foss lecture3 appendix

8/12/2019 Foss Lecture3 Appendix

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1 Appendix: the u.o.c. convergence

Letfn: R R, n = 1, 2, . . ., and f : R R be any functions.

Definition 1.

fnu.o.c. f asn , if, for any compact (bounded and closed) setK,

supxK

|fn(x) f(x)| 0 asn .

For anyfand for any t >0, define a norm ft :

ft = sup|x|t

|f(x)|,

and a norm f :

f=

m=1

2m fm1 + fm

.

Lemma 1.

fnu.o.c. f fn f 0 asn .

Indeed,

(). >0, choose m0:

m=m0+1

2m 2m0

2

Then choose n0: fn(x) f(x)m0 2m0

n n0.

|= fn f m0m=1

fn fm+ 2m0 m0

2m0+

2=.

(). Let K be a compact non-empty set. Take m0 1: K [m0, m0]. Set

gn = fn f. Then

fn f 0 |= 2m0

gnm01 + gnm0

0 |= gnm0 0 |= supxK

|gn(x)| 0.

Remark 1.

Forfto be finite, allfm have to be finite. This is always the case

iff is, say, continuous or monotone.

Remark 2.

If f is defined on a measurable subsetB R only, we letf(x) = 0

x B.

Remark 3.

Any (right- and/or left-) continuous function is determined by its val-

ues on a dense subset (e.g., by values at all rational points).

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Lemma 2.

Let allfn,n = 1, 2, . . .be non-decreasing functions andfa continuous

function.

Then the following are equivalent:

(a) fnu.o.c. f;

(b) for any dense subsetB of the real line, fn(x)f(x), x B.

Proof.

( Only!). Take anym and prove that fn fm 0. Assume, by the contrary,

that

lim supn

fn fm = C >0 (Note: C 0 : (i) B;

(ii)|f(x+) f(x)| C/4.Then

fnl(xl)f(xl) =fnl(xl)fnl(x+)+fnl(x+)f(x+)+f(x+)f(x)+f(x)f(xl)

and

limsup

l

(fnl(xl) f(xl)) 0 + 0 +C/4 + 0 = C/4.

Similarly, choose >0 : (i) B;

(ii)f(x ) f(x) C/4.Then

lim infl

(fnl(xl) f(xl)) C/4 we arrive at a contradiction!

Lemma 3.

(The triangular scheme).

LetD,d >0 and{fn} : sup|x|d |fn(x)| D n.

Then a subsequence{nr} : rationalx,

a limit limr

fnr(x) :=f(x).

Proof. Number all rational points in [d, d] in an arbitrary order: x1, x2, . . .Start

with the procedure:

Step 1. Since |fn(x1)| D n, a subsequence {nl,1}: {fnl,1(x1)} converges

(denote the limit by f(x1)).

. . .

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Stepr + 1. Assume we have defined a sequence {nl,r}: fnl,r(xi) f(xi) i=

1, . . . , r. Since |fnl,r(xr+1)| D l, |= a subsequence {nl,r+1}: {fnl,r+1(xr+1)}

converges (denote the limit by f(xr+1)).

. . .Thus, r = 1, 2, . . ., we defined a sequence {nl,r}: fnl,r(xi) f(xi) i =

1, . . . , r.

Now: setnr =nr,r.

Note: r0, {nr, r r0} is a subsequence of{nl,r0}

|= fnr(xr0)f(xr0).

Corollary 1.

(I). Assume, in addition, that C


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Theorem 1.

Letf :R R (orf :R+ R) be Lipshitz continuous. Then

(a) a measurable setB B(f) R (orR+) : (B) = 0 and

x B, f(x) = lim0

f(x+ ) f(x)

, |f(x)| C.

(b) Letf(x) = 0 forx B. Then x, t >0

f(x+t) f(x) =

x+tx

f(z)dz.

Definition 3.

Any pointxB is a regular point off.

Corollary 2.

Letf :R+ R+ be Lipshitz continuous, f(0)> 0.Assume >0 :

regular pointt 0, iff(t)> 0, thenf(t) .

Then t0 f(0)

: f(t) = 0 t t0.

Proof.

First, iff(t0) = 0, then f(t) = 0 t t0, since f(t) f(t0) =tt0

f(z)dz0.

Second, iff(t)> 0 t f(0)

, then

f

f(0)

f(0)/

0

()dz+f(0) 0.

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foss lecture3 appendix

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