Tx),\quad\forall y^*\in Y^*,\; x\in X.$$

Notice that this formula defines a linear mapping

$$(T^* y^*): X\to\mathbb{C},\quad (T^* y^*)\in X^*,$$

and thus also a mapping

$$T^*: Y^*\to X^*.$$

If we have a Hilbert space H, and a linear mapping T->H, then strictly according to the previous definition, we have T**->H*. However, the dual of a Hilbert space is isometric to the space itself, so we can identify all vectors of H* as vectors of H, and let the action of $x^*\in H^*$ be given with the inner product. So $x^*(z)$ becomes replaced by (x|z), if the $x^*\in X^*$ and $x\in X$ assumed to be identified in the isometry. The previous condition then becomes

$$(T^* y|x) = (y|Tx),$$

so the old formula, that you knew for matrices already, is obtained as a special case.

Then there is another definition, which deals with unbounded operators. If we have T:D(T)->H, with $D(T)\subset H$ being some subspace, it is possible to give some definitions $T^*:D(T^*)\to H$ so that the same inner product formula is still valid for some vectors, but this time the domain issue gets more difficult.

4. Apr 6, 2008

jostpuur

I found a theorem that says that T*(Y*) is closed if T(X) is closed, assuming that X and Y are Banach spaces and that T is continuous.