# Local Hölder continuity

Assume that $$f:[0,1]\to [0,1]$$ is Hölder continuous with the constant $$1/3$$ and assume that for every $$s\in [0,1]$$ we have $$\limsup_{x\to s}\frac{\lvert f(x)-f(s)\rvert}{\lvert x-s\rvert^{1/2}}<\infty.$$ Does this imply that $$\sup_{x\neq s}\frac{\lvert f(x)-f(s)\rvert}{\lvert x-s\rvert^{1/2}}<\infty?$$

$$\newcommand{\de}{\delta}$$ $$\newcommand{\fl}{\lfloor1/x\rfloor}$$ The answer is no. E.g., consider the function $$f$$ defined by the conditions that $$f(0):=0$$ and $$f(x)=\frac{(x-x_{n+1})^{1/2}(x_n-x)} {(x_n-x_{n+1})^{7/6}}$$ for each natural $$n$$ and all $$x\in[x_{n+1},x_n]$$, where $$x_n:=1/n$$, so that $$n=\fl$$.

Details: Let $$\begin{equation} \de_n:=[x_{n+1},x_n],\quad h_n:=x_n-x_{n+1}=\frac1{n(n+1)}, \end{equation}$$ $$\begin{equation} g(x):=g_n(x):=(x-a)^{1/2}(b-x),\quad a:=a_n:=x_{n+1},\quad b:=b_n:=x_n, \end{equation}$$ $$AB\vee CD:=\max(AB,CD)$$. Hence, $$\begin{equation} f(x)=\frac{g_n(x)}{h_n^{7/6}}=\frac{g(x)}{h_n^{7/6}} \end{equation}$$ for each natural $$n$$ and all $$x\in\de_n$$. Next, for any $$x,y,h$$ such that $$a\le x\le y=x+h\le b$$ we have \begin{align} |g(y)-g(x)|&=|(y-a)^{1/2}(b-x)-(x-a)^{1/2}(b-x) \\ &+(y-a)^{1/2}(b-y)-(y-a)^{1/2}(b-x)| \\ &=\big|(b-x)\big((y-a)^{1/2}-(x-a)^{1/2}\big)-(y-a)^{1/2}h\big| \\ &\le(b-x)\big((y-a)^{1/2}-(x-a)^{1/2}\big)\vee(y-a)^{1/2}h \\ &\le(b-x)h^{1/2}\vee(y-a)^{1/2}h \\ &\le(b-a)h^{1/2}\vee(b-a)^{1/2}h \\ &=(b-a)h^{1/2}=h_n(y-x)^{1/2}, \end{align} so that $$f$$ is locally $$1/2$$-Hölder continuous on $$(0,1]$$ and, moreover, $$|f(y)-f(x)|\le h_n^{-1/6}(y-x)^{1/2}\le(y-x)^{1/3}$$ and hence $$\begin{equation} |f(y)-f(x)|\le(y-x)^{1/3}, \tag{*} \end{equation}$$ for any $$x,y$$ such that $$x_{n+1}\le x\le y\le x_n$$.

If now $$x\in\de_m$$ and $$y\in\de_n$$ for some natural $$m$$ and $$n$$ such that $$m>n$$, then $$x_{m+1}\le x\le x_m\le x_{n+1}\le y\le x_n$$ and \begin{align} |f(y)-f(x)|&\le f(x)\vee f(y) \\ &=(f(x)-f(x_m))\vee(f(y)-f(x_{n+1})) \\ &\le(x_m-x)^{1/3}\vee(y-x_{n+1})^{1/3}\le(y-x)^{1/3}. \end{align}

So, (*) holds for all $$x,y$$ such that $$0, that is, $$f$$ is $$1/3$$-Hölder continuous on $$(0,1]$$.

Next, for each natural $$n$$ and all $$x\in\de_n$$ $$\begin{equation} 0\le f(x)\le\frac{2(b-a)^{1/3}}{3\sqrt3}\le\frac{(b-a)^{1/3}}2=\frac1{2n^{1/3}(n+1)^{1/3}} \le\frac1{(n+1)^{2/3}}\le x^{2/3}\le x^{1/2}\le x^{1/3}. \end{equation}$$ So, recalling that $$f$$ is locally $$1/2$$-Hölder continuous on $$(0,1]$$ and $$1/3$$-Hölder continuous on $$(0,1]$$, we conclude that $$f$$ is locally $$1/2$$-Hölder continuous on $$[0,1]$$ and $$1/3$$-Hölder continuous on $$[0,1]$$; moreover, it follows that $$0\le f\le1$$ on $$[0,1]$$.

However, $$f$$ is not $$1/2$$-Hölder continuous on $$[0,1]$$, because $$\begin{equation} \frac{f(a_n+h_n/2)-f(a_n)}{(h_n/2)^{1/2}}=h_n^{-1/6}/2\to\infty \end{equation}$$ as $$n\to\infty$$, which fully confirms the "no" answer.

The graphs of the functions $$x\mapsto f(x)$$, $$x\mapsto f(x)/(x-x_{n+1})^{1/2}$$, $$x\mapsto f(x)/(x-x_{n+1})^{1/3}$$ with $$n=\fl$$ are shown here, left-to-right, respectively:

• That is very nice! – Nik Weaver Oct 20 at 14:59
• @NikWeaver : Thank you for your kind comment! – Iosif Pinelis Oct 21 at 1:08