L. Maligranda, A simple proof of the Hölder and the Minkowski inequality. Amer. Math. Monthly 102 (1995), 256-259.At present I'm undecided about which approach I prefer. (Maybe it would be best to show the students both approaches.) The concrete computations are actually a bit unmotivated; they're based on clever tricks that are produced like a magician pulling rabbits out of hats. The abstract, functional analytic approach (i.e., via 12.29.g) appears more natural and more general, but it's actually not more general: If you try to replace the function tp with an arbitrary convex function of t and then try to carry through the proof of Minkowski's inequality based on 12.29.g, you'll actually find that many conditions get imposed on that convex function, and ultimately (after much computation) those conditions force you to use tp.
H. König, A simple proof of the Minkowski inequality. General Inequalities, 6 (Oberwolfach, 1990), 469, Internat. Ser. Numer. Math., 103, Birkhäuser, Basel, 1992.
Ky Fan and Irving Glicksberg, Fully convex normed linear spaces, Proc. Nat. Acad. Sci. USA 41 (11) (Nov. 15, 1955), pp. 947-953.
phi(t) = mu( [a,t) ),which is left-continuous. (By the way, the earlier part of the proof of 24.35 can be shortened slightly: Use 24.32.b to show that K is closed under finite union; this shortens the proof that K is a sigma-algebra.)
R. D. Mauldin, The set of continuous nowhere differentiable functions, Pacific J. Math. 83 (1979), 199-205.By the way, here is a related question for which I haven't yet found a good answer. Does anyone know of an explicitly constructible example of a subset of the reals that is Lebesgue measurable but not Borel measurable? (I haven't finished investigating this; I suppose I should look at some books on descriptive set theory.)
In any topological vector space, any weakly bounded set is weakly totally bounded.This follows from 19.15.g, without using the Axiom of Choice or any weak form of choice. (Here "totally bounded" is defined as in 19.14.) Next, using (UF24), we obtain this result:
(UF25.5) In any topological vector space, any weakly bounded set is weakly precompact.I'm calling that "(UF25.5)" because it would be inserted right before (UF26). We can prove without much difficulty that (UF25.5) implies (UF26). Proof: Let X* have the weak-star topology; note that that is the relative topology induced by the product topology on FX, where F is the scalar field. Let V be an equicontinuous subset of X*, and let C be its closure in FX. Then C is also equicontinuous, by 18.33.a, and we easily verify that C is a subset of X* and that C is complete (in the weak-star topology). We also verify that C is bounded; hence by (UF25.5) it is precompact; hence it is compact; hence V is relatively compact.
If a Banach space is either (i) reflexive or (ii) a separable dual, then it has the RNP.This can be proved by slight modifications of the proof that I gave in 29.26. Throughout the proof, replace the weak topology with the weak-star topology. (That's no change at all in the reflexive case.) A couple of other changes must be noted:
Every continuous linear functional on C[a,b] can be represented as the Riemann-Stieltjes integral with respect to some function of bounded variation. (Moreover, the function in BV can be chosen so that its variation is equal to the operator norm of the continuous linear functional.)A relatively short proof of that result can be given; it is similar to the 2nd, 3rd, 4th, and 5th paragraphs of the proof of 29.34 (with the correction noted above). The theorem apparently is due to F. Riesz (1909). The proof using the Hahn-Banach Theorem apparently first appeared in Banach's book (1932). Banach made the error about not dealing with the endpoint properly; I guess it got past the referees because they already knew Riesz's theorem was true. Banach's proof, along with the minor error, subsequently found its way into many other books -- for instance,