Construct a function $f$ which is $Lip-α$ but not $Lip-β$.

Question

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• Definition : Let $f$ be a continuous function, suppose $0<α\le1$ , $f\in Lip-α$ if there exists a constant $M$ s.t. $\qquad \qquad \bbox[8px,border:2px solid black] {|f(x)-f(y)|\le M|x-y|^α }$

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• if $0<α<β<1$ , Im trying to construct a function $f$ which is $Lip-α$ but not $Lip-β$.

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My approach:

• For $κ\in(0,1)$ let $f_κ(x)=x^κ,\forall x\in[0,1],$
  then, for $x>0$ $f_κ'(x)=κx^{κ-1}$ and by the Mean Value Theorem and for $0<h<1$ we have for some $ξ\in [x,x+h]$
  $f_κ(x+h)-f_κ(x)=f'_κ(ξ)\cdot h=κ\cdotξ^{κ-1} \cdot h=h^κκ(\frac{h}{ξ})^{1-κ} $.
  If $h\le x$ then $(\frac{h}{ξ})^{1-κ}\le 1$
  so that in that case we have $f_κ(x+h) -f_κ(x)\le κ\cdot h^κ$.
  If $x<h$ then $f_κ(x+h)-f_κ(x)\le (x+h)^κ\le2^κh^κ$
  So whenever $x>0$ we have $f_κ(x+h)-f_κ(x)\le 2^κh^κ$.
  For $x=0$ we have: $f_κ(0+h)-f_κ(0)=h^κ$
  So $\forall x\in[0,1]$ we have proved that $f_κ(x+h)-f_κ(x)\le 2^κh^κ$, so that $f_κ\in Lip-κ$.
  Checking the difference $f_κ(x+h)-f_κ(x)$ for $x=0$ shows immediately that this function cannot be in any higher $Lip-κ'$ $(\text{i.e: for } k'>k)$.
  
  Is there something wrong with that?