First of all I want to mention that this is about an assignment I was given at school. I don't need the straight answer as much as a few hints to get started.
To give you a context, at the moment we're beginning to study integrals.
The question I'm stuck at is to verify :
Show that $f(x) = \sqrt{x}$ is uniformly continuous on $[0,+\infty[$. The exercise then suggest that this can be accomplished by proving $f$ is Hölder continuous with exponent $\frac12$.
So I know if $f$ is Hölder 1/2 then we must have :
$$ \exists C \in \mathbb{R}, \forall (x,y) \in [0,+\infty[^2 : \left( |f(x)-f(y)| \le C.|x-y|^{1\over 2}\right)$$
Applied to $\sqrt{x}$ it gives :
$$ \exists C \in \mathbb{R}, \forall (x,y) \in [0,+\infty[^2 : \left( |\sqrt{x}-\sqrt{y}| \le C.|x-y|^{1\over 2}\right)$$
Which is
$$ \exists C \in \mathbb{R}, \forall (x,y) \in [0,+\infty[^2 : \left( {|\sqrt{x}-\sqrt{y}|\over |x-y|^{1\over 2}} \le C\right)$$
This is where I'm stuck. I think I'm missing some kind of comparison theorem relative to the derivative of $\sqrt{x}$
Please provide me with hints, thanks.
We have: $|\sqrt{x}-\sqrt{y}|=\dfrac{|x-y|}{\sqrt{x}+\sqrt{y}}=\sqrt{|x-y|}\cdot \dfrac{\sqrt{|x-y|}}{\sqrt{x}+\sqrt{y}}\leq 1\cdot |x-y|^{\frac{1}{2}}$, since $\sqrt{|x-y|} \le \sqrt{x}+\sqrt{y}$ is clearly true.