let fn(x) =x^n
a) show that the sequence {fn} converges to zero uniformily on [0,
1-d] for each d>0
b) show directly , by negating the definitio of uniform convergence
that {fn} does not converge uniformily on [0,1]

Hi fmunshi!!

You can find the answer to this problem at the following page from The
London School of Economics & Political Science website:
"Uniform convergence" (look for the examples at the second page of the
pdf document)
http://www.maths.lse.ac.uk/Courses/MA203/sec7a.pdf

For the part a) replace 1/2 with (1 - d) (0 < d < 1):
Suppose that fn(x) = x^n with x in A = [0;1-d]. Then (fn) converges
uniformly to the identically-0 function f (given by f(x) = 0 for all x
on [0; 1 - d], because, for all x in [0;1-d] we have:

 fn(x) - f(x) = x^n =< (1-d)^n

Now given E > 0, we can chose N so that (1-d)^N < E, so that for any n
> N and any x in [0,1-d] we have:

I fn(x) - f(x) I = x^n =< (1-d)^n < E

This means that gn(x) converges uniformly to the identically-0
function f.


For the part b) please see the recommended page "Uniform convergence"
(look for the examples at the second page of the pdf document).


I hope this helps you. Please ask for a clarification if you find this
answer obscure and/or incomplete. I will gladly respond your request.

Sincerely.
livioflores-ga

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Request for Answer Clarification by fmunshi-ga on 23 Jul 2003 13:56 PDT

i am having trouble seeing how the definition of uniform convergence
is negated on the interval [0,1].  I know

I fn(x) - f(x) I = x^n =< 1^n 
but how does that negate uniform comvergence

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Clarification of Answer by livioflores-ga on 23 Jul 2003 22:59 PDT

Hi!!
The bound I fn(x) - f(x) I = x^n =< 1^n  is not helpful to negate
uniform comvergence of fn.
What do you need to prove in uniform convergence is the following:
For each E > 0 exists n0 such that for ALL N >= n0 is:
Ifn(x) - f(x)I < E  FOR ALL x in the interval. This means that n0
depends only on E and not of x :  n0 = n(E)

How do you negate this ?
For example showing that given E > 0, the number n0 asociated with it
always depends on both E and x; and/or that exists E > 0 for wich
there does not exist a number n0 that satisfies FOR ALL x in the
interval the following.
Ifn(x) - f(x)I < E  for all n >= n0.
This is equivalent to show that exists E > 0 such that for each number
n0 exist x in the interval such that Ifn0(x) - f(x)I >= E.

As you have seen when we worked with the interval [0 , 1-d] (0 < d <
1) you have found a value n0 that not depends on x (it is valid for
all x in the interval), this n0 is the first natural number higher
than log(E)/log(1-d), this n0 clearly depends only on E and it is
independent of x. This shows the uniform convergence of fn(x) in the
interval [0 , 1-d] (0 < d < 1).

What happens with the interval [0,1]?
Ifn(x) - f(x)I = x^n
Given E > 0 we have that:
If x = 1 , Ifn(1) - f(1)I = 1^n - 1 = 0 < E for all n.
If 0 =< x < 1 :
Ifn(x) - f(x)I = x^n < E if and only if n.log(x) < log(E); if and only
if (log(x) is negative) n > log(E)/log(x), this shows that n0 =
n(E,x).
But we can go further. Taking 1 > E > 0 and any number n0, if 1 > x >
E^(1/n0) > 0 , then:
Ifn0(x) - f(x)I = x^n0 > E.

So we just find E > 0 such that for each number n0 exist x in the
interval such that Ifn0(x) - f(x)I >= E. Or the equivalent, we just
find that exists E > 0 for wich there does not exist a number n0 that
satisfies the following:
Ifn(x) - f(x)I < E  for all n >= n0  and FOR ALL x in the interval.

This shows that for such E > 0 it doesn´t exist n0 such that for all n
>= n0 is Ifn0(x) - f(x)I < E FOR ALL x in [0,1]. So fn does not
converge uniformly to f.

I hope this clarify the answer. Please feel free to request for more
clarifications if hey are needed.

Regards.
livioflores-ga