33.7. EXERCISES 681

3. ↑Let f (x) be as given above. Now let

f̂ (x)≡{

f (x) if x ≤ 00 if x > 0

Show that f̂ (x) is also infinitely differentiable. Now let r > 0 and define g(x) ≡f̂ (−(x− r)) f̂ (x+ r). show that g is infinitely differentiable and vanishes for |x| ≥ r.Let ψ (x) = ∏

pk=1 g(xk). For U = B(0,2r) with the norm given by

∥x∥= max{|xk| ,k ≤ p} ,

show that ψ ∈C∞c (U).

4. ↑Using the above problem, let ψ ∈ C∞c (B(0,1)) . Also let ψ ≥ 0 as in the above

problem. Show there exists ψ ≥ 0 such that ψ ∈C∞c (B(0,1)) and

∫ψdmp = 1. Now

define ψk (x)≡ kpψ (kx) . Show that ψk equals zero off a compact subset of B(0, 1

k

)and

∫ψkdmp = 1. We say that spt(ψk) ⊆ B

(0, 1

k

). spt( f ) is defined as the closure

of the set on which f is not equal to 0. Such a sequence of functions as just defined{ψk} where

∫ψkdmp = 1 and ψk ≥ 0 and spt(ψk)⊆ B

(0, 1

k

)is called a mollifier.

5. If you have f ∈ L1 (Rp) with respect to Lebesgue measure and ψk is a mollifier, showthat f ∗ ψk (x) ≡

∫Rp f (x−y)ψk (y)dmp is infinitely differentiable. Hint: First

show it equals∫Rp f (y)ψk (x−y)dmp. Then use dominated convergence theorem.

6. Let φ : R→ R be convex. This means

φ(λx+(1−λ )y)≤ λφ(x)+(1−λ )φ(y)

whenever λ ∈ [0,1]. The following picture illustrates what is about to be shown.

(y,φ(y))

φ(x)≥ φ(y)+λ y(x− y)

(a) Show that for x < y < z, φ(z)−φ(x)z−x ≤ φ(z)−φ(y)

z−y .

(b) Next show φ(z)−φ(x)z−x ≥ φ(y)−φ(x)

y−x . To do these, use convexity applied to y.

(c) Conclude φ(z)−φ(y)z−y ≥ φ(z)−φ(x)

z−x ≥ φ(y)−φ(x)y−x . In particular, φ(z)−φ(y)

z−y ≥ φ(y)−φ(x)y−x .

(Difference quotients increase from left to right.)

(d) Let λ y ≡ inf{

φ(z)−φ(y)z−y : z > y

}. Then for y ≥ x,

φ (y)−φ (x)≤ λ y (y− x)

Show that even if x > y, the same inequality holds. Thus for all x,

φ (y)+λ y (x− y)≤ φ (x)