wulf exercice - philippe bich, professor at university ... · 07.09.2018 · wulf exercice for the...
TRANSCRIPT
Wulf ExerciceFor the représentation there are two ideas
first when there are several conditions
the and also denoted Ï correspondsto intersection and the er ton union
it allows to decompose the set intosoinplier onesSecond when you have a set with
inequality conditions libre fluideTRYn'taie 13 you begin to studythe equality case in general it will
given you the boundary of the firstset and you find the set by takinga particular point the boundary being
in the set a not depending ifthe inequality is strict or not
1 A nn c R sitter A laideur n o
n Cnn C KE y o
For A1 you consider ln g c W Ntfswhich gives a unit aide centered at
O D and which is the boundary of AI
A1 is then the disc
Æ
Æ
EAAAz As AI
A is bounded because la µ c AIl lush Il E 1
Ae Az and Az are closed as preuûages ofclosed intervals by continuous functions
fa example At f YEE whereclosed
flank n 4 1 is continuous as a polynomial
idem for Az and AzThen A is closed as intersection of closedsets
A is compact because closed and bounded
in R I finite dimensional space
forthe exam you should give soûilararguments for the next questions we justgive the solution
y R
y n
B not bounded because n o c B tn 1
and k ch o Il to
B not closed because cut o c B tu 1
tout LE d go et Bhosts
B not open because 1 de B but there is
not e o l B e c B since
Hero 11 f e Bla I e but not in B
Burt compact because in R2 a subset
is compact d is bounded and closed
3
C not bounded since n e c c tn o
but Knew o
C closed C f Co toc where
closed
floyd y_et is continuous because
x et continuous and ln ul y continuous
Ü for example Co 1 e c but
there is no Eso I Bligh e CC
because Hcg 0,1 E e BUN c
and is not in c
Cnrt npact l sane argument than B
D not bounded lanice nif c D
and It ln Il a Thushrtempnot
D not open 11,1 c D tut there is
not e ol Blah e CD since
11,1 E D te o butin Blait
D closed if cnn.int ED converges to
nm yn 1g nn o
xn converges to x o but n o becauseotherwise yn would converge to
to l and not toy Thus
yn1 but abryn y
to K hosts
Thus y In no i e qu c D
Thus D closed
Exercises A
ÎÎÈÜ
fËÏÏÏ
because po c B C yo tB but
4dtf qu et B
convex if
and c Cop then y n
y ni
ntll H.nl s n a Huit because
n n convex function since
Cnil 270 Un c R
s y a a y y
Thus Amy LI Hoi De C
Exercicef1 lut tn on Ar 70 toc
fr not bounded above on A1 because
fr f to and the A1hosts
Thus Sup f141 tone 11
fr bounded below on Ar since
f1 n 70 Vu c A1 Thus
Inf f4 exists à Rne A1
Also he AI Vu 1 and
film 0 Which is a muinant
of fon At From the sequential
characterization of vif this proves
Inf film o
neAlas fzln not bounded above suice
lin feld to Thus sup fin tanato ne AL
fa x n 1 _1 canonical form
Thus fut z 1 tuer i e
1 is a minorant off ou Az
But fr111 1 Thus
Inf fin Min fr14 1ne AL REALstrictly
lut is P on Cpac
Tf Kui II I
l lena l Il
Thus