REAL ANALYSIS I

We will prove on compact metric space.

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REAL ANALYSIS I (6650)

The Minh Tran

Answer :

1 a)

Yes

Explain

We can check the polynomial

2

n 2 n

1

1 x n

n

1/ 1 x

p (x) 1

n

We need to prove converging poiwise p (x) e on [0,1]

Indeed, we need to check p (x) at 0, 1, and 0 x 1

Pr oof :

+

→

< <

-

2

n

1/(1 x )

At x 0 :

We see that

1 lim p (0) lim(1 ) e e lim p (x) e at x 0

n

1

So there exist a sequence p (x) 1 of polynomial converging

(1 x )n point wise to e e at x 0

+

+

-

2 2

2

2

n 2 n

1

1 x 2

n

(1 x )n 2

n

n

1/ (1 x )

n

1

(1 x )n

1 Because we see lim 1 e for 0 x 1

(1 x )n

1

So there exist a sequence p (x) 1 of polyn

(1 x )n

→∞

+

→∞

+

→∞

+

 +  = < <

+

+

2

1/(1 x )

omial converging

point wise to e + when 0<x<1

-

2

2

n

1

1 1 2

n

2n n

1

1 x n

n

n

Finally,we check p (x)

At x 1:

We check that

lim p (1) lim(1 ) lim (1 ) e e

1 ( Because lim(1 ) e at x 1)

2n lim p (x) e at x 1

1

So there exist a sequence p (x) 1 of

(1 x )n

+

→∞

+

→∞

+

2 1

polynomial converging

point wise to e + =e at x =1

Condition :

2

n

1

1 There exist a sequence p (x) 1 of polynomial converging

(1 x )n point wise to e + e on [0,1]

+

=

b)

Yes

Explain

We can check the polynomial

2

2

n 2 n

1

1 x n

1

1 x n

n

1/ 1 x

p (x) 1

n

We need to prove converging uniformly p (x) e on [0,1]

Pr oof :

As the proof above, we see that p (x) converging poinwise to e on [0,1]

Now, we need to check the property of uniform of p (x)

+

+

→

-

2

2 2

2

2

2

1

1 x

1

1 x

(1 x )n 2

n

1

1 x

Indeed, set p(x) e

We compute

1

(1 x )n Moreover

1

(1 x

+

+

+

→∞

+

=

+

+

2

1 x n

n

n

)n

So given 0,there is some N(which may depend on ) such that p (x) p(x)

for all 0 x 1 and all n N

Condition :

1 Thus, there exist a sequence p (x) 1 of polynomial converging

(1 x )n uniformly

+

→∞

+

2

1/(1 x )

to e + on [0,1]

2 2

2

1

1 x

n

so

1 Thus, there exist a sequence p (x) 1 of polynomial converging

(1 x )n po

+

+

2

1/(1 x )

int wise to e + on

d) Yes

2

2

2 2

2

1

1 x n

1

1 x

1

1 x

We see that p (x) converging poinwise to e on

Indeed, set p(x) e

We compute

Moreover

1

p (x) p(x) 1

(1 x )n

+

+

+

=

+



2

2

1 x n

n

1

0

1 x

n

n

1

In particular,for n fixed p (x) 1 1 as x and

(1 x )n

So given 0,there is some N(which may depend on ) such that p (x) p(x) for all x and all n N

+

→∞

+





+

2

n

1/(1 x )

Condition :

1 Thus, there exist a sequence p (x) 1 of polynomial converging

(1 x )n uniformly to e + on

+

2

2

nx

n 1

1

nx n

nx

n

n

n

ne dx

Set f (x) ne and we compute

n lim f (x) lim ne lim 0 for all 1 x 2

e ( u sing l'Hospital rule)

So f (x) is converges point wise to f 0 on [1,2]

We have f (1/ n) n / e as n

Thus,f (x) does not converges un

=

−

−

=

=

∑

∫

iformly to f =0 on [1,2]

Now, we can compute

nx

n

n 0

e

We will compute and

1

We may use the formula q where q 1

1 q

−

∞

=

=

= −  =  − 

−

∑

n

n

2 n

2

1

1

e 1

1

e

−

−

−

−

nx

2

n 1

1

Conlusion :

e

ne dx

∞

−

=

=

−

∑

∫

3

( )

{ }

n

Let f : M, be continuous for each n and suppose that f (x) 0

for all x M where f (x) f (x) for all x M and n If M is compact

Show that f converges uniformly to f(x) 0 on M

+

=



( )

n

n n

n

: Suppose S is a set and f :S , n We say that the sequence

f is uniformly convergent with limit f : S if 0, N such that, x S

n N we have f (x) f(x)

∈

Recall



Proof :

From the problem 3 and recall, we see that S≡(M,ρ) If M is compact then (M,ρ) is compact metric space We will prove on compact metric space

Indeed,

Suppose that f is continuous, for each n ∈  ,n f (x)n →0 this mean that

n

n

lim f (x) f(x) 0

→∞ = = and suppose that where f (x)n 1+ ≤f (x), xn ∀ ∈M

n

n 1 N

Given 0 , set B x M : f (x) f(x) so B is closed set

and f (x) f (x) is decrea sin g sequence so B B

( because f (x) f(x) f (x) f(x) )

We have : lim f (x) f(x) 0, x M so B Hence, N

such that B , this mean tha

+

∞

→∞

=

= ∅



∩

N

t f (x) f(x)− < ε ∀ ∈, x M (*)

Therewith f (x) f (x) is decrea sin g sequence so n N

then f (x) f(x) f (x) f(x) , x M ( according to (*) )

( ) { }

n

Thus,the sequence f (x) f(x) converges uniformly to 0

Conlution : f converges uniformly to f(x) 0 on M

−

=

4 Show that h : M, ( ρ → ) ( N, τ ) is uniformly continuous on M if and only if

( h(x ),h(y )n n ) 0 whenever (x , y )n n 0

Recall:

( ) ( )

f : M,d N, is uniformly continuous if

0, 0 which may depend on f and such that

f(x),f(y) whenever x,y M satisfy d(x,y)

Proof :

We are proving the problem 4 on two metric spaces ( M, ρ ) ( , N, τ )

( )

n

Suppose h is uniformly continuous on M

Given 0 , 0 ( depend on 3 and f )

such that h(x),h(y) , h(y),h(y ) , h(x ),h(x)

whenever x , y , x,y M satisfy (x , x) / 3, (y , y) / 3, (x,y) / 3

Because is a metric

Hence, h(x

⇒

τ

),h(y ) h(x ),h(x) h(x),h(y) h(y),h(y )

3

So h(x ),h(y ) 0

whenever x ,y x ,x x, y y,y (because is a metric)

/ 3 / 3 / 3 ( depend on 3 and f )

< ε + ε + ε = ε

-

( ) ( )

We will prove that :

0, 0, such that h(x ),h(y ) whenever x , y M satisfy x ,y depend on and f

Indeed, let 0, choose 0

We have h(x ),h(y ) 0,this mean that h(x ),h(y )

whenever x , y M satisfy x , y 0 this mean that x ,y

depend on and f

Thus, h is uniformly continuous on M

ε > δ >

5

{ }

n 1

Define M to be the of all sequences (x ) where x 0,1 for all n

Define (x ),(y ) sup x y Is M, separable ?

Explain why or why not ?

≥



Answer : ( M, τ ) is not separable

Explain :

( )

n 1

n 1

First, we can prove M, is metric space

We have

(x ),(y ) sup x y 0, x ,y

(x ),(y ) sup x y 0 x y

(x ),(y ) sup x y sup y x (y ),(x )

(x ),(y ) sup x y sup x z z y sup x z z

≥

≥

τ

y sup x z sup z y

−

( )

Thus, M, τ is metric space

-

Proof :

Indeed, we have infinite subset of M such as define :

n 1

n 1

Set Sup x y 0 so that (x ),(y ) Sup x y 0

for all (x ),(x ) , ,(x ) M ; (y ),(y ) , ,(y ) M

≥

≥

We assume that every infinite subsets of M has a limit point

so we have an inf inite set of point s in M such that (x ,y ) ,

Now,we assume that it have inf inite number of limit point x , y

in M such that (x , x ) / 3, (y ,y ) / 3

(x , y ) (x , x ) (x , y ) (y ,y )

/ 3 / 3 / 3 where 0

τ < δ τ < δ

< δ + δ + δ = δ δ >

This is contradiction that τ(x ,y )n n ≥ δ so M has no limit points

Thus, (M,τ) is not separable