This has been giving me a headache...
Let C([a;b]) be the set of all continuous real functions on the closed interval [a;b]. Then C([a;b]) is an algebra over the real numbers and further more a vector space over the reals.
Now we define a norm on this space so that
f_2 = sqrt("Integrate from a to b"(f(t)^2)dt) (This is a Riemann integral, not Lebesque afaik)
Now I'm supposed to show that with this norm this is not a Banach Space...
I've been making sequences of functions all day trying to find a cauchy sequence that converges outside of C([a;b]) but I haven't succeeded. Maybe I'm trying the wrong method to solve this but I can't think of another way.
Help?
Homework help, Banach spaces...
Moderators: gmalivuk, Moderators General, Prelates
 Torn Apart By Dingos
 Posts: 817
 Joined: Thu Aug 03, 2006 2:27 am UTC
Re: Homework help, Banach spaces...
Try to think of a sequence of continuous functions such that f_n(x) tends to 0 for x in [0,1) and f_n(x) tends to 1 for x in [1,2].

 Posts: 3
 Joined: Sat Feb 28, 2009 4:54 pm UTC
Re: Homework help, Banach spaces...
f_n(x) = (x^(2n))/(1+x^(2n)) works but I can't see how that helps.
Re: Homework help, Banach spaces...
Is that a Cauchy sequence? If so, does it converge (under your norm) to a function which is continuous on [0,2]? If it had to converge to some function, what would that function be?

 Posts: 3
 Joined: Sat Feb 28, 2009 4:54 pm UTC
Re: Homework help, Banach spaces...
Ok, I think I can finish this now.
I just need to show that it's a Cauchy sequence because the limit is not continuous.
I just need to show that it's a Cauchy sequence because the limit is not continuous.
 Torn Apart By Dingos
 Posts: 817
 Joined: Thu Aug 03, 2006 2:27 am UTC
Re: Homework help, Banach spaces...
Well, actually, there are lots of limits (in the space of all square integrable functions). You would need to show that they are all discontinuous.
Re: Homework help, Banach spaces...
Torn Apart By Dingos wrote:Well, actually, there are lots of limits (in the space of all square integrable functions). You would need to show that they are all discontinuous.
Good catch. To show that a normed vector space fails to be a Banach space you just need to show that it isn't complete. Which is to say that there is one Cauchy sequence which does not have a limit. So strictly speaking the statement that it's limit is not continuous is nonsensical. On the other hand, it shouldn't be hard to prove that if there were a limit it would be equal to 0 almost everywhere on a certain subinterval and so on.
Who is online
Users browsing this forum: No registered users and 9 guests