By N. L. Carothers
This brief direction on classical Banach house conception is a usual follow-up to a primary path on practical research. the subjects coated have confirmed invaluable in lots of modern examine arenas, resembling harmonic research, the speculation of frames and wavelets, sign processing, economics, and physics. The publication is meant to be used in a complicated issues path or seminar, or for autonomous research. It bargains a extra common creation than are available within the current literature and contains references to expository articles and recommendations for additional interpreting.
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Extra info for A short course on Banach space theory
Exercises 33 Exercises 1. Let X be a Banach space with basis (xn ). We know that the expression |||x||| = supn Pn x deﬁnes a norm on X that is equivalent to · . Show that under ||| · |||, each Pn has norm one. That is, we can always renorm X so that (xn ) has basis constant K = 1. 2. Let ( f k ) denote Schauder’s basis for C[0, 1] and let (tk ) denote the associated enumeration of the dyadic rationals. If f ∈ C[0, 1] is written as n−1 f = ∞ k=0 ak f k , prove that an = f (tn ) − k=0 ak f k (tn ).
Indeed, x = P x + (I − P)x, and P(I − P)x = P x − P 2 x = 0 for any x ∈ X . While we’re at it, notice that Q = I − P is also a projection (the range of Q is the kernel of P; the kernel of Q is the range of P). As before, though, typically there are many different projections on X with a given range (or kernel). In summary, given a subspace M of X , ﬁnding a complement for M in X is equivalent to ﬁnding a linear projection P on X with range M. But we’re interested in normed vector spaces and continuous maps.
Thus, the two spaces must coincide. Since the dyadic rationals are dense in [0, 1], it’s not hard to see that the f n have dense linear span. Thus, ( f n ) is a viable candidate for a basis for C[0, 1]. If we set pn = nk=0 ak f k , then pn ∞ = max | pn (tk )| 0≤k≤n because pn is a polygonal function with nodes at t0 , . . , tn . And if we set pm = m k=0 ak f k for m > n, then we have pm (tk ) = pn (tk ) for k ≤ n because f j (tk ) = 0 for j > n ≥ k. Hence, pn ∞ ≤ pm ∞ . This implies that ( f n ) is a normalized basis for C[0, 1] with basis constant K = 1.