If be a continuous function and if f satisfies . The prove that on

Results 1 to 15 of 20

- Jul 14th 2010, 01:32 AM #1

- Joined
- Feb 2009
- From
- Chennai
- Posts
- 148

- Jul 14th 2010, 04:33 AM #2
Maybe you can show that the set : has an accumulation point. Then

We can easily show that are in . Maybe we can construct a sequence that must be in U.

(EDIT: this is not true... I misremembered a theorem from Complex analysis,sorry)

We must show U is dense.

Didn't solve it yet myself. Also curious for a solution.

- Jul 14th 2010, 05:18 AM #3
My attempt...

I defined the following function:

Now:

Therefore:

Now:

There is so that:

Hence:

or:

, therefor have no zeros...

But:

so has a point which

And I stuck here...

Edit: (maybe therefor f(-1)f(1)=0 ==> f(1)=0=f(-1), and because the function is continuous f(x)=0 for every x in [-1,1])

- Jul 14th 2010, 06:17 AM #4
Surely this is just the zero function, by definition!

Now, note that actually (why?).

Next, you should notice that and . So, if you can prove the result for restricted to this interval then you are done. So, define,

to be the restriction of to ( for ).

Now, this function is continuous and defined over a closed interval, thus it must reach a minimum value. Let be such that for all .

This now means that , and so where and . As and multiplication by can only make it bigger we have that where and . As this means that . Note that .

Thus, it is now sufficient to prove that the maximum value of ...which I can't seem to do...(proving the maximum value of would also work as this would imply that the maximum value of .)

- Jul 14th 2010, 09:56 AM #5

- Joined
- Jun 2008
- From
- Plymouth
- Posts
- 125
- Thanks
- 5

- Jul 14th 2010, 10:05 AM #6

- Joined
- May 2010
- From
- Texas
- Posts
- 48

- Jul 14th 2010, 10:48 AM #7

- Joined
- Jun 2008
- From
- Plymouth
- Posts
- 125
- Thanks
- 5

- Jul 14th 2010, 11:01 AM #8

- Jul 14th 2010, 11:33 AM #9

- Joined
- Jun 2008
- From
- Plymouth
- Posts
- 125
- Thanks
- 5

- Jul 14th 2010, 11:35 AM #10

- Joined
- Feb 2009
- From
- Chennai
- Posts
- 148

- Jul 14th 2010, 11:43 AM #11

- Joined
- Jun 2008
- From
- Plymouth
- Posts
- 125
- Thanks
- 5

- Jul 14th 2010, 12:00 PM #12

- Joined
- Jun 2008
- From
- Plymouth
- Posts
- 125
- Thanks
- 5

- Jul 14th 2010, 12:17 PM #13

- Joined
- Aug 2009
- From
- Israel
- Posts
- 976

Fine until here..

So your conclusion is false.

and for the general case if we let be arbitrary such that , and then let we finally get:

Which gives us, well... not much (this is the same result you should've reached with your operator, you just miscalculated.)

"I turn away with fright and horror from this lamentable evil of functions which do not have derivatives" Hermite

P.S.

If then which does not change the result.

- Jul 14th 2010, 01:08 PM #14

- Joined
- Jun 2008
- From
- Plymouth
- Posts
- 125
- Thanks
- 5

- Jul 14th 2010, 01:30 PM #15

- Joined
- Aug 2009
- From
- Israel
- Posts
- 976