Let $0<a<1$. Does there exist a function $f:\mathbb{R} \rightarrow \mathbb{R}$ such that:
$$\left|f(x)-f(y)\right |\geq \left|x-y\right|^a, \;\; \forall x, y \in \mathbb{R}$$
Solution
No, there does not since if we suppose that there exists then this function would be $1-1$ (fair and square). Let $g:f(\mathbb{R}) \rightarrow \mathbb{R}$ be its inverse. Then:
$$\left | x-y \right | \geq \left | g(x)-g(y) \right |^a \Rightarrow \left | g(x)-g(y) \right |\leq \left | x-y \right |^{1/a} \tag{*} \label{*}$$
This shows that both $f$ and $g$ are continuous. More specifically, by applying Bolzano we easily see that $f$ is onto , hence $g$ is defined everywhere. $(**)$.
The equation \eqref{*} shows that $g$ is differentiable with $g'(x)=0$ , meaning that $g$ is constant, a contradiction since $f$ is not constant as it is onto.
$(**)$ We need this step because just after it we differentiate , thus we somehow need to assure that $g$ is defined in an interval (and not only in e.g isolated points)
$$\left|f(x)-f(y)\right |\geq \left|x-y\right|^a, \;\; \forall x, y \in \mathbb{R}$$
Solution
No, there does not since if we suppose that there exists then this function would be $1-1$ (fair and square). Let $g:f(\mathbb{R}) \rightarrow \mathbb{R}$ be its inverse. Then:
$$\left | x-y \right | \geq \left | g(x)-g(y) \right |^a \Rightarrow \left | g(x)-g(y) \right |\leq \left | x-y \right |^{1/a} \tag{*} \label{*}$$
This shows that both $f$ and $g$ are continuous. More specifically, by applying Bolzano we easily see that $f$ is onto , hence $g$ is defined everywhere. $(**)$.
The equation \eqref{*} shows that $g$ is differentiable with $g'(x)=0$ , meaning that $g$ is constant, a contradiction since $f$ is not constant as it is onto.
$(**)$ We need this step because just after it we differentiate , thus we somehow need to assure that $g$ is defined in an interval (and not only in e.g isolated points)
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