Is it possible for an increasing continuous function on $[0,\infty)$ not to be H$\ddot{\mathrm{o}}$lder continuous at some point?

Question

Let $f$ be a real valued function defined on the domain $[0,\infty)$, such that $f$ is strictly increasing and $f$ is continuous on $[0,\infty)$. It seems to me that, at any point $a$, in a neighborhood $U$, $f(x)-f(a)$ has to be dominated by something like $(x-a)^\alpha$, for some $\alpha>0$, at any point, i.e. I suspect that this kind of functions will be H$\ddot{\mathrm{o}}$lder continuous at any given point (maybe with different $\alpha$ s). But I cannot materialize this intuition of mine using some formal mathematical techniques.

Is it possible for $f$ not to be H$\ddot{\mathrm{o}}$lder continuous at some point? If it is not possible can anyone give me a hint to the techniques that might be helpful addressing this kind of question?

In case such an example do not exist, please do not give a full answer, but give hint to possible methods that I have to learn so that I can address this question myself. Any help is appreciated.

Edit: By the way, note that at any point where the functions is differentiable and the derivative is finite, the functions is automatically Lipscitz continuous at that point. Also, I suppose, though I am not sure, the points where the functions has one of its Dini derivatives finite, it is Lipscitz at that point. Thus the main points of interest are the points where the function is not differentiable and also the Dini derivatives blow up, which I think leave us the only possibility as the point $0$.

Answer 1

Hint: Try something like

$$f(x) = \sum_{n=1}^{\infty}c_n x^{1/n}$$

for suitable positive constants $c_n.$

Answer 2

It seems to me that these functions have to be dominated by something like $x^\alpha$ for some $\alpha > 0$

False. Take the function $f(x)=e^x$ as a counterexample.

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Also, there is no such thing as "Lipschitz continuous at some point". Lipschitz continuity is defined on a set, not at a point:

A function $f:X\to Y$ (where $X$ and $Y$ are metric spaces with metrics $d_X$ and $d_Y$ is Lipschitz continuous if there exists some $K\in\mathbb R$ such that $d_Y(f(x_1),f(x_2))\leq K\cdot d_X(x_1,x_2)$ for all pairs $x_1,x_2\in X$

Answer 3

As it's noted by @5xum, the concepts of Lipschitz continuity and Hölder continuity are related to global behaviour of the function, but I suppose you have an intuition about local properties of $ f $. So I give a hint to make a counterexample for a local version of your question.

It's possible that $ f $ is not Hölder continuous when restricted to any neighborhood of some point. To make a counterexample, you can construct a function that grows faster than any $ x ^ \alpha $ near $ 0 $. Such function can be defined using $ C ^ \infty $ functions that are not analytic.

If you think this hint in not helpful enough, feel free to ask me to elaborate more.