33.3. MIHLIN’S THEOREM 1131

Lemma 33.3.4 Let 33.3.25 hold and suppose ψ ∈C∞c (Rn \{0}). Then for each α, |α| ≤

L, there exists a constant C ≡C (α,n,ψ) independent of k such that

supx ∈Rn

|x||α|∣∣∣Dα

(ρ (x)ψ

(2kx))∣∣∣≤CC0.

Proof:

|x||α|∣∣∣Dα

(ρ (x)ψ

(2kx))∣∣∣≤ |x||α| ∑

β+γ=α

∣∣∣Dβρ (x)

∣∣∣2k|γ|∣∣∣Dγ

ψ

(2kx)∣∣∣

= ∑β+γ=α

|x||β |∣∣∣Dβ

ρ (x)∣∣∣ ∣∣∣2kx

∣∣∣|γ| ∣∣∣Dγψ

(2kx)∣∣∣

≤C0C (α,n) ∑|γ|≤|α|

sup{|z||γ| |Dγψ (z)| : z ∈ Rn}=C0C (α,n,ψ)

and this proves the lemma.

Lemma 33.3.5 There exists

φ ∈C∞c([

x :4−1 < |x|< 4]), φ (x)≥ 0,

and∞

∑k=−∞

φ

(2kx)= 1

for each x ̸= 0.

Proof: Letψ ≥ 0, ψ = 1 on

[2−1 ≤ |x| ≤ 2

],

spt(ψ)⊆[4−1 < |x|< 4

].

Consider

g(x) =∞

∑k=−∞

ψ

(2kx).

Then for each x, only finitely many terms are not equal to 0. Also, g(x)> 0 for all x ̸= 0. Toverify this last claim, note that for some k an integer, |x| ∈

[2l ,2l+2

]. Therefore, choose k an

integer such that 2k |x| ∈[2−1,2

]. For example, let k =−l−1. This works because 2k |x| ∈[

2l2k,2l+22k]=[2l−l−1,2l+2−l−1

]=[2−1,2

]. Therefore, for this value of k, ψ

(2kx)= 1

so g(x)> 0.Now notice that

g(2rx) =∞

∑k=−∞

ψ

(2k2rx

)=

∞

∑k=−∞

ψ

(2kx)= g(x).