Giga Y. Surface Evolution Equations: A Level Set Approach

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138 Chapter 3. Comparison principle

so that O ≤ Q ≤ J ≤ 2I,weobservethatA

≥ αf



(ρ)Q ≥ O.ByA

≥ O,wesee

that

| =sup{(

ξ,

η)A





; |ξ|

+ |η|

=1,ξ,η∈ R

Since

≤ α



(ρ)

J + αf



(ρ)Q ≤ αK

(ρ)J ≤ 2αK

(ρ)I

with K

(ρ)=f



(ρ)/ρ+f



(ρ), we have an estimate |A

|≤2αK

(ρ). Since J

=2J,

=2Q, JQ = QJ =2Q, A

is of the form

=2α

{(



(ρ)

)

J +(2



(ρ)

+ f



(ρ) −



(ρ)

)(f



(ρ) −



(ρ)

)Q}

≤ 2α

{(



(ρ)

)

J +(f



(ρ))

Q}≤2α

(ρ)J

with K

(ρ)=(f



(ρ)/ρ)

+(f



(ρ))

. Thus (3.4.11) implies

−(

+2αK

(ρ))I ≤



σδ

O −Y

σδ



≤ (αK

(ρ)+2α

(ρ))J.

This yields X

σδ

≤ Y

σδ

as in Case 2 of the proof of Theorem 3.1.1 (§3.4.1) and a

bound

σδ

|≤2αK

(ρ)+4α

(ρ)λ

with ρ = |x

σδ

− y

σδ

|. The right-hand side is bounded as δ → 0andβ →∞by

(3.4.23). We thus have a bound for X

σδ

independent of δ and β. By (3.4.23) and

(3.4.24) we observe that

ˆϕ

+2δx

σδ

=0, − ˆϕ

− 2δy

σδ

=ˆϕ

+2δy

σδ

=0

for suﬃciently small δ>0. Since u and v are sub- and supersolutions, we have

2β(t

σδ

− s

σδ

)+γ + F (t

σδ

, ˆu, ˆϕ

+2δx

σδ

+2δI) ≤ 0, (3.4.25)

2β(t

σδ

− s

σδ

) − γ + F (s

σδ

, ˆv, − ˆϕ

− 2δy

σδ

− 2δI) ≥ 0. (3.4.26)

By (F2) and (F4) with c

= 0, (3.4.26) with X

σδ

≤ Y

σδ

implies

2β(t

σδ

− s

σδ

) − γ + F (s

σδ

, ˆu, ˆϕ

− 2δy

σδ

− 2δI) ≥ 0. (3.4.27)

Sending δ → 0, we obtain from (3.4.25) and (3.4.27) with a bound for |X

σδ

| and

(3.4.23) that

2β(

− s

)+γ + F (t

, u

,αf



(|p

|), X

) ≥ 0, (3.4.28)

2β(

− s

) − γ + F (s

, u

,αf



(|p

|), X

) ≤ 0 (3.4.29)

3.5. Lipschitz preserving and convexity preserving properties 139

for some

, s

∈ [0,T], u

∈ R, X

∈ S

, p

∈ R

by taking a subsequence

if necessary; here we invoke the assumption that u is bounded so that

exists.

Sending β →∞so that

− s

→ 0 by (3.4.14), we now obtain from (3.4.28)

and (3.4.29) that 2γ ≤ 0since

, |p

|, |p

−1

are still bounded as β →∞.This

contradicts γ>0sowehaveprovedthatθ

≤ 0. 

Remark 3.4.3. The classiﬁcation into Case 1 and Case 2 for bounded domains

(Theorem 3.1.1) is slightly diﬀerent from that for unbounded domains (Theorem

3.1.4). If we classify into the following two cases for bounded domains with ﬁxed

α, this exactly corresponds to the classiﬁcation for unbounded domains.

Case 1. For each r>0thereisβ

→∞(as r → 0) and a maximum z

αβ

w − ϕ

αβ

in Q

∗

× Q

∗

such that |x

αβ

− y

αβ

|≤r,wherez

αβ

=(x

αβ

Case 2. There is r

> 0 such that for suﬃciently large β any maximizer z

αβ

w − ϕ

αβ

satisﬁes |x

αβ

− y

αβ

| >r

3.5 Lipschitz preserving and convexity preserving

properties

We study some general properties of solutions of the Cauchy problem for (3.1.1).

As we shall see in §4.3, for a given u

∈ BUC(R

) there exists a unique viscosity

solution u ∈ BUC(R

× [0,T)) of (3.1.1) satisfying u|

t=0

= u

, for example

under the assumption of Corollary 3.1.5. Here the space of all bounded, uniformly

continuous functions on D ⊂ R

is denoted by BUC(D). We ﬁrst observe that

global Lipschitz continuity is preserved for spatial homogeneous equations.

Theorem 3.5.1. Assume that F (x, t, r, p, X) is independent of x and r. Assume

that (F1)–(F2) and (F3) (or (F3



) with the assumption that F

N is invariant

under positive multiplications.) If u ∈ BUC(R

× [0,T)) is an (F-) solution of

(3.1.1) in R

× (0,T) with some constant L>0 satisfying

|u(x, 0) − u(y, 0)|≤L|x − y| (3.5.1)

for all x, y ∈ R

,then

|u(x, t) − u(y, t)|≤L|x − y| (3.5.2)

for all x, y ∈ R

, t ∈ [0,T).

Proof. Since F is independent of x and u,wesee

v(x, t)=u(x + h, t)+L|h|,h∈ R

is also a viscosity solution in BUC(R

× [0,T)) of (3.1.1) in R

× (0,T). By the

assumption of the initial data (3.5.1) and uniform continuity we see that u and v

satisfy (3.1.4). By (CP) we have

u(x, t) ≤ v(x, t)

140 Chapter 3. Comparison principle

u(x, t) − u(x + h, t) ≤ L|h|

for all x ∈ R

, t ∈ [0,T). A symmetric argument comparing u(x + h, t) − L|h|

and u(x, t) yields

u(x, t) − u(x + h, t) ≥−L|h|.

We thus prove (3.5.2). 

We next study whether or not concavity of initial data u

is preserved as

time develops.

Theorem 3.5.2. Assume that F = F (x, t, r, p, X) is independent of x and r and

satisﬁes (F1), (F2). Assume that F is geometric. Assume that

X → F (t, p, X) is convex on S

(3.5.3)

for all p ∈ R

\{0}, t ∈ [0,T].Letu ∈ C(R

× [0,T)) be an F-solution of (3.1.1)

in R

× (0,T) satisfying (3.5.2) with some L>0.Ifu(x, 0) is concave, then

u(x, t)+u(y, t) − 2u(z,t) ≤ L|x + y − 2z| (3.5.4)

holds for all x, y, z ∈ R

, 0 ≤ t ≤ T .Inparticular,x → u(x, t) is concave for all

t ∈ [0,T).

We shall prove Theorem 3.5.2 in several steps. Our ﬁrst observation is the

equivalence of concavity and (3.5.4) type inequality when a function is globally

Lipschitz.

Lemma 3.5.3. Assume that v

is globally Lipschitz in R

with constant L.Then

is concave if and only if the inequality

(x)+v

(y) − 2v

(z) ≤ L|x + y − 2z| (3.5.5)

holds for all x, y, z ∈ R

Proof. If v

is concave, then

(x)+v

(y) − 2v

(z)

= v

(x)+v

(y) − 2v

(z)(

x + y

)+2{v

(

x + y

) − v

(z)}

≤ 2{v

(

x + y

) − v

(z)}.

Since v

is globally Lipschitz, the right-hand side is dominated by 2L|

x+y

− z|.

We thus obtain (3.5.5).

The inequality (3.5.5) yields

(x)+v

(y)

≤ v

(

x + y

)

3.5. Lipschitz preserving and convexity preserving properties 141

by taking z =(x + y)/2. This says that v

is mid-concave. Successive use of this

inequality yields

λv

(x)+(1− λ)v

(y) ≤ v

(z),z= λx +(1− λ)y

for all λ ∈ (0, 1) of the form λ = k2

−h

with positive integers h and k.Sincev

continuous, this implies concavity of v

. 

Proof of Theorem 3.5.2. We suppress the symbol F in the proof. For M>0and

R>0letθ

and θ

be functions on R of the form

(η)=min(M,η),θ

(η)=max(−R, η).

Since F is geometric, by the invariance under change of dependent variables proved

later in §4.2.1 we see that

(x, t)=(θ

◦ u)(x, t)=min(M, u(x, t)),

(x, t)=(θ

◦ u)(x, t)=max(−R, u(x, t))

are solutions of (3.1.1) in Q = R

× (0,T) satisfying (3.5.2). Since u

is still

concave in x, we may assume that u is bounded from above in R

× (0,T). As

in the proof of Theorem 3.1.1 we may assume that u is a solution of (3.1.1) in

∗

= R

× (0,T] by taking T smaller. By replacing u by u/L we may assume

that L =1.

Our goal is to prove

u(x, t)+u(y, t) − 2u(z,t) ≤|x + y − 2z| (3.5.6)

for all x, y, z ∈ R

, t ∈ [0,T]. We take f ∈F.Forκ>0 there is a unique A(κ) > 0

such that

η ≤ A(κ)f(η)+κ =: N

(η) for all η>0

and the equality holds only at one point η = η(κ) > 0. We thus observe that

η =inf

κ>0

(η)(3.5.7)

for η>0. We set

κR

(x, s

,y,s

,z,s

)=u(x, s

)+u(y, s

) − 2u

(z,s

) − N

(|x + y − 2z|) − γs

for R>0, γ ≥ 0, κ>0; we use w

κ∞

when u replaces u

Assume that the conclusion (3.5.6) was false. By (3.5.7) there would exist

κ = κ

such that

sup{w

∞

(x, t, y, t, z, t); x, y, z ∈ R

,t∈ [0,T]} > 0.

142 Chapter 3. Comparison principle

We may replace u(z,t)byu

(z,t)forlargeR and observe that

:= lim sup

r↓0

(x, s

,y,s

,z,s

); x, y, z ∈ R

∈ [0,T],

− s

|≤r, |s

− s

| <r} > 0.

Since u is bounded from above on

Q = R

× [0,T], i.e., K := sup

u<∞,wesee

that θ

≤ 2K +2R<∞. For suﬃciently small γ>0weobservethat

:= lim sup

r↓0

(x, s

,y,s

,z,s

); (x, s

), (y, s

), (z,s

) ∈ Q,

− s

| <r,|s

− s

| <r} > 0.

We shall ﬁx R and γ such that µ

> 0 and suppress the subscript γ and the

superscript R as well as κ

.Wedeﬁne

(x, s

,y,s

,z,s

):=w(x, s

,y,s

,z,s

) − ϕ

):=β(|s

− s

+ |s

− s

),θ(β)=sup

where W =

Q × Q × Q.

We shall prove that for suﬃciently large β,sayβ>β

,if(x, s

), (y, s

(z,s

) ∈ Q satisﬁes

(x, s

,y,s

,z,s

) >µ

/2, (3.5.8)

then s

> 0, s

> 0ands

> 0. By (3.5.5) and (3.5.7) we observe that

(x)+u

(y) − 2u

(z) − N

(|x + y − 2z|) ≤ 0, x,y,z ∈ R

(3.5.9)

for all κ>0whereu

(x)=u(x, 0), u

(x)=u

(x, 0) = max(−R, u

(x)). The

left-hand side of (3.5.9) is always negative if either u

(x) ≤ R

or u

(y) ≤ R

for

= K +2R. Thus (3.5.9) is equivalent to

(x)+u

(y) − 2u

(z) − N

(|x + y − 2z|) ≤ 0, x,y,z ∈ R

(3.5.10)

for all κ>0. Since u

∗

is in BUC(R

), u

∗

is in BUC(R

× [0,T)) for every

∗

> 0 by the uniqueness of solutions. By this uniform continuity and (3.5.10)

there exists a small r

> 0 such that

sup{u

(x, s

)+u

(y,s

) − 2u

(z,s

) − N

(|x + y − 2z|)

−γs

− ϕ

); s

=0, |s

− s

|≤r

, |s

− s

|≤r

(x, s

), (y, s

), (z,s

) ∈ Q}≤µ

/2.

By the choice of R

this implies

sup {Φ

(x, s

,y,s

,z,s

); s

=0, |s

− s

|≤r

, |s

− s

|≤r

(x, s

), (y, s

), (z, s

) ∈ Q}≤µ

/2.

(3.5.11)

Since Φ

≤ 2K +2R,Φ

(x, s

,y,s

,z,s

) > 0 implies

− s

)

+(s

− s

)

≤ 2(K + R)/β. (3.5.12)

3.5. Lipschitz preserving and convexity preserving properties 143

Take β

so large that (3.5.2) with β>β

implies that |s

−s

|≤r

and |s

−s

|≤

. Then from (3.5.11) it follows that (3.5.8) implies s

> 0, s

> 0if

β>β

We shall distinguish two cases whether or not the supremum of Φ

is ap-

proximated at x, y, z with the property that x + y − 2z is close to zero. We shall

always assume that β>β

.Weset

(r, β):= sup{Φ

(x, s

,y,s

,z,s

); |x + y − 2z|≤r,

(x, s

), (y, s

), (z,s

) ∈ Q}≤θ(β).

By the deﬁnition of µ

there exists r

∗

> 0 (independent of β) such that µ

(r, β) ≥

3µ

/4forr<r

∗

Case 1. For each r ∈ (0,r

∗

)thereisβ

such that θ(β

)=µ

(r, β

)andβ

→∞

as r → 0.

We argue in a similar way to the proof of Case 1 of Theorem 3.1.4 (§3.4.2).

We ﬁrst ﬁx r ∈ (0,r

∗

). By the deﬁnition of µ

there is a sequence {a

} with

=(x

) such that

) ≥ θ(β) − 1/m and |x

+ y

− 2z

|≤r

with β = β

. We may assume that (s

) → (ˆs

, ˆs

), x

+ y

−2z

→ ˆω

for some ˆs

, ˆs

∈ [0,T]andˆω ∈ R

with |ˆω|≤r by taking a subsequence. We

now consider the function

(x, t)=u(x, t) − ϕ

(x, t),

(x, t)=Af(|x + y

− 2z

|)+β(t − s

)

+f(|x + y

− 2z

− ˆω|)+(t − ˆs

)

+ γt,A = A(κ

Let (ξ

,τ

)beamaximumofψ

over Q. Such a point exists since lim

r→∞

f(r)=

∞. By deﬁnition we observe that

) ≤ ψ

(ξ

,τ

Adding u(y

) − 2u(z

) − β(s

− s

)

to both sides yields

) − f (|x

+ y

− 2z

− ˆω|) − (s

− ˆs

)

≤ Φ

) − f (|ξ

+ y

− 2z

− ˆω|) − (τ

− ˆs

)

with b

=(ξ

,τ

). Since Φ

≤ θ, this implies

f(|ξ

+ y

− 2z

− ˆω|)+(τ

− ˆs

)

≤ θ(β) − Φ

)+f(|x

+ y

− 2z

− ˆω|)+(s

− ˆs

)

(3.5.13)

and

) ≤ Φ

)+f(|x

+ y

− 2z

− ˆω|)+(s

− ˆs

)

. (3.5.14)

144 Chapter 3. Comparison principle

The inequality (3.5.13) implies that ξ

+ y

− 2z

→ ˆω, τ

→ ˆs

as m →∞

since x

+ y

− 2z

→ ˆω, s

→ ˆs

. The inequality (3.5.14) implies that

) >µ

/2(3.5.15)

for suﬃciently large m since Φ

) → θ(β). We may assume that (3.5.15) holds

for all m.

Since (3.5.15) implies (3.5.8), we observe that τ

> 0 for all m so that

(ξ

,τ

) ∈ Q

∗

.Sinceu is a subsolution in Q

∗

and since u − ϕ

is maximized at

(ξ

,τ

) ∈ Q

∗

, we have as in the proof of Theorem 3.1.4,

+ F (τ

, ∇ϕ

, ∇

) ≤ 0at(ξ

,τ

)

if p

:= ξ

+ y

− 2z

=0and

(ξ

,τ

)=2β(τ

− s

)+2(τ

− ˆs

)+γ ≤ 0

if p

=0.Sinceξ

+ y

− 2z

→ ˆω, τ

→ ˆs

, s

→ ˆs

,wesendm to inﬁnity to

get

2β(ˆs

− ˆs

)+γ + F (ˆs

,A∇

f(|p|),A∇

f(|p|)



≤ 0atp =ˆω (3.5.16)

if ˆω =0and

2β(ˆs

− ˆs

)+γ ≤ 0

if ˆω = 0 by the deﬁnition of f ∈F.

In the same way we study the maximum of

(y,s)=u(y, s) − ϕ

(y,s),

(y,s)=Af(|x

+ y − 2z

|)+β(s − s

)

+f(|x

+ y − 2z

− ˆω|)+(s − ˆs

)

and obtain

2β(ˆs

− ˆs

)+F (ˆs

,A∇

f(|p|),A∇

f(|p|)) ≤ 0atp =ˆω (3.5.17)

if ˆω =0and2β(ˆs

− ˆs

) ≤ 0ifˆω =0.

In the same way we study the maximum of

−

(z,s

)=−u(z, s

)+ϕ

−

(z,s

2ϕ

−

(z,s

)=−Af (|x

+ y

− 2z|) − β(s

− s

)

− β(s

− s

)

−f(|x

+ y

− z +ˆω|) − (s

− ˆs

)

and obtain

β(ˆs

− ˆs

)+β(ˆs

− ˆs

)+F (ˆs

∇

f(|p|), −

∇

f(|p|)) ≥ 0atp =ˆω (3.5.18)

3.5. Lipschitz preserving and convexity preserving properties 145

if ˆω =0andβ(ˆs

− ˆs

)+β(ˆs

− ˆs

) ≤ 0ifˆω =0.Ifˆω = 0, we easily get

a contradiction: γ ≤ 0 so we may assume that ˆω = 0. We consider (3.5.16) +

(3.5.17) −2× (3.5.18) and send r → 0. Since |ˆω|≤r, this yields a contradiction:

γ ≤ 0.

Case 2. There is r

∈ (0,r

∗

) such that for suﬃciently large β,sayβ>β

for

some β

, the inequality θ(β) >µ

,β)holds.

We set µ

= µ(r

,β)forβ>β

.Forδ>0 we introduce

βr

(a)=Φ

(a) − δ(|x|

+ |y|

+ |z|

)

with a =(x, s

,y,s

,z,s

). By deﬁnition

sup

βδ

↑ θ as δ ↓ 0.

Thus there is δ

= δ

(β) > 0 that satisﬁes

sup

βδ

> max(µ

,µ

/2) for 0 <δ<δ

since µ

≤ θ. Clearly, Ψ

βδ

attains a maximum at some points

ξ =(ˆx, ˆs

, ˆy, ˆs

, ˆz, ˆs

)

∈ W =

Q×Q×Q depending on β and δ. We suppress the dependence with respect

to β and δ for simplicity of notation but we shall later send δ to zero ﬁrst and β

to inﬁnity. Since (3.5.8) is fulﬁlled at

ξ,weseethat ˆs

is positive for i =1, 2, 3. By

the deﬁnition of µ

we observe that

−1

(2(K + R)/A(κ

)) ≥|ˆx +ˆy − 2ˆz| >r

;(3.5.19)

the left inequality follows from the fact that Ψ

βδ

is bounded from above by 2(K +

R)andΨ

βδ

(

ξ) > 0. This fact also yields the bound of δ(|ˆx|

+ |ˆy|

+ |ˆz|

)soin

particular we have

δ(|ˆx| + |ˆy| + |ˆz|) → 0asδ → 0. (3.5.20)

Since s

> 0forδ<δ

(β), i =1, 2, 3, the function

βδ

= u(x, s

)+u(y, s

) − 2u(z,s

) − δ(|x|

+ |y|

+ |z|

) − ψ,

ψ = N

(|x + y − 2z|)+ϕ

)

attains its maximum at

ξ =(ˆx, ˆs

, ˆy, ˆs

, ˆz, ˆs

)overQ

∗

×Q

∗

×Q

∗

for δ<δ

(β). We

now apply Theorem 3.3.3 with k =3,N

= N

= N +1, Z

= Z

∗

;  = 2 with projection P

and P

deﬁned by P

(x, s

,y,s

,z,s

)=(x, y, z)

and P

= I − P

. As in Case 2 of the proof of Theorem 3.1.1, for each λ

> 0there

exists X, Y, Z ∈ S

such that

−(

+ |B

|) ≤ I ≤

⎛

⎝

XO O

OY O

OO−Z

⎞

⎠

≤ B

+ λ

, (3.5.21)

146 Chapter 3. Comparison principle

(γ +

+2δˆx, X +2δI) ∈ P

2,+

u(ˆx, ˆs

), (3.5.22)

(

+2δˆy, Y +2δI) ∈ P

2,+

u(ˆy, ˆs

), (3.5.23)

(−

, −

(

+2δˆz),

(Z − 2δI)) ∈

2,−

u(ˆz, ˆs

), (3.5.24)

with

⎛

⎝

⎞

⎠

, (3.5.25)

where

denotes ∇

ψ evaluated at

ξ and so on. We shall ﬁx λ

.NotethatX, Y, Z

may depend on δ and β.

We shall derive a bound for X, Y, Z. As in Case 2 of the proof of Theorem

3.1.4 a direct calculation of (3.5.25) yields

= A{



(ρ)

S +(f



(ρ) −



(ρ)

)R},A= A(κ

S =

⎛

⎝

II− 2I

−2I −2I 4I

⎞

⎠

,R=

⎛

⎝

p ⊗ pp⊗ p −2p ⊗ p

−2p ⊗ p −2p ⊗ p 4p ⊗ p

⎞

⎠

with p =ˆx +ˆy − 2z, ρ = |p|.Since

(

ξ,

η,

ζ)S

⎛

⎝

⎞

⎠

= |ξ + η − 2ζ|

(

ξ,

η,

ζ)R

⎛

⎝

⎞

⎠

= |ξ + η − 2ζ,p|

, ξ,η,ζ ∈ R

so that O ≤ R ≤ S(≤ 6I), we observe that

(ρ)S ≥ B

≥ Af



(ρ)R ≥ O (3.5.26)

with K

(ρ)=f



(ρ)/ρ + f



(ρ). We thus have an estimate

|≤AK

(ρ)|S| =6A(κ

(ρ)(3.5.27)

since |S| =6.SinceS

=6S, R

=6R, RS = SR =6R, as in Case 2 of the proof

of Theorem 3.1.4

=6A

{(



(ρ)

)

S +((f



(ρ))

− (



(ρ)

)

)R}

≤ 6A

(ρ)S (3.5.28)

3.5. Lipschitz preserving and convexity preserving properties 147

with K

(ρ)=(f



(ρ)/ρ)

+(f



(ρ))

. Applying (3.5.26)–(3.5.28) to (3.5.21) we

obtain

−(

+6AK

(ρ))I ≤

⎛

⎝

XO O

OY O

OO−Z

⎞

⎠

≤ A(K

(ρ)+6λ

(ρ))S.

This inequality implies that X + Y ≤ Z since

(

ξ,

ξ)S

⎛

⎝

⎞

⎠

= |ξ + ξ − 2ξ|

=0.

Moreover, it provides a bound for X, Y and Z,

|X|, |Y |, |Z|≤6AK

(ρ)+12A

(ρ)λ

(3.5.29)

with ρ = |ˆx +ˆy − 2ˆz|. By (3.5.19) the right-hand side of (3.5.29) is bounded as

δ → 0andβ →∞.

By (3.5.19) and (3.5.20) we observe that

+2δˆx =0,

+2δˆy =0, −

− 2δˆz =0

for suﬃciently small δ>0. Since u is a solution, the relation (3.5.22)–(3.5.23)

yields

2β(ˆs

− ˆs

)+γ + F (ˆs

+2δˆx, X +2δI) ≤ 0, (3.5.30)

2β(ˆs

− ˆs

)+F (ˆs

+2δˆy, Y +2δI) ≤ 0. (3.5.31)

Since X + Y ≤ Z, the relation (3.5.24) with degenerate ellipticity yields

β(ˆs

− ˆs

)+β(ˆs

− ˆs

)+F (ˆs

, −

(

+2δˆz),

(X + Y − 2δI)) ≥ 0. (3.5.32)

Since

= −

/2=Af



(ρ)p/ρ, sending δ → 0 in the inequality (3.5.30) +

(3.5.31) −2× (3.5.32) yields

γ + F (

,µ,X)+F(s

,µ,Y ) − 2F (s

,µ,(X + Y )/2)) ≤ 0

with µ = Af



(|p|)p/|p| with some X,Y ∈ S

, p ∈ R

, s

∈ [0,T]. Since

X,Y are bounded as β →∞by (3.5.29) and |p|

−1

, |p| is bounded as σ →∞by

(3.5.19), sending β →∞implies

γ + F (˜s, ˜µ,

X)+F (˜s, ˜µ,

Y ) − 2F (˜s, ˜µ, (

X +

Y )/2) ≤ 0

by (3.5.12) for some ˜s ∈ [0,T],

Y ∈ S

,˜µ ∈ R

.SinceX → F (t, r, X)is

convex in X, this yields γ ≤ 0 which is a contradiction. We have thus proved

≤ 0 to get (3.5.6). 