Proving Hölder Continuity of f(u)=-1/(1+u)^2

[incognitO]

Hello guys, I am trying to prove that the function

f(u)=-\frac{1}{(1+u)^2}
is Hölder continuous for -1<u \le 0 but I am stuck. Here is what I have done:

If |u_1-u_0| \le \delta then

\left|-\frac{1}{(1+u_1)^2}+\frac{1}{(1+u_0)^2}\right| \le \left|\frac{(u_1+u_0)+2}{(1+u_1)^2(1+u_0)^2}\right||u_1-u_0| \le \frac{2|u_1-u_0|}{(1+u_1)^2(1+u_0)^2}

and I don't know how to continue... Any suggestions?

 

[Galileo]

If u_0 and u_1 are between -1 and 0, what can you say about the bounds on 2/[(1+u_0)^2(1+u_1)]^2?

 

[incognitO]

I don't understand what you mean.

The worst behavior of the bound is when u_1 and u_0 are close to -1, but that should reflect in the bounding constant H(u_0) right?

Let me put it this way (maybe I am saying stupid things but, that woulndt be new :P )

If -a <u \le 0 and 0<a<1, then

|f(u_1)-f(u_0)| \le \frac{2}{(1-a)^4}|u_1-u_0|

so it is Hölder as long as I am far from -1 right?

Now, what i have to do is instead of finding a bound depending on a is find a bound depending on u_0 right?


EDIT:

In fact,

|f(u_1)-f(u_0)| \le \frac{2}{(1-a)^3}|u_1-u_0|.

 

[incognitO]

How about this...

|f(u_1)-f(u_0)| \le H(u_0,u_1) |u_1-u_0|

where

H(u_0,u_1)=\left\{\begin{array}{cc} \dfrac{2}{(1+u_0)^3} & \hbox{if} \quad u_0<u_1 \\ & \\ \dfrac{2}{(1+u_1)^3} & \hbox{if} \quad u_0>u_1 \end{array}\right.

So f(u) \in C^{0,1} in -1<u_0,u_1 \le 0, just not uniformly Hölder...

If you were my analysis teacher, would you flunk me?