Integral inequalities for infimal convolution and Hamilton-Jacobi equations

0.34 MB·English

by  Patrick J. Rabier

INTEGRAL INEQUALITIES FOR INFIMAL CONVOLUTION
AND HAMILTON-JACOBI EQUATIONS
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PATRICKJ.RABIER
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This paper isdedicated to the memory of Jean Jacques Moreau

n
a
J

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1 Abstract. Let f,g : RN → (−∞,∞] be Borel measurable, bounded below
and suchthat inff+infg≥0. Weprove that withmf,g :=(inff−infg)/2,
] the inequality||(f−mf,g)−1||φ+||(g+mf,g)−1||φ ≤4||(f(cid:3)g)−1||φ holds in
A every Orliczspace Lφ,wheref(cid:3)g denotes theinfimalconvolution off and g
F andwhere||·||φ istheLuxemburgnorm(i.e.,theLp normwhenLφ=Lp).
Although no genuine reverse inequality can hold in any generality, we
h. also prove that such reverse inequalities do exist in the form ||(f(cid:3)g)−1||φ ≤
at 2N−1(||(fˇ−mf,g)−1||φ+||(gˇ+mf,g)−1||φ),wherefˇandgˇaresuitabletrans-
formsoff andgintroducedinthepaperandreminiscentof,yetverydifferent
m
from,nondecreasingrearrangement.
[ Similar inequalities are proved for other extremal operations and applica-

tions are given to the long-time behavior of the solutions of the Hamilton-
1
Jacobi andrelatedequations.
v
3
1 1. Introduction
5
If f,g : RN → (−∞,∞], the infimal convolution f(cid:3)g : RN → [−∞,∞], first
4
0 introduced by Fenchel [9] and Moreau [24], [25], [26], is defined by the formula
.
1 (f(cid:3)g)(x):= inf (f(x−y)+g(y)).
0 y∈RN
5
Since then, this operationandits extension to generalvector spaces have found an
1
evergrowingvarietyofapplications,includingconvexfunctions[13], [29],extension
:
v ofLipschitz functions [12],solutionsofthe Hamilton-Jacobiequations[2],[20],[31]
i
X andmuch more(even a proofof the Hahn-Banachtheorem[11]). In fact, there are
bynowseveralthousandspublicationsusing infimalconvolutioninareasasdiverse
r
a as image processing, economics and finance, information theory, probabilities and
statistics, etc. For a glimpse into some of these problems, see the excellent recent
survey by Lucet [21].
In this paper, we investigate the mathematical properties of infimal convolution
in a new direction, by exploring the existence of integral inequalities involving f,g
andf(cid:3)g.Theremarkthatf(cid:3)g=0wheneverf ≥0andg ≥0areintegrablecould
cast serious doubts on the value of this program,but they are quickly dispelled by
the rebuttal that no similar triviality arises from the integrability of f−1 and g−1.
Here and everywhere else, f−1 := 1/f, g−1 := 1/g, etc. This notation will not be
used to denote any set-theoretic inverse.
1991 Mathematics Subject Classification. 26D15,46E30,35F25,49L25.
Key words and phrases. Brunn-Minkowski inequality, enclosing ball, Hamilton-Jacobi equa-
tions,infimalconvolution, Orliczspace,rearrangement.
1
2 PATRICKJ.RABIER
Omitting technicalities to which we shall return shortly, the first batch of in-
equalitieswillrelatethe(Luxemburg)norm||(f(cid:3)g)−1|| inanyOrliczspaceL ,to
φ φ
the norms ||(f −z)−1|| and ||(g+z)−1|| for a suitable constantz independent of
φ φ
φ, to be defined in due time. The only restrictions are that f and g must be Borel
measurable, bounded below and that f(cid:3)g ≥ 0. The proofs depend crucially upon
(a slightly weaker form of) the Brunn-Minkowski inequality.
ThesettingofOrliczspacesinsteadofjusttheclassicalLpspacesintroducesonly
mildadditionaltechnicalities,ismorenaturalinmanyrespectsand,asweshallsee
intheexamplesofSection7,isusefulinsomeapplications. Itdoesnotevenrequire
any knowledge of Orlicz spaces beyond the definitions of Young functions and of
the Luxemburg norm, which will both be reviewed.
Thisbeingsaid,asimplespecialcaseassertsthatiff,g ≥0areBorelmeasurable
and inff =infg (see Theorem 3.4 for a full and much more general statement)
(1.1) ||f−1|| +||g−1|| ≤4||(f(cid:3)g)−1|| ,
p p p
for every 1 ≤ p ≤ ∞, where ||·|| is the norm of Lp := Lp(RN). The constant
p
4 is best possible among all constants independent of p, as is readily seen when
f =g =1 and p=∞.
The Borel measurability requirement has to do with the measurability of f(cid:3)g,
without which (1.1) cannot make sense. Curiously, we were unable to find a dis-
cussion of the measurability properties of the infimal convolution in the classical
literature, but the evidence points to the fact that f and g Lebesgue measurable
does not suffice for the measurability of f(cid:3)g. Indeed, as is well-known, the strict
epigraph of f(cid:3)g is the (vector, also called Minkowski) sum of the strict epigraphs
of f and g and Sierpin´ski [30] showed, almost a century ago, that the sum of two
Lebesgue measurable sets need not be Lebesgue measurable. In contrast, the sum
of two Borel sets is always Lebesgue measurable (but not always a Borel set). See
Section 2 for further details.
A peculiar feature of (1.1) and of more general similar inequalities is that only
the left-hand side is unchangedby modifications of f and g on null sets, as long as
Borel measurability and inff =infg >−∞ are preserved.
Mostofthe paperisactuallydevotedtoperhapsmoreimportant-anddefinitely
more delicate- reverse inequalities which, in a simpler world, would read
(1.2) ||(f(cid:3)g)−1|| ≤C(||f−1|| +||g−1|| ),
p p p
withC >0independent off andg insomesuitableclassofnonnegativefunctions.
Unfortunately, the main obstacle to (1.2) is that no remotely general converse of
the Brunn-Minkowski inequality holds in any form, even for convex sets. Such a
converse is actually trivially true for Euclidean balls, but a direct application of
this remarkonly yields (1.2) for a narrowsubclass ofradially symmetric functions.
To take advantage of the converse of the Brunn-Minkowski inequality for balls
in a much broader setting, we introduce a new function transform, strongly remi-
niscent of, yet very different from, nonincreasing rearrangement. The difference is
that the upper level sets are rounded before being rearranged, the rounding being
performed by using the concept of enclosing ball (see Section 4).
To each function f : RN → [−∞,∞] (no measurability needed), the aforemen-
tioned transform associates a measurable radially symmetric function fˆ, which in
turn produces another measurable radially symmetric function fˇ:=−(−fˆ). In the
special case when f,g ≥ 0 are Borel measurable and inff = infg (see Theorem
INEQUALITIES FOR INFIMAL CONVOLUTION 3
6.1 for a full and much more generalstatement), the reverseinequality in Lp reads
(compare with (1.1) under the same assumptions)
(1.3) ||(f(cid:3)g)−1|| ≤2N−1(||fˇ−1|| +||gˇ−1|| ).
p p p
Such an inequality breaks down completely if fˇand gˇ are replaced with f and
g, respectively, even if both functions are radially symmetric. For example, if
N =1,f(x)=x2+1 and g(x)=x2+1 when x∈/ Q,g(x)=1 if x∈Q, then f and
g are Borelmeasurable andinff =infg =1. But (f(cid:3)g)−1 =1/2is in no Lp space
with p<∞, whereas f−1 and g−1 =f−1 a.e. are in all of them. (In this example,
it turns out that fˇ=f but gˇ=1. ) This example also shows that, unlike in (1.1),
neither side of (1.3) is independent of modifications of f or g on null sets that do
not affect Borel measurability or inff =infg.
When not trivial (i.e., fˇ = f a.e.), the explicit calculation of fˇ is generally
not possible. Nevertheless, the inequality (1.3) is useful because some simple and
generalconditionsaboutf andgensurethefinitenessoftheright-handside(Lemma
6.3). There is certainly more to be discoveredin that regard.
The proofs of the inequalities involve two other classical extremal operations
(f ⊼g)(x):=supmin{f(x−y),g(y)} and (f ⊻g)(x):=infmax{f(x−y),g(y)}.
y y
Either of these operations fully determines the other, but both notations will be
useful. In general, f ⊻g = −(−f)⊼(−g) and, for nonnegative functions, f ⊻g =
(f−1⊼g−1)−1willbeimportant. Wealsoproveinequalitiessimilarto(1.1)and(1.3)
for the operations ⊼ and ⊻ (both being often referred to as “level sum” operations
in the literature). In fact, a good part of the work will consist in proving integral
inequalities for ⊼, from which those for (cid:3) and ⊻ will be derived.
In the last section, the inequalities are used to obtain Lp (and other) estimates
for the inversesof solutions of the Hamilton-Jacobiequations and variants thereof.
Throughout the paper, µ denotes the N-dimensional Lebesgue measure and,
N
without a qualifier, measurability always means Lebesgue measurability.
2. Background
The purpose of this short section is to review the basic properties of the opera-
tionsmentionedintheIntroduction,tosetthenotationusedinfuturesectionsand
to settle basic measurability issues.
Recall that if X and Y are subsets of RN, their sum X+Y is defined by
{x+y :x∈X,y ∈Y} if X 6=∅ and Y 6=∅,
X +Y :=
(cid:26) ∅ if X =∅ or Y =∅.
The following key lemma is well-known. The “proof” below merely makes the
connection with the deep property behind it.
Lemma2.1. IfX and Y areBorel subsets of RN, their sumX+Y is measurable1.
Proof. In Euclidean space (any dimension), the continuous image of a Borel set
is Lebesgue measurable; see Federer [8, p. 69]. Since X ×Y is a Borel subset of
RN ×RN and the addition is continuous on RN, the result follows. (cid:3)
1EvenaSuslinset,butnotnecessarilyaBorelset.
4 PATRICKJ.RABIER
Giventwofunctions f,g :RN →[−∞,∞]andξ ∈R, call F+ andG+ the upper
ξ ξ
level sets
(2.1) F+ :={x∈RN :f(x)>ξ}, G+ :={x∈RN :g(x)>ξ}
ξ ξ
and call W+ the corresponding upper level set of f ⊼g :
ξ
(2.2) W+ :={x∈RN :(f ⊼g)(x)>ξ}.
ξ
It is a standard elementary property that
(2.3) W+ =F++G+.
ξ ξ ξ
By Lemma 2.1 and since f ⊻g =−(−f)⊼(−g), it follows at once from (2.3) that:
Lemma 2.2. If f,g : RN → [−∞,∞] are Borel measurable, then f ⊼g and f ⊻g
are measurable.
If f :RN →[−∞,∞] we set
(2.4) M :=supf, m :=inff.
f f
The next relations are elementary, but important
(2.5) Mf⊼g =min{Mf,Mg}, mf⊼g ≥min{mf,mg},
Mf⊻g ≤max{Mf,Mg}, mf⊻g =max{mf,mg}.
We now turn to infimal convolution. Given a function f : RN → (−∞,∞], we
denote by E :={(x,ξ)∈RN ×R: f(x)<ξ} the strict epigraph of f. It is also a
f
simple well-known property that if g :RN →(−∞,∞] is another function, then
Ef(cid:3)g =Ef +Eg.
Since a function is Borel measurable (measurable) if and only if its strict epigraph
is a Borel set (measurable), it follows from Lemma 2.1 that
Lemma 2.3. If f,g : RN →(−∞,∞] are Borel measurable, then f(cid:3)g is measur-
able.
For future use, we also note that if z ∈R,
(2.6) f(cid:3)g =(f −z)(cid:3)(g+z).
3. First integral inequalities
The Brunn-Minkowski inequality (see e.g. Gardner’s survey [10]) asserts that if
X,Y are nonempty measurable subsets of RN and if X +Y is measurable, then
µ (X +Y)1/N ≥ µ (X)1/N +µ (Y)1/N. Obviously, it fails if X or Y is empty
N N N
and the other has positive measure. We shall only need the less sharp form
(3.1) µ (X +Y)≥µ (X)+µ (Y),
N N N
if X,Y and X +Y are measurable and X 6=∅,Y 6=∅.
Lemma 3.1. If f,g :RN →[0,∞] are measurable and M =M (possibly ∞; see
f g
(2.4)) and if f ⊼g is measurable, then
(3.2) f + g ≤ f ⊼g.
ZRN ZRN ZRN
INEQUALITIES FOR INFIMAL CONVOLUTION 5
Proof. Set M := M = M ≤ ∞. If M = 0, then f = g = f ⊼g = 0 and (3.2) is
f g
trivial. In what follows, M >0.
Since f ≥ 0, it is well-known (see (2.1)) that f = ∞µ (F+)dξ. By us-
RN 0 N ξ
ing F+ = ∅ when ξ ≥ M, this reads f = MRµ (F+)Rdξ. Likewise, g =
ξ RN 0 N ξ RN
Mµ (G+)dξ, so that f + g =R M(µ (RF+)+µ (G+))dξ. R
0 N ξ RN RN 0 N ξ N ξ
R By (2.2) and (2.3) anRd since fR⊼g is mReasurable by hypothesis, F++G+ =W+
ξ ξ ξ
is measurable for every ξ. If ξ < M, then F+ 6= ∅ and G+ 6= ∅ by definition of
ξ ξ
M and so, by (3.1), µ (F+)+µ (G+) ≤ µ (F+ +G+) = µ (W+). Therefore,
N ξ N ξ N ξ ξ N ξ
f + g ≤ Mµ (W+)dξ ≤ ∞µ (W+)dξ (by (2.5), the second inequality
RN RN 0 N ξ 0 N ξ
iRsevenaRnequalitRy). Now, ∞µ (WR +)dξ = f⊼g sincef⊼g ≥0andtheproof
0 N ξ RN
is complete. R R (cid:3)
Of course, (3.2) does not follow from a pointwise inequality. The condition
M =M cannotbe dropped. For example, if f >0 and g =0, then f⊼0=0 and
f g
(3.2) fails.
Lemma 3.1 is just the stepping stone for much more generalinequalities. Recall
that in the theory of Orlicz spaces, a nonconstant function φ : [0,∞] → [0,∞]
is called a Young function if φ(0) = 0 and φ is nondecreasing, convex and left
continuous([27];seealso[1],[18]forasimplifiedtreatmentlimitedtoN-functions).
In particular, φ(∞)=∞.
Remark 3.1. If φ is a Young function and h : RN → [0,∞] is measurable, the
monotonicity of φ shows at once that φ(h) is measurable.
If φ is a Young function, the corresponding Orlicz space L consists of all the
φ
measurable functions h on RN such that φ(λ|h|) < ∞ for some λ > 0 (this
RN
makes sense by Remark 3.1). It is a (compRlete) normed space for the Luxemburg
norm ||·|| defined by
φ
(3.3) ||h|| :=inf r >0: φ(r−1|h|)≤1 .
φ
(cid:26) ZRN (cid:27)
Since the right-hand side of (3.3) is finite if and only if h ∈ L , it will always be
φ
understood that ||h|| = ∞ when h is measurable and h ∈/ L . Thus, h ∈ L is
φ φ φ
equivalent to ||h|| <∞. Furthermore, it is readily checked that
φ
(3.4) ||h|| ≤||k|| if |h|≤|k|
φ φ
and, by the left-continuity of φ and monotone convergence2, that if h∈L ,
φ
(3.5) φ(||h||−1|h|)≤1.
ZRN φ
If φ(τ) := τp for some 1 ≤ p < ∞, then ||h|| = ||h|| . On the other hand,
φ p
||h|| = ||h|| when φ is the indicator function of [0,1] (φ = 0 in [0,1] and ∞
φ ∞
outside).
Lemma 3.2. If φ is a Young function and if h : RN → [0,∞], then (see (2.4))
M =φ(M ).
φ(h) h
2Thisisofcourseawell-knowninequality.
6 PATRICKJ.RABIER
Proof. It is plain that h ≤ M implies φ(h) ≤ φ(M ), so that M ≤ φ(M ).
h h φ(h) h
It only remains to show that φ(M ) ≤ M , which is trivial if M = 0. We
h φ(h) h
henceforth assume M >0.
h
By the monotonicity of φ and φ(0) = 0, there is τ ∈ [0,∞] such that φ = ∞
1
on (τ ,∞] and that φ < ∞ on [0,τ ). Specifically, τ = sup{τ ≥ 0 : φ(τ) < ∞}.
1 1 1
If τ ∈ (0,∞), then φ(τ ) may be finite or infinite. We split the proof into three
1 1
cases.
(i) M >τ .Ifso,τ <∞andthenφ(M )=∞.Theset{x∈RN :h(x)>τ }
h 1 1 h 1
is not empty and φ(h(x))=∞ for every x in that set. Thus, {x∈RN :φ(h(x))=
∞}=6 ∅, so that M =∞=φ(M ).
φ(h) h
(ii) M =τ =∞. Then, φ(M )=φ(∞)=∞. If τ >0 is finite, ∅=6 {x∈RN :
h 1 h
h(x) > τ} ⊂ {x ∈ RN : φ(h(x)) ≥ φ(τ)}(by the monotonicity of φ). As a result,
M ≥φ(τ). By letting τ →∞=τ and since lim φ(τ)=φ(∞)=∞ by the
φ(h) 1 τ→∞
left continuity of φ, it follows that M =∞=φ(M ).
φ(h) h
(iii) 0<M ≤τ .IfM =∞,thenτ =∞and(ii)aboveapplies. Assumenow
h 1 h 1
M <∞.Foreveryε>0,S :={x∈RN :h(x)>M −ε}6=∅.Ifεissmallenough,
h h
then M −ε>0 and S ⊂{x∈RN :φ(h(x))≥φ(M −ε)} by the monotonicity of
h h
φ. Hence, M ≥φ(M −ε). Since φ is left continuous, M ≥φ(M ). (cid:3)
φ(h) h φ(h) h
From Lemma 3.1 and Lemma 3.2, we obtain:
Lemma 3.3. Let φ : [0,∞] → [0,∞] be a Young function. If f,g : RN → [0,∞]
are Borel measurable and if M =M (possibly ∞), then f ⊼g is measurable and
f g
(3.6) max{||f|| ,||g|| }≤||f ⊼g|| .
φ φ φ
Proof. By Lemma 2.2, f ⊼g is measurable and so, by Remark 3.1, φ(f),φ(g) and
φ(f⊼g)aremeasurable. Since φ(min{f(y),g(x−y)})=min{φ(f(y)),φ(g(x−y))}
by the monotonicity of φ, we infer that sup φ(min{f(y),g(x − y)}) = (φ(f) ⊼
y
φ(g))(x). By Lemma 3.2 with h(y) := min{f(y),g(x−y)}, the left-hand side is
φ((f⊼g)(x)),sothatφ(f⊼g)=φ(f)⊼φ(g).Inparticular,φ(f)⊼φ(g)ismeasurable.
SinceM =M ,thenM =M ,onceagainbyLemma3.2. Thus,fromthe
f g φ(f) φ(g)
aboveandfromLemma3.1withf andg replacedwithφ(f)andφ(g),respectively,
(3.7) φ(f)+ φ(g)≤ φ(f ⊼g).
ZRN ZRN ZRN
If r > 0, then r−1(f ⊼g) = r−1f ⊼r−1g and Mr−1f = Mr−1g. Thus, (3.7) for
r−1f and r−1g yields φ(r−1f) ≤ φ(r−1(f ⊼g)), so that ||f|| ≤ ||f ⊼g||
RN RN φ φ
by (3.3). Likewise, ||g|R|φ ≤||f ⊼g||φ aRnd (3.6) follows. (cid:3)
Ofcourse,whenL =L1,Lemma3.1yieldsthestronger||f|| +||g|| ≤||f⊼g||
φ 1 1 1
but (3.6) is optimal when φ is arbitrary (let f =g =1 and L =L ).
φ ∞
We are now in a position to prove our first main integral inequality for infimal
convolution. Recall once more the notation (2.4).
Theorem 3.4. Suppose that f,g : RN → (−∞,∞] are Borel measurable, that
m ,m ∈R and that m +m ≥0. Set
f g f g
(3.8) m =(m −m )/2.
f,g f g
Then, f −m ,g+m and f(cid:3)g are measurable and nonnegative and
f,g f,g
(3.9) ||(f −m )−1|| +||(g+m )−1|| ≤4||(f(cid:3)g)−1|| ,
f,g φ f,g φ φ
INEQUALITIES FOR INFIMAL CONVOLUTION 7
for every Young function φ.
Proof. The measurability of f(cid:3)g was established in Lemma 2.3. Next, inf(f −
m )=inf(g+m )=(m +m )/2≥0,whencef(cid:3)g =(f−m )(cid:3)(g+m )≥0
f,g f,g f g f,g f,g
by(2.6). Thisalsoimplies(f−m )−1 ≥0,(g+m )−1 ≥0andsup(f−m )−1 =
f,g f,g f,g
2(m +m )−1 = sup(g+m )−1. Therefore, the inequality (3.6) is applicable in
f g f,g
the form
||(f −m )−1|| +||(g+m )−1|| ≤2||(f −m )−1⊼(g+m )−1||
f,g φ f,g φ f,g f,g φ
and (3.9) follows from 0 ≤ h−1⊼k−1 = (h⊻k)−1 ≤ 2(h(cid:3)k)−1 when h and k are
nonnegative, from f(cid:3)g =(f −m )(cid:3)(g+m )≥0 and from (3.4). (cid:3)
f,g f,g
It was noted in the Introduction that the constant 4 in (3.9) is already best
possible when L =L∞.
φ
Remark 3.2. Theorem 3.4 gives a simple necessary condition for the existence of
solutions of infimal convolution equations (see [23], [21] and thereferences therein):
Supposethat h≥0is measurableandthatg is Borelmeasurableandboundedbelow.
If h−1 ∈L for some Orlicz space L and ||(g−m +m /2)−1|| > 4||h−1|| (in
φ φ g h φ φ
particular, if (g − m + m /2)−1 ∈/ L ), the equation f(cid:3)g = h has no Borel
g h φ
measurable solution f. Indeed, if f exists, then m = m − m ∈ R and (3.9)
f h g
cannot hold.
Ifz 6=0isaconstant,thereisnosimplepointwiserelationshipbetween(f−z)⊻
(g+z) and f ⊻g. As a result, the method of proof of Theorem 3.4 does not yield
a variant of (3.5) or (3.9) with f(cid:3)g replaced with f ⊻g. However, if f,g ≥0, then
0≤f ⊻g ≤f(cid:3)g and such a variantcan be obtained as a straightforwardcorollary
of Theorem 3.4:
Corollary 3.5. Suppose that f,g :RN →[−∞,∞] are Borel measurable, that g ≥
0 and that f 6≡∞ and g 6≡∞, so that 0≤m +m <∞, where f :=max{f,0}.
f+ g +
Then, f −m ,g+m (see (3.8 )) and f⊻g are measurable and nonnegative
+ f+,g f+,g
and
(3.10) ||(f −m )−1|| +||(g+m )−1|| ≤4||(f ⊻g)−1|| ,
+ f+,g φ f+,g φ φ
for every Young function φ.
Proof. Since g ≥ 0, it follows that 0 ≤ f ⊻g = f ⊻g ≤ f (cid:3)g. By Lemma 2.2,
+ +
f ⊻g =f ⊻g is measurable. Therefore, the corollary follows from (3.4) and from
+
Theorem 3.4 for f and g. (cid:3)
+
The constant 4 is also best possible in (3.10) (among constants independent
of φ): If f = 1 and g = ℓ > 1 is constant, the inequality for the L∞ norm is
4(ℓ+1)−1 ≤ 4ℓ−1. By letting ℓ → ∞, it follows that, in the right-hand side, 4
cannot be lowered.
4. The radial transforms fˆand fˇ
The proofof Lemma 3.1 showsthat the existence ofa converseof the inequality
(3.2), that is,
f ⊼g ≤C f + g ,
ZRN (cid:18)ZRN ZRN (cid:19)
8 PATRICKJ.RABIER
with C > 0 independent of f and g would require µ (F++G+) ≤ C(µ (F+)+
N ξ ξ N ξ
µ (G+)) for everyξ >0. However,aspointed outinthe Introduction, no converse
N ξ
ofthe Brunn-Minkowskiinequality orits weakerform(3.1)holds inanygenerality.
The transforms defined in this section will enable us (in the next section) to
take advantage of the fact that such a converse trivially exists when X and Y are
Euclidean balls. The thought that this case is so special that it cannot have any
broad value would result in a serious oversight.
By a classicaltheoremof Jung [8, p. 200],[17], everynonempty bounded subset
X of RN is containedina unique closedball B with minimal diameter amongall
X
closed balls containing X, called the enclosing ball of X. If X is unbounded, no
closedball contains X and we set B :=RN. Lastly, if X =∅, every singleton{x}
X
satisfiesthe“minimaldiameter”requirement,whenceuniqueness,butnotexistence,
is lost. For definiteness, we arbitrarily set B := {0}. Evidently, X ⊂ B in
∅ X
all cases. Jung’s theorem also provides the estimate diam(X) ≤ diam(B ) ≤
X
2N/(N +1)diam(X), but we shall only make use of the (trivial) first one.
p
Remark 4.1. An easily overlooked aspect of enclosing balls is that X ⊂Y implies
only µ (B )≤µ (B ) but not B ⊂B , unless N =1.
N X N Y X Y
The following property of enclosing balls will be important.
Lemma4.1. IfX isanondecreasingsequenceofsubsetsofRN,thenµ (B )=
n N ∪Xn
limµ (B )=supµ (B ).
N Xn N Xn
Proof. Set X := ∪X . If X is unbounded, then B = RN and so µ (B ) = ∞.
n X N X
Since X := ∪X and X ⊂ X , the diameter of X and, hence, that of B ,
n n n+1 n Xn
tends to ∞. Accordingly, limµ (B )=∞.
N Xn
Suppose now that X is bounded, so that B is a ball. Since X ⊂ X im-
X n
plies µ (B ) ≤ µ (B ) and since µ (B ) is nondecreasing, it is plain that
N Xn N X N Xn
limµ (B ) ≤ µ (B ). To prove the converse, call r ≥ 0 the radius of B .
N Xn N X n Xn
The sequence r is nondecreasing and bounded above (by the radius of B ) and
n X
soit hasa limit r ≥r foreveryn. As aresult, limµ (B )is the measureofany
n N Xn
ball with radius r.
Next, call x the center of B . By a simple contradiction argument, the se-
n Xn
quence x is bounded (since x might not be in B -see Remark 4.1- this is not
n n X
totally trivial). After extracting a subsequence, assume that x → x ∈ RN. Ev-
n
ery y ∈ X is in X for n large enough. Since B ⊂ B(x ,r), it follows that
n Xn n
y ∈ B(x,r). Thus, X ⊂ B(x,r), whence µ (B(x,r)) ≥ µ (B ) by definition
N N X
of B . Since r = limr amounts to µ (B(x,r)) = limµ (B ), it follows that
X n N N Xn
limµ (B )≥µ (B ). (cid:3)
N Xn N X
Given any function f : RN → [−∞,∞], we now proceed to constructing a
measurable radially symmetric function fˆ:RN →[−∞,∞] whose upper level sets
Fˆ+ have measure equal to µ (B ) for every ξ. The construction follows that of
ξ N F+
ξ
the nonincreasing rearrangementof f.
Lemma 4.2. The function µ (B ) is nonincreasing and right-continuous on
N F+
ξ
[−∞,∞].
INEQUALITIES FOR INFIMAL CONVOLUTION 9
Proof. If ξ < η, then F+ ⊂ F+, so that µ (B ) ≤ µ (B ). For the right
η ξ N Fη+ N Fξ+
continuity, let ξ ց ξ, so that F+ = ∪F+ and then, by Lemma 4.1, µ (B ) =
n ξ ξ N F+
n ξ
limµ (B ). (cid:3)
N F+
ξn
Call ρ+(ξ) the radius of B . Since ρ+(ξ) is proportional to µ (B )1/N, it
f F+ f N F+
ξ ξ
follows from Lemma 4.2 that ρ+ is nonincreasing and right-continuous. Therefore,
f
(4.1) γ+(t):=inf{ξ :ρ+(ξ)≤t},
f f
is a nonincreasing and right-continuous function on [0,∞) and
(4.2) {t≥0:γ+(t)>ξ}=[0,ρ+(ξ)).
f f
Indeed,whenf ≥0andρ+(ξ)isreplacedwithµ (F+),γ+becomesthenonincreas-
f N ξ f
ingrearrangementoff andthese propertiesfollowuniquely fromthe monotonicity
andright-continuityofµ (F+); see for instance [33, pp. 26-27]. We alsopointout
N ξ
that in most modern expositions, the nonincreasing rearrangementof a function f
is defined to be that of |f|. This has not always been the case (see Day [7] or Lux-
emburg[22])andthemonotonicityandright-continuitypropertiesofnonincreasing
rearrangementsare independent of whether f or |f| is used in their definition.
We now set
(4.3) fˆ(x):=γ+(|x|).
f
Some basic properties of fˆare summarized in the next theorem.
Theorem 4.3. Given f : RN → [−∞,∞], the function fˆhas the following prop-
erties:
(i) fˆis measurable and fˆ=f a.e. if and only if f(x) is a.e. equal to a nonincreas-
ing function of |x|. If also f(x) is a right-continuous function of |x|, then fˆ=f.
(ii) µ (Fˆ+) = µ (B ) for every ξ ∈ [−∞,∞], where Fˆ+ denotes the upper ξ-
N ξ N F+ ξ
ξ
level set of fˆ.
(iii) M ≤ M and m ≥ m (in particular, f ≥ 0 ⇒ fˆ ≥ 0). Furthermore,
fˆ f fˆ f
M =esssupfˆand m =essinffˆ.
fˆ fˆ
(iv) (f +zˆ)=fˆ+z for z ∈R and (f(c·)ˆ)=fˆ(c·) for c∈R\{0}.
(v) (cfˆ)=cfˆfor every c≥0.
(vi) If h:RN →[−∞,∞] and h≤f, then ˆh≤fˆ.
(vii) If f is bounded below on bounded subsets and lim f(x) = −∞ and if
|x|→∞
h : RN → [−∞,∞] satisfies h(x) ≤ f(x) for |x| large enough, then hˆ(x) ≤ fˆ(x)
for |x| large enough. Furthermore, if f(x) is a strictly decreasing function of |x|
and if h(x) ≤ f(x) when x ∈/ B for some open ball B centered at the origin, then
ˆh(x)≤fˆ(x) (=f(x) by (i)) for every x∈/ B.
Proof. (i) The measurability of fˆfollows at once from the monotonicity of γ+ and
f
the necessity of the given conditions for fˆ = f a.e. is obvious. Conversely, if
f(x) = γ(|x|) with γ : [0,∞) → [−∞,∞] nonincreasing, the upper level sets of f
are balls centered at the origin (possibly RN) and ρ+ is the distribution function
f
of γ, so that γ+ = γ∗, the nonincreasing rearrangement of γ. Since γ∗ = γ except
f
perhaps at the countably many points of discontinuity of γ, it follows that fˆ= f
10 PATRICKJ.RABIER
a.e. Clearly, this remains true if f(x) = γ(|x|) a.e. If γ is right-continuous, then
γ∗ =γ and fˆ=f.
(ii) Just notice that, by (4.2) and (4.3), Fˆ+ is the open ball with center 0 and
ξ
radius ρ+(ξ).
f
(iii) With no loss of generality, assume M < ∞. Since γ+ is nonincreasing,
f f
maxγ+ =γ+(0)=inf{ξ :ρ+(ξ)=0}.Ifξ ≥M ,thenF+ =∅,whenceB ={0}
f f f f ξ F+
ξ
andsoρ+(ξ)=0.Thus,maxγ+ =inf{ξ :µ (F+)=0}≤M . Onthe otherhand,
f f N ξ f
by (4.3), maxγ+ = M . This shows that M ≤ M . Furthermore, M = esssupfˆ
f fˆ fˆ f fˆ
by (4.3) and the right-continuity of γ+.
f
That m ≥m is obvious if m =−∞, or if m =∞ (for then f =fˆ=∞). If
fˆ f f f
m ∈ R and ξ < m , then ρ+(ξ) = ∞. Thus, γ+ ≥ m by (4.1) and so fˆ≥ m ,
f f f f f f
whence m ≥m . That m =essinffˆfollows from the monotonicity of γ+.
fˆ f fˆ f
(iv) Since {x∈RN :f(x)+z >ξ}=F+ , it follows that ρ+ (ξ)=ρ+(ξ−z),
ξ−z f+z f
whichinturnyieldsγ+ =γ++z,i.e.,(f+zˆ)=fˆ+z.Theproofsthatf(c·ˆ)=fˆ(c·)
f+z f
if c∈R\{0} is equally straightforward.
(v) Since this is trivial when c = 0, assume c > 0. Then, ρ+(ξ) = ρ+(ξ/c),
cf f
whence γ+ =cγ+, i.e. (cfˆ)=cfˆ.
cf f
(vi)Ifh≤f,then(withaself-explanatorynotation)H+ ⊂F+andsoµ (B )≤
ξ ξ N H+
ξ
µ (B ). Hence, ρ+(ξ)≤ρ+(ξ) and, by (4.1), γ+ ≤γ+, so that ˆh≤fˆ.
N F+ h f h f
ξ
(vii) Choose an open ball B centered at the origin such that h(x) ≤ f(x) when
x ∈/ B. Since f is bounded below on bounded subsets, inf f is finite and, if
B
ξ < inf f, then B ⊂ F+ for every ξ ≤ ξ . By (vi), hˆ is increased when h is
0 B ξ 0
increased. Thus, if it can be shown that ˆh ≤ fˆafter increasing h, this inequality
also holds before h is increased. In particular, we may increase h on B so that
ξ < inf h and then B ⊂ H+ for every ξ ≤ ξ . Thus, B ⊂ H+ ∩F+ for every
0 B ξ 0 ξ ξ
ξ ≤ ξ . On the other hand, {x ∈/ B : h(x) > ξ} ⊂ {x ∈/ B : f(x) > ξ} since h ≤ f
0
onRN\B. Altogether,if ξ ≤ξ , then H+ ⊂F+ andso ρ+(ξ)≤ρ+(ξ). As a result,
0 ξ ξ h f
(4.4) γ+(t):=inf{ξ :ρ+(ξ)≤t}≤inf{ξ ≤ξ :ρ+(ξ)≤t}≤
h h 0 h
inf{ξ ≤ξ :ρ+(ξ)≤t}.
0 f
Since lim f(x) = −∞, the level set F+ is bounded, whence ρ+(ξ ) < ∞.
|x|→∞ ξ f 0
0
Choose any t ≥ ρ+(ξ ). If ξ > ξ , then ρ+(ξ) ≤ ρ+(ξ ) ≤ t by the monotonicity
f 0 0 f f 0
of ρ+, so that γ+(t) := inf{ξ : ρ+(ξ) ≤ t} = inf{ξ ≤ ξ : ρ+(ξ) ≤ t}. By (4.4),
f f f 0 f
γ+(t) ≤ γ+(t). Since this is true for every t ≥ ρ+(ξ ), it follows that hˆ(x) ≤ fˆ(x)
h f f 0
when |x|≥ρ+(ξ ).
f 0
To complete the proof, assume in addition that f(x) is a strictly decreasing
function of |x|. We show that hˆ(x)≤fˆ(x) when x∈/ B. If B =∅, the resultfollows
from (v). From now on, assume B 6=∅ (hence B 6={0} as well since B is open).
Bythemonotonicityoff in|x|,inf f <inf f ift>1.ItfollowsthatF+ ⊂tB
tB B ξ
0
ift>1andξ aboveiscloseenoughtoinf f.Thus,B ⊂tB,sothatρ+(ξ )≤tρ
0 B F+ f 0
ξ0
where ρ is the radius of B. Fromthe above, ˆh(x)≤fˆ(x) when |x|≥tρ and, hence,
when |x| ≥ ρ by first letting t → 1 (which gives only |x| > ρ) and next using the