Chapter 8 sequence sand series

Question 1

Definition: pointwise convergence

Answer 1

We say that the sequence (f_n) converges pointwise to a function f:[a,b] TO R

If for each t in [a,b] we have

Limit as n tends to infinity of f_n(t) =f(t)

Ie if the sequence converges pointwise to f for each of those ts and ε bigger than 0 there exists N∈N st |f_n (t) - f(t)| is less than ε whenever n is bigger than or equal to N.

N depends on both t and ε!!!

Question 2

Definition uniform convergence:

Answer 2

Let f_n : [a,b] to R.

We say that the sequence (f_n) converges uniformly to a function f:[a,b] to R
When for all ε bigger than 0 we have N∈N st |f_n (t) - f(t)| is l
Less than
ε for all n bigger than or equal to N and t in [a,b]

N DOESNT depend on t
Only on ε!!!
Uniform convergence of a sequence to a function implies pointwise convergence of the same function- stronger!

Question 3

Proposition 8.1.6: equivalent statements for uniform convergence and supremum

Answer 3

Consider a sequence of functions f_n:[a,b] to R.
Let f:[a,b] to R.
Then the following are equivalent:
• the sequence (f_n) converges uniformly to f
•let M_n = sup{|f_n (t) -f(t)| t∈[a,b]}. Then M_n tends to 0 as n tends to infinity

This conditions on the M_n s is the easiest way to prove uniform convergence

Question 4

Theorem: uniform limit theorem

Answer 4

Let f_n : [a,b] to R be continuous for each n∈N. suppose the sequence (f_n) converges uniformly to a function f:[a,b] to R. Then f is continuous.

Proof I

Question 5

Theorem 8.3.1: uniform convergence theorem

Answer 5

Let f_n: [a,b] to R be a CONTINUOUS FUNCTION for n∈N.
Suppose the sequence (f_n) converges uniformly to a function f.

Then 
(limit as n tends to infinity
Of integral over [a,b] of f_n (t) .dt)
= 
(integral over [a,b] of f(t) .dt)

This allows us to swap limits and integral signs

Question 6

Corollary to uniform convergence theorem 8,3,1

Answer 6

Consider differentiable functions f_n :[a,b] to R.
Suppose (f_n) converges pointwise to a function f and the sequence of derivatives (f_n ‘) converges uniformly to a function g. Then f is differentiable and f’=g.

Thus under suitable conditions
We can swap limits and differentiation.

Question 7

Definition: uniformly continuous

Answer 7

Let f:[a,b ] to R. We say f is uniformly continuous on [a,b] if for all
εbigger than 0 we have δ bigger than 0 such that if |x-y| is less than δ then | f(x) -f(y)| is less than ε for all x,y ∈[a,b]
—–
Delta depends on epsilon only not on chosen point . For a given epsilon the same delta has to work across while interval.

Question 8

If F is uniformly continuous then f is continuous

True or false

For f :R to R

Answer 8

tRUE

Question 9

If f is continuous then f is uniformly continuous

f: R to R

Answer 9

FALSE

example.

Only if on [a,b]

f: R to R
f(x) = x^2 is continuous but not uniformly continuous
Ie for x and y in reals with |x-y| = c bigger than 0

|f(x) -f(y)| = |x^2 -y^2| = |(x-y)(x+y)| = c|x+y|

So if x,y bigger or equal to 1/c then This is bigger than 1. Taking epsilon =1 there is no delta in delta epsilon condition for convergence. Thus f is not uniformly continuous.

Question 10

Theorem 8.4.3 continuous bs uniformly continuous

Answer 10

Let f: [a,b] to R be continuous then f is uniformly continuous

Only true for [a,b]

Question 11

Theorem 8.4.4

Riemann integrabke and continuity

Answer 11

Let f:[a,b] to R be continuous

Then f is Riemann integrable

Question 12

Definition 8.5.1: partial sums and convergence pointwise and uniformly

Answer 12

Let (f_n) be a sequence of functions f_n :[a,b] to R

We consider the sequence of partial sums (s_n) where s_n:[a,b] to R is defined by

s_n (t) = f₁(t) + f₂(t) +…+ f_n (t).

We say that the series ( Σ from n=1 to infinity of f_n)

Converges pointwise if
For each t ∈[a,b] the series sum of f_n(t) converges

Uniformly summable or converges uniformly of sequence of the partial sums converges uniformly

Question 13

Theorem 8.5.2 uniformly summable sequence and continuity

Answer 13

Let (f_n) be a uniformly summabke sequence of continuous functions f_n :[a,b] to R.

Then the function f:[a,b] to R is given by

f(t) =
Σ from n=1 to infinity of
f_n(t)
(exists) and is continuous

Question 14

Theorem 8.5.3 **

Sequence of differentiable functions and series uniform and pointwise

Answer 14

Let (f_n) be a sequence of differentiable functions f_n :[a,b] to R, such that the series

Σ from n=1 to infinity of f_n converges POINTWISE to a function f, and the series
Σ from n=1 to infinity of f’_n is UNIFORMLY summable.

Then f is differentiable and
f’(t) = Σ from n=1 to infinity of f’_n(t)

For all t in [a,b].

Question 15

Theorem 8.5.4 WEIERSTRASS M-TEST

Answer 15

Let f_n from [a,b] to R be a sequence of functions.

Suppose we have a SUMMABLE sequence of real numbers (M_n) such that
| f_n (t) | is less than or equal to M_n for all n and all t in [a,b].

Then the sequence (f_n) is UNIFORMLY summmable.

Further for each t in [a,b] the series
Σ from n=1 to infinity of f_n(t)
Converges ABSOLUTELY.

Question 16

Proposition 8.8.5:

Continuous function that is differentiable everywhere?

Answer 16

There is a function f:R to R which is continuous everywhere but differentiable nowhere

Proof: sketch proof f_n(t) = 1/10^n

By weierstrass M test show that the series sum from n =1 to infinity of f_n converges
…
Applying uniform limit theorem to show f is continuous
Check f is not differentiable anywhere to give that required limit doesn’t exist

Question 17

Example 8.1.2:
Define f_n : [0, 2pi] to R by f_n (t) = cos(t/n)

show that the sequence (f_n) converges pointwise and determine the limit function

Answer 17

Observe that t/n converges to 0 as n tends to infinity

And

That cos is a continuous function

So for each t in [0,2pi] we have that cos(t/n) converges to cos0 =1 as n tends to infinity.

Thus the sequence (f_n) has pointwise limit the constant function f:[0,2pi] to R with f(t) =1 for all t in [0,2pi]

Question 18

Example 8.1.3
Define f_n : [0, 1] to R by f_n (t) = t^n

show that the sequence (f_n) converges pointwise and determine the limit function

Answer 18

For t less than 1 f(t) = the converges to 0 as n tends to infinity

For t =1 1^n =1 for all n so f(1) tends to 1 as n tends to infinity

So the sequence has pointwise limit f where

f(t) =
{ 0 if t less than 1
{ 1 if t=1

Question 19

Example 8.1.4: define f_n:[0,2] to T by f_n (t) = t^n. Show that the sequence (f_n) does not converge pointwise.

Answer 19

Not that if t is bigger than 1 then t ^n converges to infinity as n tends to infinity

If t is bigger than 1 then the limit of f_n(t) as n tends to infinity does not exist and (f_n) has no pointwise limit

Question 20

Pointwise and uniform convergence and continuous

Answer 20

Uniform convergence preserves continuity. Pointwise convergence does not preserve continuity.

Question 21

Example: 8.1.7 for n ≥ 2

Define f_n: [0,2pi] to R
By

f_n (t) = (1-2n) /(1-n) • sint

Show sequence converges pointwise and determine limi function f then show that (f_n) converges uniformly to f.

Answer 21

f_n(t) = ((1/n) -2)/((1/n)-1)• sint tends to 2sint

As n tends to infinity
so (f_n) converges pointwise to f where f(t) = 2sint

Then |f_n(t) -f(t)|≤ | (1-2n)/(1-n). - 2 | •|sint| ≤ | (1-2n)/(1-n). - 2 | for all t and

Thus if M_n = sup{ |f_n(t) -f(t) | t in [0,2pi]} then

0 ≤ M_n ≤. 1/(1-n). And since rhs converges to 0 M_n converges to 0 as n tends to infinity.

So f_n converges uniformly to f.

(1-2n)/(1-n). - 2 | = 1/(1-n) tends to 0 as n tends to infinity.

Question 22

Example 8.2.2: let f_n :[0,1] to R be given by f_n(t) = t^n show hat the sequence (f_n) does not converge uniformly

Answer 22

Solution: previous example pointwise limit is
f(t) = { 0 if t less than 1 or 1 if t=1

If (f_n) did converge uniformly the limit this would have to be this function but is clearly not continuous so the sequence does not converge uniformly

Question 23

Example 8.3.2. Let f_n : [0,1] to R for n in the naturals be defined by

f_n (t) = nt(1-t^2) ^n
Show (f_n) converges pointwise to the 0 function

Calculate the limit as n tends to finity of integral from 0 to 1 of f_n(t).dt

Does the sequence converge uniformly?

Answer 23

By binomial expansion….

2) substitute u=t^2 we can show that the integral is equal to n/ (2(n+1))

So taking limits = 0.5

3) from previous part this limit does not equal 0 which is the integral of the limit as n tends to infinity of f_n . Dt

Thus is cannot converge uniformly by the uniform convergence theorem (a contradiction with it)

Question 24

Example 8.3.4
Define
f_n: [0,2pi] to R by
f_n(t) = (1/n ) • sin(n^2t)show that converges uniformly to the zero function but (f’_n) does not converge pointwise

Answer 24

Observe that |f_n (t)| ≤ 1/n for all t in [0,2pi] and this converges to 0 as n tends to infinity.

Thus the sequence converges uniformly to 0 to zero function

But f’_n (t) = ncos(n^2t) and the sequence (f’_n) does not converge pointwise

Let alone uniformly.

Question 25

Example 8.4.2

Define f:R to R by f(x) = x^2 certainly f is continuous show that f is not uniformly continuous

Answer 25

Let x,y in R and suppose |x-y| =c >0 then
|f(x) -f(y)| = |x^2 -y^2| = |(x-y)(x+y)| = c|x+y|
so if x,y ≥ 1/c then
|f(x)-f(y)| > 1
taking epsilon =1 in the definition of uniform continuity we see that there is no such delta bigger than 0 such that | x-y| less than delta ensures that |f(x)-f(y)| is less than 1. Thus f is not uniformly continuous