Some things in linear algebra are easier to see in infinite dimensions, i.e. in Banach spaces. Distinctions that seem pedantic in finite dimensions clearly matter in infinite dimensions.
The category of Banach spaces considers linear spaces and continuous linear transformations between them. In a finite dimensional Euclidean space, all linear transformations are continuous, but in infinite dimensions a linear transformation is not necessarily continuous.
The dual of a Banach space V is the space of continuous linear functions on V. Now we can see examples of where not only is V* not naturally isomorphic to V, it’s not isomorphic at all.
For any real p > 1, let q be the number such that 1/p + 1/q = 1. The Banach space Lp is defined to be the set of (equivalence classes of) Lebesgue integrable functions f such that the integral of ||f||p is finite. The dual space of Lp is Lq. If p does not equal 2, then these two spaces are different. (If p does equal 2, then so does q; L2 is a Hilbert space and its dual is indeed the same space.)
In the finite dimensional case, a vector space V is isomorphic to its second dual V**. In general, V can be embedded into V**, but V** might be a larger space. The embedding of V in V** is natural, both in the intuitive sense and in the formal sense of natural transformations. We can turn an element of V into a linear functional on linear functions on V as follows.
Let v be an element of V and let f be an element of V*. The action of v on f is simply fv. That is, v acts on linear functions by letting them act on it.
This shows that some elements of V** come from evaluation at elements of V, but there could be more. Returning to the example of Lebesgue spaces above, the dual of L1 is L∞, the space of essentially bounded functions. But the dual of L∞ is larger than L1. That is, one way to construct a continuous linear functional on bounded functions is to multiply them by an absolutely integrable function and integrate. But there are other ways to construct linear functionals on L∞.
A Banach space V is reflexive if the natural embedding of V in V** is an isomorphism. For p > 1, the spaces Lp are reflexive.
Suppose that X is a Banach space. For simplicity, we assume that X is a real Banach space, though the results can be adapted to the complex case in the straightforward manner. In the following, B(x0,ε) stands for the closed ball of radius ε centered at x0 while B◦(x0,ε) stands for the open ball, and S(x0,ε) stands for the corresponding sphere.
Let Q be a bounded operator on X. Since we will be interested in the hyperinvariant subspaces of Q, we can assume without loss of generality that Q is one-to-one and has dense range, as otherwise ker Q or Range Q would be hyperinvariant for Q. By {Q}′ we denote the commutant of Q.
Fix a point x0 ≠ 0 in X and a positive real ε<∥x0∥. Let K= Q−1B(x0,ε). Clearly, K is a convex closed set. Note that 0 ∉ K and K≠ ∅ because Q has dense range. Let d = infK||z||, then d > 0. If X is reflexive, then there exists z ∈ K with ||z|| = d, such a vector is called a minimal vector for x0, ε and Q. Even without reflexivity condition, however, one can always find y ∈ K with ||y|| ≤ 2d, such a y will be referred to as a 2-minimal vector for x0, ε and Q.
The set K ∩ B(0, d) is the set of all minimal vectors, in general this set may be empty. If z is a minimal vector, since z ∈ K = Q−1B(x0, ε) then Qz ∈ B(x0, ε). As z is an element of minimal norm in K then, in fact, Qz ∈ S(x0, ε). Since Q is one-to-one, we have
QB(0, d) ∩ B(x0, ε) = Q B(0, d) ∩ K) ⊆ S(x0, ε).
It follows that QB(0,d) and B◦(x0,ε) are two disjoint convex sets. Since one of them has non-empty interior, they can be separated by a continuous linear functional. That is, there exists a functional f with ||f|| = 1 and a positive real c such that f|QB(0,d) ≤ c and f|B◦(x0,ε) ≥ c. By continuity, f|B(x0,ε) ≥ c. We say that f is a minimal functional for x0, ε, and Q.
We claim that f(x0) ≥ ε. Indeed, for every x with ||x|| ≤ 1 we have x0 − εx ∈ B(x0,ε). It follows that f(x0 − εx) ≥ c, so that f(x0) ≥ c + εf(x). Taking sup over all x with ||x|| ≤ 1 we get f(x0) ≥ c + ε||f|| ≥ ε.
Observe that the hyperplane Q∗f = c separates K and B(0, d). Indeed, if z ∈ B(0,d), then (Q∗f)(z) = f(Qz) ≤ c, and if z ∈ K then Qz ∈ B(x0,ε) so that (Q∗f)(z) = f(Qz) ≥ c. For every z with ||z|| ≤ 1
we have dz ∈ B(0,d), so that (Q∗f)(dz) ≤ c, it follows that Q∗f ≤ c/d
On the other hand, for every δ > 0 there exists z ∈ K with ||z|| ≤ d+δ, then (Q∗f)(z) ≥ c ≥ c/(d+δ) ||z||, whence ||Q∗f|| ≥ c/(d+δ) . It follows that
||Q∗f|| = c/d.
For every z ∈ K we have (Q∗f)(z) ≥ c = d ||Q∗f||. In particular, if y is a 2-minimal vector then
(Q∗f)(y) ≥ 1/2 Q∗f ||y||….
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