2
$\begingroup$

I am curious to know why we don't normally talk about Hölder spaces with Hölder exponent $\beta$ $\ge$ $1$. I tried finding the answer and all i found out was that the function assumes constant. But i don't know why ? It would be good to get a proper explanation.

3 Answers 3

3

Suppose $|f(x)-f(y)|\le C|x-y|^{1+\alpha}$ for some positive constants $C, \alpha$ and for all $x$, $y$. Then we have $$\biggl|{ f(x)-f(y) \over x-y }\biggr|\le C|x-y|^\alpha,\quad x\ne y.$$ Since $\alpha>0$, we have, for $x$ fixed, $\lim\limits_{y\rightarrow x} |x-y|^\alpha=0$. This implies that $f$ is differentiable and that $f'(x)$ is zero for all $x$. Thus, $f$ is a constant function.

  • 0
    nice, well explained :)2012-05-13
3

For $\beta=1$, it's the set of Lipschitz continuous function, which is not uninteresting.

But for $\beta>1$, if $f$ is such that $|f(x)-f(y)|\leq C|x-y|^{\beta}$ for an universal constant $C$, we have for a fixed $x$ that $$|f(x)-f(0)|=\left|\sum_{j=0}^{n-1}f\left(\frac{(j+1)x}n\right)-f\left(\frac{jx}n\right)\right|\leq C\sum_{j=0}^{n-1}\left(\frac{|x|}n\right)^{\beta}=C|x|^{\beta}\frac 1{n^{\beta-1}}$$ hence $f(x)=f(0)$ for all $x$ and $f$ is constant.

  • 0
    @ David can you explain me the expression which equals to $|f(x)-f(0)|$.2012-05-13
  • 0
    A $j$ was missing, it's a telescopic sum.2012-05-13
  • 0
    @ David can you explain me what is special about your way of formulation ? Its interesting but it would be interesting to know the speciality of this formulation :)2012-05-13
  • 0
    @Ananda I don't understand what you mean by special.2012-05-13
  • 0
    @ Davide i understood few things as Gaston explained below, that ur proof requires only well defined sum and a norm whereas the other proof by David Mitra requires convexness ( which i am still not understanding why ?) I just wanted to understand the aspects of your proof and compare between the two proofs :)2012-05-13
1

This is only true when the function take values in a normed vector space and domain is an convex set. If it holds Davide Giraudo answer can apply.

  • 0
    @ Gaston can you explain a bit more why ? Why should the domain be connected and normed ? :)2012-05-13
  • 0
    For Davide Giraudo proof, you only need a norm and a well defined sum beetween values, in other words, if you have two elements f(x), f(y) you need that (f(y)-f(x))/2 exists, for this you need only closure under sum and convexness. Connected isn't enough sorry, I make a mistake. But convex space implies connected space.2012-05-13
  • 0
    Thanx for the explanation , can you tell me what do u mean by closure here ? And what would happen if it was not convex ? tnx :)2012-05-13
  • 0
    If is not convex, probably one of terms in linear combination between f(x) and f(y) doesnt exist in space. Closure meaning in this case is related to binary operation "+". If a and b are elements of set, then a+b it is, then we say that the space is closed under "+".2012-05-13