1726 CHAPTER 54. FUNCTIONAL ANALYSIS APPLICATIONS

Now let δ =Cηβ so η =C−1/β δ1/β and ηβ−1 =C

1−β

β δ(β−1)/β . Thus for all x ∈ H,∣∣∣∣∣∣(−A)−β x

∣∣∣∣∣∣≤ 1Γ(β )

(δ

β||x||+2C

1−β

β δ(β−1)/β

∣∣∣∣A−1x∣∣∣∣) .

Let ε = δ

βΓ(β ) =δ

Γ(1+β ) . Then the above is of the form

∣∣∣∣∣∣(−A)−β x∣∣∣∣∣∣ ≤ ε ||x||+2

C1−β

β

Γ(β )(εΓ(1+β ))(β−1)/β

∣∣∣∣A−1x∣∣∣∣

≤ ε ||x||+2C1−β

β (εΓ(1+β ))(β−1)/β∣∣∣∣A−1x

∣∣∣∣because Γ is decreasing on (0,1) . I need to verify that for β ∈ (0,1) ,

Γ(1+β )(β−1)/β

is bounded. It is continuous on (0,1] and so if I can show limβ→0+ Γ(1+β )(β−1)/β exists,then it will follow the function is bounded. It suffices to show

limβ→0+

β −1β

lnΓ(1+β ) =− limβ→0+

lnΓ(1+β )

β

exists. Consider this. By L’Hospital’s rule and dominated convergence theorem, this is

limβ→0+

∫∞

0 ln(t) tβ e−tdtΓ(1+β )

= limβ→0+

∫∞

0ln(t) tβ e−tdt

= limβ→0+

∫∞

0ln(t)e−tdt.

Thus the function is bounded independent of β ∈ (0,1) . This shows there is a constantC which is independent of β ∈ (0,1) such that for any x ∈ H,∣∣∣∣∣∣(−A)−β x

∣∣∣∣∣∣≤ ε ||x||+Cε(β−1)/β

∣∣∣∣A−1x∣∣∣∣ . (54.3.28)

Now let y ∈ D(A) = D((−A)) and let x = (−A)y. Then the above becomes∣∣∣∣∣∣(−A)−β (−A)y∣∣∣∣∣∣≤ ε ||(−A)y||+Cε

(β−1)/β ||y||

I claim that(−A)−β (−A)y = (−A)1−β y.

The reason for this is as follows.

(−A)β (−A)1−β y = (−A)y

and so the desired result follows from multiplying on the left by (−A)−β . Hence∣∣∣∣∣∣(−A)1−β y∣∣∣∣∣∣≤ ε ||(−A)y||+Cε

(β−1)/β ||y||