Solution to the Exercises of Lecture 10

Solution to the Exercises of Lecture 10

University of Salerno February 20, 2015

Exercise 10.1: For this exercise letS be the strip S :={z∈C;−1/2<Rez <1/2}.

Leth:S7→Cbe continuous and holomorphic onS. Assume thathis bounded on∂S, and that there existsα < πsuch that

|h(z)| ≤ce^eα|Imz| (z∈S).

Show thathis bounded bykhk∂S.

Solution: Forn∈N, the functionψn(z) :=e^{−n1(eiβz+e−iβz)}, withα < β < π, is continuous onSand holomorphic onS. By the maximum principle one obtains

kψnhk_S

k ≤ kψnhk∂S_k,

whereS_k :={z∈S:|Imz|< k}, k∈N. Observe thatkψ_nhk_{{z∈∂Sk:|Imz|=k}}→ 0 (k→ ∞). In fact,

|ψnh(z)| ≤c

e^{−n1(eiβz+e−iβz)} e^eα|Imz|

=c e^{−n1(e−βImz+eβImz) cos(β|Rez|)}e^eα|Imz|

≤c e^{−cos(β/2)n (e−βImz+eβImz)}e^eα|Imz|

Hence,

kψnhk_{{z∈∂Sk:|Imz|=k}}≤c e^{−cos(β/2)n (e−βk+eβk)}e^eαk

=c e^{−cos(β/2)n e−βk}e^eβk(−^cos(β/2)_n +e^(α−β)k)

For big k, we have that e^(α−β)k < ε. Thus, choosing ε < ^cos(β/2)_2n , so that e^(α−β)k−^cos(β/2)_n <−ε, we have that, in the limit ask → ∞, the right-hand side of the inequality tends to 0.

Therefore,

kψ_nhk_S ≤ kψ_nhk_∂S. Lettingn→ ∞we obtain the assertion.

Exercise 10.2: Let (Ω, µ) be a measure space. Show that

k1

2(f +g)k^p_p+k1

2(f−g)k^p_p ¹_p

≤2^−p1 kfk^p_p+kgk^p_p¹_p

for allf, g∈L_p(µ), 2≤p≤ ∞.

Solution: For 2≤p≤ ∞consider

Tp:Lp(µ)×Lp(µ)−→Lp(µ)×Lp(µ) (f, g) 7→

1

2(f+g),1 2(f −g)

.

We endowLp(µ)×Lp(µ),2≤p≤ ∞, with the norms

k(f, g)kp×p = (kfk^p_p+kgk^p_p)^1p, f, g∈Lp(µ),2≤p <∞, k(f, g)k_∞ = max(kfk_∞,kgk_∞), f, g∈L_∞(µ).

We first estimate the norm of the operatorT onL2(µ)×L2(µ) =L2(µ⊗µ) and L∞(µ)×L∞(µ) =L∞(µ⊗µ). For p= 2 one has

kT(f, g)k2×2 = 1

2kf+g, f−gk2×2

= 1

2 kf+gk²₂+kf −gk²₂¹₂

= 1

2 kfk²+ 2(f|g) +kgk²+kfk²−2(f|g) +kgk²¹₂

= 1

√2(kfk²₂+kgk²₂)¹²

= 2⁻¹²k(f, g)k_2×2. And forp=∞, one obtains

kT(f, g)k_∞ = 1

2kf+g, f−gk_∞

= 1

2max (kf+gk_∞,kf−gk_∞)

≤ 1

2max (kfk∞+kgk∞,kfk∞+kgk∞)

= 1

2(kfk∞+kgk∞)

≤ max(kfk_∞,kgk_∞)

= k(f, g)k∞.

Letp∈(2,+∞),p0=∞,p1= 2,M0 = 1,M1= 2⁻¹² andτ= ²_p ∈(0,1), then by Riesz-Thorin Theorem, see Corollary 10.6, one has

kT(f, g)kp≤M_τk(f, g)kp, f, g∈S(Ac) (1) withM_τ=M₀^1−τM₁^τ = 2^−1p and

A_c={A⊂Ω, measurable:µ(A)<∞}.

By density ofS(Ac) onLp(µ), (1) holds for allf, g∈Lp(µ). Therefore,

k1

2(f+g)k^p_p+k1

2(f−g)k^p_{p p}¹

≤2^−1p kfk^p_p+kgk^p_p¹_p

, f, g∈L_p(µ).

Exercise 10.3: (a) Letp∈(1,∞),r∈[0,∞). Show that r= inf

α∈Q∩(0,∞)

1

pα^1−pr^p+ (1−1 p)α

. (2)

(b) Let (Ω, µ) be a measure space, and letS∈ L(L₂(µ)) be sub-Markovian and substochastic. Show thatS isLp-contractive for allp∈(1,∞).

(c) Let (Ω, µ) be a measure space, and let S ∈ L(L2(µ)) be sub-Markovian, and assume that there existsc >0 such that ¹_cS is substochastic. Show thatS interpolates to an operatorSp∈ L(Lp(µ)) withkSpk ≤c^1p, for 1< p <∞.

Solution: (a) Letp⁰be the conjugate ofp, i.e. _p¹+_p¹0 = 1 and letα∈Q∩(0,∞).

Applying Young’s inequality

ab≤ 1 pa^p+ 1

p⁰b^p0

witha=α^1−pp randb=α^p−1p , one obtains r=ab ≤ 1

p(α^1−pp r)^p+

1−1 p

α^p−1p _p−1^p

≤ 1

pα^1−pr^p+

1−1 p

α.

Passing to the infinimum, one gets r≤ inf

α∈Q∩(0,∞)

1

pα^1−pr^p+ (1−1 p)α

.

On the other hand, consider α_n = E(nr)

n ∈ Qfor n ≥1, where E(nr) is the integer part ofnr. Since lim

n−→∞α_n=r, it follows that

n−→∞lim 1

pα^1−p_n r^p+

1−1 p

αn

=r.

Finally,

r= inf

α∈Q∩(0,∞)

1

pα^1−pr^p+ (1−1 p)α

.

(b) Letu∈L2(µ)∩Lp(µ) andAbe a measurable subset of Ω with finite measure.

Then, by (2) (withr=|u(x)|χA(x)) , it follows that

|u|χA≤1

pα^1−p|u|^pχA+ (1−1 p)αχA

forα∈Q∩(0,∞). So, by the positivity and theL^∞-contractivity ofS, one has S(|u|χA) ≤ 1

pα^1−pS(|u|^pχA) + (1−1 p)αSχA

≤ 1

pα^1−pS(|u|^pχA) + (1−1 p)α.

So, passing to the infinimum over α∈Q∩(0,∞), one obtains S(|u|χA)≤(S(|u|^pχA))^p1.

Let now (A_n)_n∈Nbe an increasing sequence of measurable sets with finite mea- sure such that ∪n∈NA_n = Ω. Then, |u|χAn −→ |u| and |u|^pχ_An −→ |u|^p as n −→ ∞. Since S ∈ L(L₂(µ)) and S can be extended to a bounded linear operator onL₁(µ) (using the fact thatS is substochastic), it follows that

n−→∞lim S(|u|χA_n) =S|u|and lim

n−→∞S(|u|^pχA_n) =S|u|^p. Finally one has

S(|u|)≤(S(|u|^p))^1p, ∀u∈L2(µ)∩Lp(µ). (3) Therefore, by using theL1-contractivity ofSand the fact that|Su| ≤S|u|, one has

Z

Ω

|Su|^pdµ≤ Z

Ω

(S|u|)^pdµ≤ Z

Ω

S|u|^pdµ≤ Z

Ω

|u|^pdµ.

Thus,

kSuk^p_p≤ kuk^p_p, ∀u∈L2(µ)∩Lp(µ).

By the density ofL₂(µ)∩L_p(µ) inL_p(µ), the above estimate shows thatScan be extended to a contractive operator overL_p(µ).

(b) Letu∈L₂(µ)∩L_p(µ), then from (3) one has (S(|u|))^p≤S(|u|^p). Thus, Z

Ω

|Su|^pdµ≤ Z

Ω

(S|u|)^pdµ≤ Z

Ω

S|u|^pdµ≤c Z

Ω

|u|^pdµ.

Therefore,

kSuk^p_p≤ckuk^p_p, ∀u∈L2(µ)∩Lp(µ).

By the density of L₂(µ)∩L_p(µ) inL_p(µ), it follows thatS interpolates to an operatorS_p∈ L(L_p(µ)) andkS_pk ≤c^1/p for all 1< p <∞.

Exercise 10.4: Let the hypotheses be as in Exercise 9.5, and additionally b∈C¹(Ω). Assume thatω∈Ris such that div b(x)≤ω for allx∈Ω.

(a) Show that kT(t)uk1 ≤e^ωtkuk1 for u∈L2(Ω)∩L1(Ω) and t ≥0, where T(·) is theC0-semigroup generated by the operator −A.

(b) Compute estimates for kTp(t)k in terms of ω := sup divb fort ≥0, 1≤ p <∞, whereTp(·) is the interpolated semigroup onLp(Ω), analogous to Theorem 10.15(b).

Solution:

(a) In order to show that kT(t)uk₁ ≤ e^ωtkuk₁, we show that e^−ωtT(t) is substochastic. To do so we use the characterization in Theorem 10.12.(d).

Hence we have to show thataω((u−1)⁺, u∧1)≥0, whereaωis the form associated withA+ωI.

Let us considerV =H₀¹(Ω) and u∈C_c¹(Ω). Observe that if u∈C_c¹(Ω) thenu∧1 ∈ V, cf. [1, Proposition 4.11, p.113], and also (u−1)⁺ ∈ V, sinceu−u∧1 = (u−1)⁺.

Integrating by parts and using the fact that divb(x)≤ω, we have aω((u−1)⁺, u∧1) =

Z

Ω

∇(u−1)⁺· ∇(u∧1)dx+ Z

Ω

b· ∇(u−1)⁺(u∧1)dx +

Z

Ω

ω(u−1)⁺(u∧1)dx

≥ Z

Ω

(u−1)⁺(ω−div b)(u∧1)dx− Z

Ω

(u−1)⁺b· ∇(u∧1)dx

= Z

Ω

(u−1)⁺(ω−div b)(u∧1)dx≥0.

(b) We know that kT(t)k_∞ ≤ 1 (Exercise 9.5(b)) and kT(t)k₁ ≤ e^ωt with ω= sup div(b). Applying Exercise 10.3.(c) we have

kT(t)kp≤e^ωtp, t≥0.

References

[1] E.M. Ouhabaz, Analysis of Heat Equations on Domains, London Math.

Soc. Monographs, Vol. 31. Princeton Univ. Press 2004.