Continuous Function and lipschitz continuity

Question

Let $C(a,b)^\alpha$ the set of Hölder function continuous.

If $\alpha<\beta$ ,under this condition I must prove that $C(a,b)^\alpha \subset C(a,b)^\beta$ or $C(a,b)^\beta \subset C(a,b)^\alpha$.

Case 1 If $0<\alpha,\beta<1$.

Let $f\in C(a,b)^\alpha$ the $|f(x)-f(y)|\leq K|x-y|^\alpha, \forall x,y\in (a,b)$.

But under this condition I have not been able to establish a condition in the interval $(0,1)$ the function $f(x)=\sqrt x$, and $\alpha=0.1$ and $\beta=0.2$ the inequality is not preserved. The other case is a little easy.

I would like to get a suggestion for the prove.

Answer 1

The idea of proof:

$i)$ Suppose that $(a,b)\subset(0,1)$;

Let $f\in C^\beta(a,b) \Rightarrow |f(x)-f(y)|\leq K|x-y|^\beta$ if $x,y\in (a,b)$, then we have to $0<\alpha,\beta<1$, and $(a,b)\subset(0,1)$, from this it follows: $K|x-y|\beta \leq K|x-y|\alpha$, this implies that $|f(x)-f(x)| \leq K|x-y|^\alpha$.

Hence $f\in C^\alpha(a,b)$ .

Answer 2

When $b-a \le 1$, the conclusion is always $C(a,b)^{\beta} \subset C(a,b)^{\alpha}$, given that $0< \alpha < \beta< +\infty$.

${\bf Fact.}$ For any $c \in (0,1)$, $c^{\beta}\le c^{\alpha}$ whenever $0< \alpha < \beta< +\infty$.

You can apply this fact into your inequality and show that whenever $f\in C(a,b)^{\beta}$, it is true that $f\in C(a,b)^{\alpha}$.