Here is my own problem.
Let f be a function on (-oo,+oo) such that for any interval I, f is unbounded on I.
*I have been able to show that f is discontinous everywhere.
Now I want to find an example of such a function.
* shows how difficult it might be.
Or show it does not exist.
As we can construct a sequence of nested closed intervals I_n, and our
function is unbounded on all of them it must be infinite (undefined on
(-infty, infnty)) at the common point of the I_n's. This implies that it is undefined
everywhere so does not exist!
RonL
How about this for an idea:
1. let f(x)=0 for x irrational
2. Arrange the rationals [0,1) in the order 0, 1/2, 1/3, 2/3, 1/4, 2/4, 3,4,...
3. Then if r_n is the n-th rational in this sequence then put f(r_n)=n
4. Put f(x) = f(frac(x)), frac(x) the fractional part of x.
Conjecture: F as defined above has the property you are looking for
(and if it does not it can be modified slightly to have)
RonL
I did not check it yet. Because I am about to leave for class.
But, remember the important point I mentioned. IT MUST be discontinous everywhere. It provides an easy check.
Note, if in your function it turns out that f(0)=0 then we have a problem. It is continous at 0.
I was discussing this with somebody else today he gave the following construction.
f(x)=0 for irrationals.
f(a/b)=b for rationals where gcd(|a|,b)=1 and a!=0
f(0)=1.
I think that works.
Because small intervals contain rationals with large denominators.
I think we should not forget this function, because it might be useful for disproving certain conjectures about functions.
Look at this this is like the sort of thing that I had in mind.
Looking t what I suggested I now note that it needs the refinment (to make
my f single valued):
f(x) = 0 for x irrational.
for x rational f(x) = n, where x=r_n and n is the least n for which
this is true.
RonL