# Thread: Metric spaces - continuity / compact

1. ## Metric spaces - continuity / compact

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... :-)

2. 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.

3. 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 :-)

4. Originally Posted by christina182
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??

5. Originally Posted by christina182
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!

6. 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.