33.3. MIHLIN’S THEOREM 1133

(∫ 22−k

2−2−kρ

n−1−2Ldρ

)1/2

≤C (n,L,φ ,C0) tn/2−L2k(L−n/2).

Now estimate 33.3.26 in another way. The support of γk is in Sk, a bounded set, and soF−1γk is differentiable. Therefore,∫

|x|≥2t

∣∣F−1γk (x−y)−F−1

γk (x)∣∣dx =

∫|x|≥2t

∣∣∣∣∣∫ 1

0

n

∑j=1

D jF−1γk (x−sy)y jds

∣∣∣∣∣dx

≤ t∫|x|≥2t

∫ 1

0

n

∑j=1

∣∣D jF−1γk (x−sy)

∣∣dsdx

≤ t∫ n

∑j=1

∣∣D jF−1γk (x)

∣∣dx

≤ tn

∑j=1

(∫ (1+∣∣∣2−kx

∣∣∣2)−L

dx

)1/2

·

(∫ (1+∣∣∣2−kx

∣∣∣2)L ∣∣D jF−1γk (x)

∣∣2 dx

)1/2

≤C (n,L) t2kn/2n

∑j=1

(∫ (1+∣∣∣2−kx

∣∣∣2)L ∣∣D jF−1γk (x)

∣∣2 dx

)1/2

. (33.3.31)

Now consider the jth term in the last sum in 33.3.31.

∫ (1+∣∣2−kx

∣∣2)L ∣∣D jF−1γk (x)∣∣2 dx≤

C (n,L)∫

∑|α|≤L 2−2k|α|x2α∣∣D jF−1γk (x)

∣∣2 dx=C (n,L)∑|α|≤L 2−2k|α| ∫ x2α

∣∣F−1 (π jγk)(x)∣∣2 dx

(33.3.32)

where π j (z)≡ z j. This last assertion follows from

D j

∫e−ix·y

γk (y)dy =∫

(−i)e−ix·yy jγk (y)dy.

Therefore, a similar computation and Plancherel’s theorem implies 33.3.32 equals

=C (n,L) ∑|α|≤L

2−2k|α|∫ ∣∣F−1Dα (π jγk)(x)

∣∣2 dx