9.7. EXERCISES 235

Note that the proof of this theorem depends on two manipulations, the interchange ofthe order of integration and Holder’s inequality. Note that there is nothing to check in thecase of double sums. Thus if ai j ≥ 0, it is always the case that(

∑j

(∑

iai j

)p)1/p

≤∑i

(∑

jap

i j

)1/p

because the integrals in this case are just sums and (i, j)→ ai j is measurable.

9.7 Exercises1. Establish the inequality ∥ f g∥r ≤ ∥ f∥p ∥g∥q whenever 1

r = 1p +

1q .

2. Let (Ω,S ,µ) be counting measure onN. Thus Ω =N and S =P (N) with µ (S) =number of things in S. Let 1≤ p≤ q. Show that in this case,

L1 (N)⊆ Lp (N)⊆ Lq (N) .

Hint: This is real easy if you consider what∫

Ωf dµ equals. How are the norms

related?

3. Consider the function, f (x,y) = xp−1

py + yq−1

qx for x,y> 0 and 1p +

1q = 1. Show directly

that f (x,y)≥ 1 for all such x,y and show this implies xy≤ xp

p + yq

q .

4. Give an example of a sequence of functions in Lp (R) which converges to zero in Lp

but does not converge pointwise to 0. Does this contradict the proof of the theoremthat Lp is complete?

5. Let φ : R→ R be convex. This means φ(λx+ (1− λ )y) ≤ λφ(x) + (1− λ )φ(y)whenever λ ∈ [0,1]. Verify that if x < y < z, then φ(y)−φ(x)

y−x ≤ φ(z)−φ(y)z−y and that

φ(z)−φ(x)z−x ≤ φ(z)−φ(y)

z−y . Show if s ∈ R there exists λ such that φ(s) ≤ φ(t)+λ (s− t)for all t. Show that if φ is convex, then φ is continuous.

6. ↑ Prove Jensen’s inequality. If φ : R→ R is convex, µ(Ω) = 1, and f : Ω→ R is inL1(Ω), then φ(

∫Ω

f du)≤∫

Ωφ( f )dµ . Hint: Let s =

∫Ω

f dµ and use Problem 5.

7. B(p,q) =∫ 1

0 xp−1(1− x)q−1dx,Γ(p) =∫

∞

0 e−tt p−1dt for p,q > 0. The first of theseis called the beta function, while the second is the gamma function. Show a.) Γ(p+1) = pΓ(p); b.) Γ(p)Γ(q) = B(p,q)Γ(p+q).

8. Let f ∈Cc(0,∞) and define F(x) = 1x∫ x

0 f (t)dt. Show

∥F∥Lp(0,∞) ≤p

p−1∥ f∥Lp(0,∞)

whenever p > 1. Hint: Argue there is no loss of generality in assuming f ≥ 0and then assume this is so. Integrate

∫∞

0 |F(x)|pdx by parts as follows:∫

∞

0 F pdx =show = 0︷ ︸︸ ︷xF p|∞0 − p

∫∞

0 xF p−1F ′dx. Now show xF ′ = f −F and use this in the last integral.Complete the argument by using Holder’s inequality and p−1 = p/q. The measureis one dimensional Lebesgue measure in this problem.