Uniform continuity

1. Jul 24, 2007

daniel_i_l

1. The problem statement, all variables and given/known data
Prove that if y>=x>=0:
a) $$y^2 arctan y - x^2 arctan x >= (y^2 - x^2) arctan x$$

b) $$\ | \ y^2 arctan y - x^2 arctan x \ | \ >= (y^2 - x^2) arctan x$$

c) use (b) to prove that x^2 arctan(x) isn't UC in R.

2. Relevant equations

3. The attempt at a solution

a) We have to prove that $$y^2 ( arctan y - arctan x ) >= 0$$
And since arctan(y) - arctan(x) >= 0 for all y>=x and y^2 > 0 this is true.
b)Since $$y^2 arctan y - x^2 arctan x >= 0$$ for all y>=x>=0 then this is obviously true from a.

c)If we choose Epsilon (E) = 1/2 , Lambda (L) > 0 and y = x+L then
(y^2 - x^2)arctan(x) = (2xL - L^2)arctanx and the limit of that at infinity is infinity. So we can find N>0 so that for every x>N
(y^2 - x^2)arctan(x) = |y^2 arctan(y) - x^2 arctan(x)| > E

Now my questions are:
1) a & b seemed too trivial -are those the right are answers?
2)Is (c) right?

Thanks.

Last edited: Jul 24, 2007
2. Jul 24, 2007

neutrino

How are a and b different?

3. Jul 24, 2007

daniel_i_l

Oops, in b you need to prove that
|y^2 arctan(y) - x^2 arctan(x)| >= (y^2 - x^2)arctanx
the same as (a) just with the absolute value on the right.
Thanks.

4. Jul 24, 2007

ZioX

These look suspiciously like mean value theorem applications.

5. Jul 24, 2007

ZioX

These look suspiciously like mean value theorem applications.

Edit: Sorry for me being a dolt.

a and b are both correct. The reasoning is thusly: Suppose y>x>=0

We have $arctan(y)\ge\arctan(x)$ since $f(x)=\arctan(x)$ is monotonically increasing ($f' \ge 0$ on that particular domain). Therefore $y^2\arctan(y)-x^2\arctan(x) \ge 0$. Since $arctan(y)\ge\arctan(x)$ we have $y^2\arctan(y)-x^2arctan(x) \ge y^2\arctan(x)-x^2\arctan(x)=(y^2-x^2)\arctan(x)$.

b follows since both sides are positive, hence their absolute values are the same.

you haven't properly finished c. "(y^2 - x^2)arctan(x) = |y^2 arctan(y) - x^2 arctan(x)| > E" is dead wrong, as the equality does not hold. You actually want to be doing a proof by contradiction. Suppose it was U.C. on the positive reals. Let E=1/2. Suppose L is the delta that works, ie for |y-x|<L then |f(y)-f(x)|<E=1/2. Since y>x then y=x+a for some constant a>0. Etc Etc and look for your contradiction. Your idea was right, that for large enough x, irregardless of distance between x and y, the difference between f(x) and f(y) exceed 1/2.

I would do a finishing statement, saying something along the lines of "therefore, there does not exist any L>0 such that when |y-x|<L, |f(y)-f(x)|<1/2, ie, f is not UC on the positive reals.

A cople of notes, use \ge and \le for your greater/less then or equal to signs. Classically we use the definitions of $|x-y|<\delta, |f(x)-f(y)|<\epsilon$, which is no big deal.

Last edited: Jul 24, 2007
6. Jul 24, 2007

daniel_i_l

| "(y^2 - x^2)arctan(x) = |y^2 arctan(y) - x^2 arctan(x)| > E" is dead
| wrong, as the equality does not hold.
But isn't it enough to prove that there's some N>0 so that for all x>N it holds?
Thanks.