R ENDICONTI
del
S EMINARIO M ATEMATICO
della
U NIVERSITÀ DI P ADOVA
M ICOL A MAR
V IRGINIA D E C ICCO
The uniqueness as a generic property for some one-dimensional segmentation problems
Rendiconti del Seminario Matematico della Università di Padova, tome 88 (1992), p. 151-173
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Article numérisé dans le cadre du programme Numérisation de documents anciens mathématiques
http://www.numdam.org/
for Some One-Dimensional Segmentation Problems.
MICOL AMAR - VIRGINIA DE
CICCO(*)
ABSTRACT - We
give
auniqueness
resultconcerning
the minimizers of the func- tionalproposed by
Mumford and Shah in order tostudy
theproblem
ofimage segmentation
inComputer
VisionTheory.
Our result concerns the modelcase in dimension one. It is easy to see that the
uniqueness
of this minimumproblem
does not hold, but we state that it is a«generic
property» in thesense that for «almost all» the
grey-level
functions and the parameters of theproblem,
the minimumpoint
isunique.
1. Introduction.
Given a
function g
EL’(0),
with f2 an open bounded subset ofR’,
and three realnumbers oc, fi,
y E(0, + ~ ],
let us consider the function- alwhere
Su
is thejumping
set of the function u andHn -1
is the n - 1 Hausdorff measure on R n.We can associate to the
following minimizing problem
where the minimum is taken on a suitable class of functions.
In the case n =
2,
the functional defined in(1.1)
wasproposed by
Mumford and Shah in
[11],
in order togive
a mathematicaldescription
to a
problem
ofimage segmentation
inComputer
VisionTheory.
(*) Indirizzo
degli
AA.: S.I.S.S.A., Strada Costiera 11, 34014 Trieste Italia.In
[11]
and[12],
Mumford and Shahconjectured
that has mini-mizers,
whosediscontinuity
set5~
ispiecewise
smooth. In[2]
Ambro- sioproved
the existence of the solution of(1.2)
in thegeneral
case of thespace dimension n % 1. Some results about the
regularity
ofSu
can befound in
[7].
In
[6]
DalMaso,
Morel and Solimini studied theparticular
casen =
2, giving
a constructiveproof
of the existence.Further results about this
problem
can be found in[ 1 ], [3]
and
[7].
Moreover,
we recall also that in[14],
the one-dimensional case has beenconsidered;
inparticular
thesmoothing properties given by
theformulation
(1.1)
of thesegmentation problem
have been studied.It is
possible
also to consider the functionaland the associated
problem
We
point
out that(1.3)
can be considered aparticular
case of(1.1),
inwhich we restrict our attention to the
piecewise
constant functions orequivalently
in which weput
« _ + 00 and~3
= 1. In the case of n =2,
a constructive methodprovides
the existence of minimizers forproblem (1.4),
asproved
in[9]
and[10].
Thegeneral
case n ~ 1 is studied in[5]
by Congedo
and Tamanini.However it is not
possible,
ingeneral,
to say that the minimizers for theseproblems
areunique.
To this pourpose, let us consider thesimple
case n =
1,
S~ _[0, 1]
and the function g:[0, 1] -~ R, g E L °° ([0, 1])
de-fined
by g(x)
= X[112, 11(X), where XE
is the characteristic function of the setE;
then the minimumproblem (1.4)
for that ghas,
asunique
sol-ution,
the function ul for 0 y1/4
and the function U2 =1/2
for y >
1/4,
butfor y
=1/4
both functions ul and U2 are solutions.From these
arguments,
one couldexpect
thatgiven
afunction g
EE L 2 (S~),
there isuniqueness
for these minimumproblems except
for a«small» set
(possibly countable)
of values of theparameter
y.Unfortunately,
this is not the case ingeneral,
as thefollowing
coun-terexample
shows.Let
g(x)
= 2/3] +2/(2/3,
n and consideragain
theproblem (1.4)
associated to this function g; then it is easy to prove thatfor y
>1/2
theunique
solution is ul = 1 and for 0 y1/6
theunique
solution is U2 == g, but for all the interval
1/6
y1/2
we have two solutions u3 ==
(3/2)~1/3, i]
and u4 =(1/2)
X[O, 2/3] +2X(2/3,
n andfinally for y
=1/6
thefunctions u2 , U3 and u4 are solutions and
for y = 1/2
the functions Ul, U3 and u4 are solutions.Actually,
we will see that for every non constantfunction g
E([0, 1])
it will bepossible
tofind y
E(0, + (0)
such that theproblems (1.2)
and(1.4)
have more than one solution.On the other
hand, fixed y
E(0, + 00),
we canfind g
EL 2 ([Og 1])
such that(1.4)
has more minimizers. In fact it isenough
totake, for
in-stance, g
=(1
+ X(0,1/2) + X(1/2, 1), and to observe thatFg (g)
(g), where 9
is the mean valueof g
on[0, ].
The sameproperty
holdsalso for
problem (1.2) (see Corollary
3.5 and Remark3.6).
These
arguments
lead us to observe that the best we canhope
is theuniqueness
for these minimumproblems only
if we restrict the func-tions g
or the values of theparameter y
to suitable«large»
subsets ofL 2 (~)
and I~ +respectively.
The aim of this paper
is, indeed,
togive
arigorous proof
of this fact forproblems (1.2)
and(1.4),
in dimension n = 1.The main result
is,
infact,
that for every ybelonging
to R +unique-
ness for
(1.2)
and(1.4)
is ageneric property
ofg E L 2 ([o, 1]).
Moreover,
for ageneric
gbelonging
toL 2 ([0, 1]), uniqueness
for(1.2)
and(1.4)
is ageneric property
ofy E R + .
To prove these
results,
weadapt
anargument
of G. Vidossich in[15]
to oursituation, following
the outline of Carriero and Pascali in[4].
More
precisely, given
a > 0and (3
>0,
we construct a countable sub- setm° ofL~([0, 1]),
dense inL 2 ([0, 1])
and a countable subset r of R+,
such that for every g E
No
and for every y E R +~1’, problem (1.2)
rela-tive
to g
has aunique
solution.Then, by
means ofNo,
for every y E R+,
we can construct a dense
GJ-subset
3lltof L ([0, 1]),
such that when the datum g is chosen in thecorresponding problem (1.2)
hasonly
oneminimizer.
Really
this result can beimproved by constructing
a denseG¿-subset
which works for all theparameters y
of a countable subset contained in R + . On the otherhand,
we can construct a denseG-subset
of
L~([0, 1])
such thatwhen g belongs
to thisset, problem (1.2)
isuniquely
solvableif y belongs
to thecomplement
of a countable subset rg in R +depending
on g.Similar
arguments
are used to obtainanalogous
results forproblem (1.4).
Since the
complement
of aG,-subset
ofL 2 ([0, 1])
is a set of first cat-egory and ru is countable it is clear now what we meant
by «large»
or«generic»
in theprevious
informal discussion. We observethat,
fromthis
point
ofview,
our results are in line with thegenericity
resultsof [4], [8], [13]
and[15].
In
particular,
the setJK°
will be constructedby
means of a suitableclass of
piecewise
constant functions. In order to find thisclass,
we willstudy
in detail theproperties
of thesolution,
and inparticular
its formand its
discontinuities,
when g ispiecewise
constant.The paper is
organized
as follows: in the second section we reformu-late the
problem
in a suitable way to the one-dimensional case, whichpermits
us to reduce(1.2)
and(1.4)
to thestudy
ofsimpler problems,
with fixed
jump term;
in the third section we state somepreliminary
results about the form of the solutions of
(1.2)
and(1.4)
and their con-tinuous
dependence
on the datum g;finally
section 4 contains the main theorems.Acknowledgments.
We would like to thank Prof. Dal Maso for sug-gesting
us thisproblem
and for the valuable attention that he gave usduring
the research.A
particular
thank also to Prof. E. Pascali and Prof. G. Vidossich for theirhelpful
advices.2. Formulation of the
problem.
We will write
L 2,
L 00 instead ofL 2 ([0, 1]),
L 00([0, 1]).
We will de- noteby
R + the subset ofstrictly positive
real numbers.In the
following, for j e N,
apartition ~, _ (b~ ) s ± o
of[0,1]
will be identified with a subset... , bj + 1 ~
of[0,1]
such that 0 =b° b1
...
bj+1 = l.
Fixed j
EN,
we denoteby x)
the space of all the functions uon
[0, 1]
such that there exists apartition ~, _ (bs ) s ± o
of[0, 1]
such thatthe restriction of u to
belongs
tobs + 1 )),
for everys =
0, ..., j. Therefore,
we defineH1 = U Xjl.
Forevery j
E N we con-sider also the subset
x)
ofXi composed by
the functions which haveexactly j jumps.
Moreover,
we denoteby S
the space of all thepiecewise
constantfunctions on
[0, 1].
It is easy to see that each function u E s can be writ- ten in the formU E
b$ + 1 > withf3s E R
fors = 0, ..., j
andj EN.
s =~’
Finally Su
is the set ofjump points
of the function ubelonging
to ~C Ior s and # is the
counting
measure on R.Given
g E L 2
andy E R +,
we consider thefollowing
functionalwhere 1 = #
(Su );
it is easy to see that the functionaldepends only
on uand not on its
representation.
Moreover we consider the functionaland the associated
problems:
we note that
(2.1)
is obtainedby (1.1)
with a= {3
= 1.We note that all the results we are
going
to prove still hold for finite«
and fi
different from1,
because it ispossible
to reduce thegeneral
functional to our case.
We observe that the existence for this
problems
will be discussed in thefollowing.
We
point
out that the results we aregoing
to obtain forproblem (2.3)
cannot be deriveddirectly
from those forproblem (2.2),
but sincethe method is the same in both cases, we treat
explicitly only problem (2.2), remarking,
when it is necessary, the differences and the analo-gies
withproblem (2.3).
Given g
EL 2,
for every r E R + we definewe shall see later that the minimum is achieved.
Moreover,
we consider the functional G 9 :[0, + 00] I
definedby
For
every j
E N we consider theproblem
The existence for this minimum
problem
followsby
the usualcompact-
ness
property
of the sequences ofpartitions
andby
the standard direct methods of calculus of variationapplied
on each subinterval of[o,1].
It is clear that
but,
in this case, the minimum is notalways
achieved.Let us define the non
empty
subset N9 of N of theintegers j
for whichthe value
M9
is attained on at least a function which hasexactly j jumps.
Moreover,
it can beeasily
seenthat j
E Ng if andonly
if the mini-mum of G on
Xi
is achievedand,
in this case,For
every j
E N and for every yE R +,
let us define nowSince
(y) ~ yj and y
>0,
it follows that for every -r E l~ + there exists the min I(y).
Moreover we aregoing
to prove thatIn
fact, given U
EX’
with #= j,
we haveand
taking
the infimum withrespect
to u E it follows thatThe
opposite inequality
is trivial.In order to prove that such an infunum is
attained,
we fix y E R + and we choosejo
E N such thatmin
=7%£ (y);
thisimplies
thatthere exists uo e
XJo
suchthat
hence the infimum in
(2.10)
is attained on uo andactually
it is a mini-mum, moreover
(2.9)
holds.We note that # =
jo ;
in fact if # = 1jo
we haveFg (uo) =
= G 9
(uo)
+yl
G 9 +yjo
and this contraddicts(2.10).
Therefore uo EE
x)
andjo
E This proves thatfor every y e
R +,
andthat,
if(y)
= min(y)
for some y e R+,
thenj E N
every minimizer uo of
problem (2.6)jo
hasexactly jo jumps,
i.e.u E K1j0.
This leads us to define the subset JQ of NY in the
following
way:
(see figure 1).
REMARK 2.1. It is clear that
(2.11)
can be rewritten asMoreover, if j
EJg,
every minimizer u ofproblem (2.6)j
hasexactly j jumps,
i.e. u EWl
IWe observe
that, by (2.8)~ , mjg (-r)
has a lineardependence
on y, henceby (2.12)
is a concave function(see figure 1).
Finally
wepoint
out that the sequence(MJ)j E N
isdecreasing
sinceXi c Xk for j
k. Inparticular, if j
EJg,
it isstrictly decreasing;
infact if
by
contradictionj,
k E Jgwith j
1~ andM9
=M~ ,
then for everyYER+
which
implies k g
J g.We
introduce,
forevery j
EJg,
the nonempty
subsets of Rand
(see figure 1).
REMARK 2.2. It is clear that
r9
is a(possible degenerate)
intervalof
R +,
since it can be rewritten asand is a concave function. Moreover the
intervals ’ j with j
EE J 9 are non
overlapping,
since theangular
coefficient of7%/
isstrictly
increasing with j,
andro
is unbounded. Hence for every and every ybelonging
to the interior ofPq
we havem 9
Given two consecutive
elements j and j’
ofJ9,
theequality m9 (y) =
=
m9 (y)
is satisfied for at most one y ER + ; finally
we note that Fg is the set of all theendpoints
of the intervals hence it is a discrete count- able subset of R + and theonly possible
accumulationpoint
is thepoint
y = 0
(see figure 1).
PROPOSITION 2.3. Fixed
y E R +,
we consider the nonempty
sub-set of Jg
Then is attained on u E
X’
if andonly
if thereexists j
EJ 9
suchthat u is a minimum
point
of theproblem
which defines Inparticu- lar,
if there exists aunique j
EJg
and if theproblem (2.6)j
hasunique-
ness, then also the
problem (2.2)
hasuniqueness.
PROOF.
If j = # (Su)
then u EXi. Hence,
if is attained on u,then
which
implies that j
EJ~
and thatM9
is attained on u.Viceversa,
andM9
is attained on u,then j
= #(Su) (see
Re- mark2.1 ).
The conclusion followsby
COROLLARY 2.4.
(i) If y
thenproblem (2.2)
has more thanone solution.
(ii)
If fora y belonging
to the interior ofy9 with j
e JQproblem
(2.2)
has more than onesolution,
then forall y belonging
to the interior ofrjg problem (2.2)
has notuniqueness.
PROOF.
(i)
Followsby
the definition of 1,9 andr9.
(ii)
If ul , u2 are minimizers ofF9 ,
thenby Proposition
2.3they
are minimizers of
G 9,
that is for everys = 1, 2 G 9 (us ) = M 9 .
Since forevery y
belonging
to the interior of we havewith j fixed,
it follows that ul , U2 are minimizers ofF.,g. 0
PROPOSITION 2.5. Let
Q= 10 be
a fixedpartition
of[0, 1]
andH1Q.
be the subset ofXi
constitutedby
the functions u =I 0 .8s (X)x(b" bu 1)(X)
with.8s
EH bs
+1»
for every s =0, 1, ... , j.
Then the functional G g defined in
(2.5)
hasexactly
one minimizer onH1Q.
PROOF. The existence is standard. For the
uniqueness
it is suffi-cient to observe that
xo is
a linearsubspace
of~C
and the functional G 9 isstrictly
convex onof
PROPOSITION 2.6. Fixed a
partition
of[0, 1]; if g
is a func-k
tion of the
type g = ~
P 1+where «i
E R for every i =0, ..., k,
then
fixed j
> k we have thatmg (y)
forevery y E R +,
... ,
1~~
and Fg is finite.PROOF. Since for
every j a
k g E 9 it follows thatM9
= 0. As-sume
by
contradictionthat, given j
> k thereexists y
E R + such that(Y)
=m9 (y),
thenBut,
if we take u = g, we have that Fl(g) ~ yk
which is a valuestrictly
less than and this is not
possible.
This
implies
also that forevery j
> k we obtainthat j g
then Jgis finite and is contained
and, by (2.13)
and Remark2.2,
rg is the set ofpoints y
E R + such that =mg (y)
=m9 (y), where j
and j’
are two consecutive elements Therefore it follows that is finite and contains at most kpoints.
REMARK 2.7. We may
analogously
define the minimumproblem
moreover we can consider the functional s ~
[0,
definedby
and the associated
problem
With suitable
modifications,
we can also introduce the definitions ofJ9, r9
andr9
relative to theproblem (2.3).
REMARK 2.8. In order to
explain
betterthe previous definitions,
we
give
an easyexample,
relative to the functional in which we em-phasize
thoseconcepts.
Let
g(x)
=~(1/3,2/3)(x).
An easy calculation shows that G 9 has oneminimizer ul on
hil
and one minimizer u4 onx) ,
and that Gg has twominimizer U2 and u3 on
~, i . Moreover,
ul is the minimizer of on~C1
for and u4 is the minimizer of on~C1 for y,
wherey
= o.l 1. Thispermits
us to construct thefollowing graph:
3.
Preliminary
results.In this section we state some results
concerning
the form of a sol- ution of the minimumproblem.
Inparticular,
wegive
anexplicit
for-mula,
in terms of g, for a minimumpoint
EX’
of theproblem (2.6)~ ,
and for a minimum
point ©
E 8 of theanalogous problem
for the func- tional without the derivativeterm,
and westudy
where such a mini-mum
point
canjump
and the continuousdependence
of it on the datumg. Moreover we
investigate
the nonuniqueness:
inparticular
we showhow it is
possible, fixed g
EL 2 (or y
E R+ ),
toconstruct y
E R +(or
g Ee L respectively)
for whose minimumproblems
have nonunique-
ness.
REMARK 3.1. It is easy to see
that, when j
EJg,
a minimizer u ofproblem (2.6)j
must be of the formwhere Q=
is apartition
of[0,1]
and for every s =0, ... , j /3s
isthe solution of the Euler
equation
in the subinterval(bs , bs + 1 )
of[0, 1],
i.e.
where
and cosh is the
hyperbolic
cosine.Finally, recalling
the definition of Jg and Remark2.1,
for every s =0, ... , j -
11 )
~fis +
1(bs + 1 ).
REMARK 3.2. In the case of
problem (2.3),
the situation is evenmuch
easier,
so thatfor j
Eig
a minimizer function has the formwhere
~, _ (b,Y,"
is apartition
of[0, 1]
and for s =0, ..., j
finally
we have 1 for every s =0, ... , j -
1.REMARK 3.3. and the solution
u(x) =
= li Øs(x)
b8+ 1)(x)
ofproblem (2. 6)~ ,
we have that for everys
= 0, ...,j
and every functionif for some s E
10, ...,jl
we have that[38
~In
fact, by Proposition
2.5 the functional G 9 defined in(2.5)
is strict-ly
convex onwhere Q = (bs)j+1s=0,
hence G 9 has aunique
minimizeron
Xi§i.
In the
following
two corollaries we will showthat, arbitrarily
fixedthe non constant
datum g
EL 2 or
theparameter y
ER +,
it ispossible
tochoose the
parameter Y E R +
or the datumg E L 2 respectively
suchthat
problems (2.2)
and(2.3)
have nonuniqueness.
COROLLARY 3.4. For every non constant function
(i)
thereexists -
E R + such thatFg
has more than one mini-mizer ;
(ii)
thereexists y E R +
such thatF9
has more than one mini-mizer.
PROOF. Let us fix a non constant
function g e L . (i)
Let uo be theunique
solution of theequation
with the Neumann conditions
u’(0)
= 0 =~’(1);
thenby
definitioni i
Mo
=)2 dx
+j(uo - g)2 dx. By (3.2), since g
is notconstant,
also uois
not constant; moreover 1]),
hence there exists b~(0,1)
such that
Uo (b) ~
0. Let nowv(z) = [30 (x)
b~(x)
+/3i (x)
X(b, 1)(x),
where[30
and[31
are the solutions of theequation (3.2)
with the Neumann con-ditions
[3ó (0)
= 0 =[3ó (b)
and[31 (b)
= 0 =[31 (1) respectively.
We observe that
clearly
uo does notsatisfy
the Neumann conditionsin
[0, b]
and in[b,1 ],
hence it follows that i.e.This
implies that, if -
=min Tg0, then -
>0; since y by Corollary
2.4
(i)
we obtain that F9 has at least two minimizers.. Since g is not
constant,
thenfor
a proper choice of b we havewhich
implies,
after somecalculation,
Now,
the samearguments
used in(i) give that y = min Tg0
isstrictly
po- sitive andP’
has at least two miinimizers.COROLLARY 3.5. For every there exists
g E L 2
such thatFj
has more than one minimizer.PROOF Let us
fix g E L 2 ; by Corollary
3.4(i)
there exists such that F9 has more than one minimizer. We can find a E R + such=
y£
thendefining v
=ylu and g = ý;, g,
it follows that(u)
=F~ (v).
Thisimplies that,
if ul , ... , ul are minimizers forF9 ,
th n
V, x- u
... ,ylui
are minimizers forREMARK 3.6. It is clear that
Corollary
3.5 can beproved
with thesame
reseating technique
also for the functionalF9.
Moreover the
previous proof
shows that thereexists gee 00 (or g piecewise constant)
such that has more than one minimizer.We want to
study
now the continuousdependence
of the solution u ofproblem (2.2)
on the datum g. We will prove that thisdependence
holds when
problem (2.2)
hasuniqueness,
and in this case it is a directconsequence of the one-dimensional case of the results of
compactness
and lowersemicontinuity
of Ambrosio in[1].
LEMMA 3.7. Let
(gn)
be a sequence of functions inL 2
such that gn - gstrongly
inL 2.
Fix y
E R + and assume thatproblem (2.2)
for isuniquely
solv-able
by ii
E Let(ün)
be a sequence of functions in such that for every n E NF9n (ii
takes the minimum value.strongly
in
L 1.
PROOF.
By
the convergence of gnto g
inL 2,
it follows thatIIgnllL2
~CI.
·Since
iin
is aminimizer,
it is easy toverify
that(~~) 5 F:n (ic) ~
~ C2 ,
whereC2 depends
onCI
and theH-norm
of ii.This
implies
that there exists a constantC3 depending
onCI
andC2
such that 2 + 5
C3 ;
moreover #(S, ) 5 C2 ,
henceby
a com-pactness
result due to Ambrosio(see [1]
Theorem2.1),
there exists asubsequence
such thatstrongly
inL 1, with -u E=- X 1.
To show that u coincide
with ii,
weapply again
Theorem 2.1 in[ 1 ] obtaining,
after somecalculation,
This shows that u is a solution of
(2.2)
forhence, by uniqueness,
u = u and all the sequence
Un
converges to u. 0To conclude this
section,
we want to showthat,
when thedatum g
EE
L 2
ispiecewise constant,
a solution ofproblem (2.6)j (and
hence a sol-ution of
problem (2.2))
canjump only where g jumps.
We note that this fact had
already appeared
in theexamples report-
ed in the
introduction;
ingeneral
this kind of behaviour is a feature of the minimumpoints independently
of the choiceof g, if g
ispiece-
wise constant.
for every
partition 0 =
of[0, 1]
we consider the set of the functions g of thetype
with ai
E R for every i =0,..., k;
we remark that is a linear sub- spaceof £2.
k
LEMMA 3.8. Let be a function
belonging
toX,,.
Fixed j E
and let u be a solution of(2.6)j
of thetype
where
Q=
is apartition
of[0, 1 ].
ThenPROOF. We recall that
by Proposition
2.6 Jg is contained in~0, ...,1~~.
We argue
by
contradiction.Suppose
that there exist..., j}
and i E
10,
... ,1~~
such thatbs
E(ai , ai , 1).
First of all we observethat, since j
EJg, f3s - I ~3~ (b,),
hence we may assume thatf3s -
1(bs)
>>
~3s (bs) (the
other casefollowing by analogous arguments)
and we canconsider the
following
two cases:(The
third is similar to the firstone).
For every 0 E
bs -
ai we define a function Ue:[0, 1]
- Rby
In the case
(1)
we note that there exists 6 > 0 such that for every x suchwe have
Hence when - 8 we obtain
This contradicts the
hypothesis
that u is a minimizer of G 9.In the case
(2)
we may consider thefollowing
two subcases:(The
last(bs) -
«i oci -(bs)
can be studiedsimilarly
to the(2)J.
If
(2)a
issatisfied,
then there exists 3 > 0 such that for every x withwe have
,~~ _ 1 (x) - «i > ai - ,~~ (b~)
and~~ _ 1 (x) >
> B3 (bs).
Hence for every e d we obtainagain
0. Nowwe consider the case where
(2)b
is satisfied. First we remark that from(3.1)
for every x E(bs,
we haveand so
then since c, =
,~s (bs ),
we can conclude that{3~ (x)
0 for every x EE
ai + 1 ). Therefore p,
is astrictly decreasing
function on(bs,
Now for every 0 ~ ai + 1 -
bs
we define a function v,~ :[0, 1]
-Rby
We
point
out that there exists d > 0 such that for every xwith bs
xbs
+ d wehave!3s -
>!3s (x);
hence for every n 8,using (2)b
and thefact that
!3; (x)
0implies
that!3s (x) !3s (bs),
we obtainThis contradiction concludes the
proof.
COROLLARY 3.9.
Let g, k and j
as in Lemma 3.8and y
e R + . Thenthe minimizers of the functional
F§ with j iumps
are at More-over the minimizers of
9
are at mostk
p
J
PROOF.
By Proposition 2.5,
for a fixedpartition
of[o,1 ],
G 9 hasexactly
one solution u eby
thepreceeding
lemma thepartitions corresponding
to a minimizer of(2.6)j
must becontained in the
partition P = (ai)k+1i=0
and hencethey
can be at most The conclusion follows since Jg c10,
...,k} (see Proposition 2.6)
and
REMARK 3.10. It is clear that if we
repeat step by step
the argu- ments used in Remark3.3,
Lemma 3.7 and Lemma 3.8cancelling
outthe term with the derivative in the functional
F.9,
we obtain the sameresults also for
F9.
4. Main results.
In Theorem 4.3 we will prove that for «almost
all» g e ~1"C~
we haveuniqueness
forproblem (2.6)~ ,
for each0 ~ j ~ k;
but to obtain this re-sult we need before the
following
lemmas.LEMMA 4.1. Assume that 0 ~ a2o
ail -
a i2 andthat g
=Let us consider for m =
2, 3
thefollowing
functionalsAssume that for
m = 2,3
um are minimumpoints
for(4. 1 )m
onthen
PROOF. Since U2 is a minimum
point
for(4.1 )2,
it follows thatMoreover,
if we had thatthen it should be U3 = g on
(a~ , a23 ),
that means U3 = 0 on(ai2’ a23 );
butby
Remark 3.1 it is notpossible,
since Ug is a minimumpoint
for(4.1 )3 .
Hence it is clear that
and this
inequality together
with(4.3) gives (4.2).
This concludes theproof.
LEMMA 4.2. Let us fix a
partition P = (ai)k+1i=0
of[o,1] and
for anychoice of
(ao ,
... ,ak)
let us define afunction g = 2
>belonging
to3K,. Let j
e0
...k
and letQ= =o
betwo different
partitions
of[o,1]
such that ~P. Let us define twofunctions
Q, R: 1--~ I~ by
and
where
xQ and ~CR
are defined as inProposition
2.5. ThenQ
and Raretwo different
polynomial
functions and the setis an open set dense in
PROOF.
Clearly Q
and R arepolynomial
functions ofdegree
2 in thek + 1 variables «o , ... , «k . The
proof
isaccomplished
if we prove that theequality Q(ao, ... , «z ) = R(ao, ..., ak)
is notidentically
satisfied.Since
q
is different from there exists 1belonging to (0, ..., j )
suchthat qm = rm for every m E
~0,
...,11
and rl + 1; we suppose that rl + 1.By hypothesis
there existio , i2 , i3
EfO,
... ,kl
such thatql = ri =
ajo,
ql + 1 = ai2 and I =ai3.
Let us take now «o , y ... , 9 ak where a4 = 1 and «m = 0 forio.
Thenby
Lemma 4.1 withail
= ai,, + 1we have that
«k)
... ,«k).
This concludes theproof.
THEOREM 4.3. Given a
partition P = (ai)k+1i=0
of[0, ],
there exists a subset~’iZ~
of3l1, ,
which is dense in~~
withrespect
to the L2_topology,
and such that for every g E
MOO problem (2.6)j
has aunique solution,
forevery j
E Jg.PROOF. Let e = be a
partition
of[0,1]
andlet g
E~’1Z~.
Letj
EJ ; by Proposition
2.6 we have that0 ~ j
k.If j
=k,
then for everyg E
X,,
theproblem (2.6)k
has theunique
solution g.Now we define
where the set is defined
by (4.4),
andby
Lemma 4.2 andby
Baire’stheorem,
is an open set dense in I and hence3K)
is dense inMo
withrespect
to the L2-topology.
Moreoverfor every g E
3K)
theproblem (2.6)j
hasuniqueness,
forevery j
EE In fact
let g = E
be a functionbelonging
toJ4.
Then
(xo...., «k)
E for with #Q =
# [R= j
+ 2 andfor every
0 ~ j
k. We supposeby
contradiction that there existjo
EE Jg and two different
solutions u, v
EXJo
of theproblem (2.6)~0 (see
Re-mark
2.1).
letQ and s1
thepartitions
associated to u and v;by Proposi-
tion 2.5
and
are different andby
Lemma 3.8Q and s1
are containedin T. Hence from the definition of must be different from
R(«o ,
I ... ,«k), where Q
and R are defined as in Lemma 4.2. But since u and v are minimizers of theproblem (2.6)~0,
we have thatQ(«o , ... , «k)
=Mlo
=R(oco,
... , this contradiction concludes theproof.
THEOREM 4.4. There exists a countable set
Nlo
dense inL 2
and acountable set 1-’ in R + such that for every g E
JKo and y
E R +~ I’ prob-
lem
(2.2)
admits aunique
solution.PROOF. For
every k E N
we consider thepartition
rPk == {0, 1/k, 2/1~,
... ,1 ~
of[0, ];
henceby
Theorem 4.3 there exists a set3Ki
dense in
Mok
such that for every g E3Ki problem (2.6)j
has aunique
sol-ution for
every j
E Jg.By
thedensity
of characteristic functions inL 2
the set
is dense in Moreover
by
theseparability of L ,
there exists a count- able set which is dense inL 2.
Let us consider r =U rg,
where 1’g is defined
by (2.13). Since, by
Remark2.2,
Fg is a countableset and
3K°
iscountable,
then 1-’ is a countable set.Now, fixed g
E~o and y
E I~ +B 1’
theuniqueness
for theproblem (2.2)
followsby
theuniqueness
forproblem (2.6)~ , by Proposition
2.3and
by
the definition of r..In the
following
theorem wegive
a«genericity»
result: we establishthat the
uniqueness
of the solution toproblem (2.2)
is ageneric property.
THEOREM 4.5. Let us assume that there exists a countable set
~o,
which is dense in
L 2,
and a countable set 1’ in R + such that for every g EW9
and for every yB r problem (2.2)
has aunique
solution.Then for every y E R + there exists a
Gd-set 3K§f
dense inL 2
such thatfor
each g
E3K§f
the solution ofproblem (2.2)
isunique.
PROOF. Let
W9
be as in the statement of the theorem. Wefix y
Eand g e L
and defineWe observe that
S(g) ;e 0,
since as we have seen, there exists at least onesolution of
(2.2).
Let us define D:
£2 ~ [0, + 00] by
This definition
implies
that(2.2)
has aunique
solution if andonly if D(g)
= 0.Now,
we aregoing
to prove that the function D is continuous in thepoints
of the set3K°.
Let us
fix 1
ENo
and suppose that there exist n e N and a sequence( fz)
inL 2
suchthat fk
convergesto f
in theL 2-topology
andThis
implies
that there are two sequences(Vk)
and(Uk)
inS( fk)
suchthat
On the other
hand,
sincef
Eby hypothesis
there exists aunique
solution of the
problem
Therefore from Lemma 3.7 we can conclude
that Vk and Uk
converge to u fstrongly
inL 1;
but this contradicts(4.5).
Hence forevery f E 3K°
andn E N there exists an open
neighborhood Uf of f in
the L2_topology
suchthat
D( g) 1 /n
forall g
EUf .
At this
point,
if we denote Un= f E Ufn,
we have that Un is an open subset ofL 2
withrespect
to the L2_topology.
Then let us define ==
n Un ;
is aGrset
inL 2 and, by construction,
contains3K° ;
so,by
nEN
hypothesis, 3K§f
is dense inL 2.
Moreover we observethat,
fixed g EE
3K§f ,
for each nE N,
gbelongs
to Un and soD(g)
= 0.This proves the theorem when Let now
y E 1’
and fixYo E R +
B I’.
Then there exists « > 0 such that «Yo = y.By
the firstpart
of the
theorem,
we know that for every g Emtio
theproblem
has
only
one solution.Multiplying
thisexpression by
«,defining v
== and
taking
into account that# (Su ) _ # (Sv )
we obtainthat,
iff E fl
=~1CY ,
then theproblem
has
only
one minimizer. SinceyÇ
isclearly
a denseG,-set
theproof
is
accomplished.
COROLLARY 4.6. If
ro
is a countable subset ofR +,
then thereexists a dense
Grset 3K)
such that for every g e and for every y ero problem (2.2)
hasuniqueness.
PROOF. It is
enough
to define3k)
=y fl e ro 3K)
and to observe thatby
Baire’s lemma the countable intersection of dense
Grset
is still a denseGs-set.
In the