Prove Uniform Continuity: y^2 arctan y - x^2 arctan x

[daniel_i_l]

Homework Statement

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.

Homework Equations

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.

 

[neutrino]

How are a and b different?

 

[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.

 

[ZioX]

These look suspiciously like mean value theorem applications.

 

[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, regardless 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.

 

[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.