# Hölder Continuity

1. Dec 12, 2005

### 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 dont know how to continue... Any suggestions?

Last edited: Dec 12, 2005
2. Dec 13, 2005

### Galileo

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

3. Dec 13, 2005

### incognitO

I dont 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 im 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 im 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,

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

Last edited: Dec 13, 2005
4. Dec 13, 2005

### incognitO

$$|f(u_1)-f(u_0)| \le H(u_0,u_1) |u_1-u_0|$$
$$H(u_0,u_1)=\left\{\begin{array}{cc} \dfrac{2}{(1+u_0)^3} & \hbox{if} \quad u_0<u_1 \\ & \\ \dfrac{2}{(1+u_1)^3} & \hbox{if} \quad u_0>u_1 \end{array}\right.$$
So $f(u) \in C^{0,1}$ in $-1<u_0,u_1 \le 0$, just not uniformly Hölder...