For (a) use what my analysis professor would call "the world's most trivial integral inequality". That is if on then . Combine this with the regular definition of continuity and the given metric to so what is continuous.
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... :-)
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 :-)