Metric spaces - continuity / compact

christina182 · #1 $Old$ June 20th, 2009, 06:01 PM

Hi, I'm doing the following assignment:

We consider the metric space M = C([0,1], R) of continuous, real functions on [0,1]. M has the usual uniform metric:
d_M(f,g)=sup{lf(x)-g(x)l l x E [0,1]}

(a) Show that I:M-->R given by:
I(f) = integral (from 0 to 1) f(x)*dx
is continuous.

(b) Show that I from (a) attains a max and a min on the closure of the unit ball:
Closure of K(0,1) = {f E M l d_M(0,f) =< 1}
Here 0EM is the nil-function and I don't have to show that the closure of the unit ball is as suggested.

(c) Show that f_n E M given by
f_n(x) = x^n, x E [0,1]
for n=1,2, ... defines a series (f_n) in the closure of K(0,1) that does not have a convergent sub-series in M and conclude that the closure of K(0,1) is not compact.

In (a) I think that an argument regarding I^(-1)(f) is sufficient. But I guess just saying that I^(-1)(f) = M, which is a closed set is too superficial?

In (b) the problem is that the closure of K(0,1) is not compact and hence I cannot use the usual theorems...

Any suggestions will be deeply appreciated... :-)

putnam120 · #2 $Old$ June 20th, 2009, 08:07 PM

For (a) use what my analysis professor would call "the world's most trivial integral inequality". That is if $|f|\le M$ on $[a,b]$ then $\int_a^bf(x)dx\le M(b-a)$ . Combine this with the regular definition of continuity and the given metric to so what $I$ is continuous.

christina182 · #3 $Old$ June 21st, 2009, 02:08 PM

Hi, thanx for the reply. I've done (a) with the epsilon-delta method (and I find that delta = epsilon establishes continuity).

In (b), we look at a situation where f(x) <=0.

The integral from

1
/
l f(x) dx
/
0

attains it's maximum at 1. And this is the case when f(x) = 1 and hence F(x) = x.

I guess that the integral attains it's min at -1. And this is the case when f(x) = -1 and hence F(x) = -x. I find it quite confusing that the maximum of I is the value of an integral, I mean, what is it that I should find - a value of x that min/max I or a function f (which is what I have done)?

Anyways, how do I show that the max/min I propose is actually max/min ... Can I do a proof by contradiction, if so, how do I get started?

In (c) I'm still lost... I mean, for
x<1: f_n(x) --> 0 for n --> infinity.
x=1: f_n(x)=1 --> 1 for n--> infinity.
And as {0,1} is a part of M i don't understand why we get into trouble at all :-)

xalk · #4 $Old$ June 27th, 2009, 08:36 AM

Quote:

Originally Posted by christina182 $View Post$

Hi, thanx for the reply. I've done (a) with the epsilon-delta method (and I find that delta = epsilon establishes continuity).

In (b), we look at a situation where f(x) <=0.

The integral from

1
/
l f(x) dx
/
0

attains it's maximum at 1. And this is the case when f(x) = 1 and hence F(x) = x.

I guess that the integral attains it's min at -1. And this is the case when f(x) = -1 and hence F(x) = -x. I find it quite confusing that the maximum of I is the value of an integral, I mean, what is it that I should find - a value of x that min/max I or a function f (which is what I have done)?

Anyways, how do I show that the max/min I propose is actually max/min ... Can I do a proof by contradiction, if so, how do I get started?

In (c) I'm still lost... I mean, for
x<1: f_n(x) --> 0 for n --> infinity.
x=1: f_n(x)=1 --> 1 for n--> infinity.
And as {0,1} is a part of M i don't understand why we get into trouble at all :-)

In (a), how did you establish continuity by putting δ=ε, show some work if you like

In (b) your definition is not very clear

In (c),do you mean sequence and not series??

Enrique2 · #5 $Old$ June 28th, 2009, 05:45 AM

Quote:

Originally Posted by christina182 $View Post$

Hi, thanx for the reply. I've done (a) with the epsilon-delta method (and I find that delta = epsilon establishes continuity).

In (b), we look at a situation where f(x) <=0.

The integral from

1
/
l f(x) dx
/
0

attains it's maximum at 1. And this is the case when f(x) = 1 and hence F(x) = x.

I guess that the integral attains it's min at -1. And this is the case when f(x) = -1 and hence F(x) = -x. I find it quite confusing that the maximum of I is the value of an integral, I mean, what is it that I should find - a value of x that min/max I or a function f (which is what I have done)?

Anyways, how do I show that the max/min I propose is actually max/min ... Can I do a proof by contradiction, if so, how do I get started?

In (c) I'm still lost... I mean, for
x<1: f_n(x) --> 0 for n --> infinity.
x=1: f_n(x)=1 --> 1 for n--> infinity.
And as {0,1} is a part of M i don't understand why we get into trouble at all :-)

Hi Christina182.

Your argument for c) is correct!! Just observe that the limit (for ANY subsequence) function f(x)=0 0<=x<1 and f(1)=0 is NOT continuous!

Jose27 · #6 $Old$ June 29th, 2009, 08:04 PM

For (b): Using the mean value theorem for integrals we get that $I(f)=f(\zeta)$ for some $\zeta \in [0,1]$ and so we get $\vert \int_0^1 f(x)dx \vert = \vert f(\zeta) \vert \leq 1$ for all $f \in B:= \{ f \in M : \Vert f \Vert _{\infty} \leq 1 \}$
so $I(B)$ is bounded by $1$ . Let $m:= \sup_{f \in B} I(f)$ and so $m \leq 1$ . Now just let $g(x)=m$ then $g \in B$ and $I(G)=m$ so $I$ attains it's maximum in $B$ . The case for the minimum is the same.