19.9. EXERCISES 521

x→∫Rm

f (x,y)dνx (y) is α measurable (19.13)

and ∫Rn+m

f (x,y)dµ =∫Rn

(∫Rm

f (x,y)dνx (y))

dα (x). (19.14)

If ν̂x is any other collection of Radon measures satisfying 19.13 and 19.14, then ν̂x = νx

for α a.e. x.

Proof: By Theorem 10.14.12 and the above lemmas, there exist unique Borel mea-surable α,νx such that 19.14 holds for all nonnegative Borel measurable functions f .This is because the Borel sets are contained in the product measurable sets. Now onecan use the Riesz representation theorem on functionals f →

∫Rn+m f (x,y)dµ and f →∫

Rm f (x,y)dνx (y) along with regularity of these measures obtained from the Riesz rep-resentation theorem to extend and obtain the same result for f only µ measurable. ■

19.9 Exercises1. Let X be a finite dimensional normed linear space, real or complex. Show that X is

separable. Hint: Let {vi}ni=1 be a basis and define a map from Fn to X ,θ , as follows.

θ (∑nk=1 xkek)≡∑

nk=1 xkvk. Show θ is continuous and has a continuous inverse. Now

let D be a countable dense set in Fn and consider θ (D).

2. Let α ∈ (0,1]. We define, for X a compact subset of Rp,

Cα (X ;Rn)≡ {f ∈C (X ;Rn) : ρα (f)+∥f∥ ≡ ∥f∥α< ∞}

where ∥f∥ ≡ sup{|f (x)| : x ∈ X} and

ρα (f)≡ sup{ |f (x)−f (y)||x−y|α

: x,y ∈ X , x ̸= y}.

Show that (Cα (X ;Rn) ,∥·∥α) is a complete normed linear space. This is called a

Holder space. What would this space consist of if α > 1?

3. Let {fn}∞n=1 ⊆Cα (X ;Rn) where X is a compact subset of Rp and suppose

∥fn∥α≤M

for all n. Show there exists a subsequence, nk, such that fnkconverges in C (X ;Rn).

We say the given sequence is precompact when this happens. (This also shows theembedding of Cα (X ;Rn) into C (X ;Rn) is a compact embedding.) Hint: You mightwant to use the Ascoli Arzela theorem.

4. Suppose f ∈C0 ([0,∞)) and also | f (t)| ≤Ce−rt . Let A denote the algebra of linearcombinations of functions of the form e−st for s sufficiently large. Thus A is dense inC0 ([0,∞)) . Show that if

∫∞

0 e−st f (t)dt = 0 for each s sufficiently large, then f (t) =0. Next consider only | f (t)| ≤ Cert for some r. That is f has exponential growth.Show the same conclusion holds for f if

∫∞

0 e−st f (t)dt = 0 for all s sufficiently large.This justifies the Laplace transform procedure of differential equations where if theLaplace transforms of two functions are equal, then the two functions are consideredto be equal. More can be said about this. Hint: For the last part, consider g(t) ≡e−2rt f (t) and apply the first part to g. If g(t) = 0 then so is f (t).