# Uniformly Lipschitz and bounded variation

I found an interesting question in a textbook (Apostol analysis chapter 6 problem… 2?) and was wondering if anyone could help me figure this out, or at least help adjust my thinking.

A function $f$ is called uniformly Lipschitz of order $\alpha>0$ on $[0,1]$ if $|f(x)-f(y)|<M|x-y|^\alpha$ for all $0\le x,y\le1$ and for some fixed $M$.

(a) Prove that $\alpha>1$ implies $f$ is constant, where $\alpha=1$ implies $f$ is of bounded variation.

(b) Find an $f$ that is uniformly Lipschitz on $[0,1]$ for some $\alpha<1$ but not of bounded variation.

(c) Find an $f$ of bounded variation on $[0,1]$ that satisfies no uniformly Lipschitz condition.

My progress:

(a) I solved, if $\alpha>1$ then $\left|\frac{f(x+h)-f(x)}h\right|<M|h|^{\alpha-1}\to0$ as $h\to0$, so $f'=0$. For $\alpha=1$, replace the RHS with $M$, so $f'$ is bounded so it is of bounded variation.

(b) I am not sure. The only functions not of bounded variation I can think of are things involving $\sin(1/x)$ which I have no conceivable way to prove are uniformly Lipschitz. In particular I guess that $x\sin(1/x)$ works but have no idea how to prove its Lipschitz if it even is.) Maybe $\sqrt{x}\sin(1/x)$ would be better because then there is a conceivable was to connect that to $\alpha=1/2<1$, plus we proved its not Lipschitz in a previous exercise.

(c) I haven't been able to find anything. If it satisfies no Lipschitz condition then I am pretty sure that probably $f'$ needs to be unbounded at some point. I'm pretty sure the way to approach this problem is by finding some monotonic function that satisfies no Lipschitz condition since that would imply it is of bounded variation. It can't be something like $\sqrt{x}$ (chosen because of the vertical tangent at $x=0$) since that would probably satisfy like $\alpha=1/2$. There are only a few classical continuous functions with vertical asymptotes, like $x^a$ ($0<a<1$) or $\sqrt{1-x^2}$, where the second is basically symmetric with the first one.

I also tried to construct it with the following reasoning: translate so that $f(0)=0$, then make $\left|\frac{f(x)}{x^\alpha}\right|$ unbounded for all $\alpha>0$. The smallest function that comes to mind that dominates all polynomial growth is $e^x$, which would give the construction of $f(x)=e^{1/x}$. But if $f(0^+)=+\infty$, then there will be no way to define $f(0)$ so that $f$ is strictly monotonic. Also there is a property that all functions of bounded variation need to be bounded too. So, this probably won't work.

Anyways thank you for the help.

• I solved part 3. I realized the problem with functions like $f(x)=x^k$ for $0<k<1$ were not strong enough since we can always just take $\alpha=k$ and kill it, so I took $f(x)=x^x$ for $x\neq0$ and $f(0)=1$ (which is bounded and decreasing on $[0,1/e]$ which means we can scale) and then noted that $f(\epsilon)/\epsilon^\alpha$ is unbounded for $\epsilon\to0^+$ (in particular $\epsilon<\min\{1/e,\alpha/2\}$ gives $f(\epsilon)/\epsilon=\epsilon^{-\alpha}-\epsilon^{\alpha/2}$) – Cody Johnson Aug 14 '16 at 16:13