Another approach to Cauchy PeanoTheorem using

Ascoli's Theorem

Piecewise Linear Approximation

let R Eto a Tota XExo b Kotb slopeM W

M supplfit X as before xotb fg sMayassumeMZlaswealyneedanupperbd xo 1,4

I 1

Define xob

W Ct HER lx xok.ly t tol to a to tota

By symmetry projCW onto t axis is toastota'surn

for some a'Eco AI

Note that f CCCR f CCCW

f is uniformly continuous on WsinceWisclosede

bounded

HE 0,7 8 0 such that

F Cti XD Ctz Xz EW with

Iti tako and la Xiao

we have Ifctz XD fCtiXDIs E

On the half interval Eto tota's

choose to s t s ta s s the total

with Iti ti i k F f i i k se.ge aoxo

FftxaDefine a function ketti on Ito tota's

gfigh

l ketto Xo to

2 keIff gis linear with slope f Itai Xi i

l ti

where Xi can be determinedsuccesinely by

is x detenuiedbyKel to is linearpasewig

through tojo and with slope f Ito XD

Ciii Note that IfctgxoskMiXi Xo I E M It tot

c Hi xD EW CR andhence fit xD welldefined

is then Xz determined by Betty is linear

pasewig through taxi and with slopefitsxD

ius Swiilarly Hitoko l IfitishkM we have

1 2 Xol EM Itz tot

tz xD EW CR and f Ita Xa welldefined

Kotb

g 1443 graph of keltXo1

Xob i

to a to a to t t total tot aun d

E tzM

And so on the function kelt is defined on Ito total

Note that

4 he is piecewise linear

G1 theft ke I E M It St t t seEtoTota'T

By slopes HitixisKM on eachsubinterval

Ike is equicontinuous as subset of CEto tota's

ke's is also uniformly banded to totab

In fact it is convex and theends points Cti xD with

Xi Recti belongs to W we have ft kits C W

by piecewise linearity As WCR thefts x Kb

and hence Reits IEHd t b V tCEtoTota's andVero

Hence Ascoli'sTheorem wiplies that the is precompact

In particular the sequence I Kenshihas a

convergent subsequence that in Cftotota'se

with tenets felt C Eto Tota'S as l to

To show Bits satisfies the differentequation we

first show that he is an approximatedsolution

for any E 0

Consider t C to Tota'S and t f to E Qb k

Then 75 1,3 h suckthat For E the o e corresponding

tj stet8 o

Using It tjykltj tj.lk Ey we have

1kelt keltj.is EMIt tj iIsOHence

I f ftp.kectj iD fftkeltD sE

Since he is piecewise linear

hefts fctj ykectj.is by ourconstruction

Henie

feldt fct kectD ISE ft C Ito totabiltgtystab

As ketto Xo kett is an approximated

solution.to

Irpdd f tix on Eto tota's

Xlt D Xo

in the sense that ddkf fit he remainder

CtoF 0

with 11remainderHSE

Integrating the ODE we have

kelt ketto t fat Kelssds SgtkdDds

Xo t kids ds

kelt Xo fetoffs kecsddskf.tok'ecss fls.kecsDlds sea

In particular if we denote 9e kate lie E he o

thenI gfts xo fttfcsgelsddsfeaef.tt

ss

Hence

1k xo

ftfcskcsdds.IEklts xo SfffCshcsDds gftitXotSfffCsgcesDdsl

I geits xofeffcs.gecsbdslsllk

gjho ffflfcsgecsD ffs.HNdstKne

Since 11g hills o and f is uniform continuity

t

15945 fcskiss Idso as j to

Therefore by letting j to we have

kits XotfetoffskGDd5t te Etotota's

date fit kits htt CEto tota

facto Xo

Similarly argument I tem t EEto asto

satisfying date f Ct Ect htt CEto a to

htt Xo

Note that by construction

dd ftD fCto Xo dateCto

Hence

yhtt t E Eto Tota's

kits t E Eto a to

is C Eto a tota and solve the EVPxx

Remarks i This proof doesn't need the Picard Lindelof

Them

c'is The spirit of this proof is more in line

with solving the IUP numerically

Lili's The 1st proof solve approximated problems

the 2nd proof solve the original problem

approximately

4.2 Baire CategoryTheorem

Def let I d be a metric space A set E in E

is dense if t X C I and E 0

Bean E 0

Notes i d Easy to see that E is denseE E

Ci's E is dense in EE d

et If E discretemetric then for Os Est

X E E Beca 2 3 Therefore L is

dense in I uirplies E E i.e I

is the only dense set in CI discrete

GL In HR standardmetric Q IRI Q I

are dense

GI Weierstrass approximation theorem implies

the set of all polynomials P fams

a dense set in C CEO I do

Det het CI d be a metric space A subset

ECE is called nowhere dense if its closuredoes not contain any metric ball

ie E has empty interier

G i 21 1 z l g l znowhere dense

in IR However

Oh has empty interior but OTER has

nonempty interim so 0h isnot nowheredense