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Let $f$ be a Lipschitz function defined on a ball $B$ in $\mathbb{R}^n$, and assume that the following property on the Lipschitz constant (restricted to smaller balls) holds:

$$ \lim_{r\to 0}\sup_{|x-y|\leq r} \frac{|f(x)-f(y)|}{|x-y|}=0. $$

Can we assert that $f$ is constant? Intuitively, this should be the case, since this limit should tend to the supremum of the derivative. However, I was not able to fill in the details.

MathFreak
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    The statement is not true on a disconnected domain like $ { 0,1 } $. Therefore, you'll need to use the connectedness (maybe path connectedness) of your domain. Consider $x $ and $ y $ points in your domain and a path between them. Let $ \varepsilon > 0 $ be given. Subdivide the path small enough that you can use the condition on the subdivision to show that $ d(f(x),f(y)) < \varepsilon $. – Jake Mirra Aug 30 '23 at 12:35
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    @JakeMirra I'm late to the party but path connectedness of the domain of $f$ is not enough, if your path has infinite length the argument will break. (consider e.g., Van Koch snowflake) Unless you additionally are assuming domain is open, in which case it's all fine. Essentially what you need is to be rectifiably connected – M W Aug 31 '23 at 00:14

3 Answers3

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My thanks go to @Bruno B for pointing out an error in my argument.

Let's fix $x$ and consider the limit as $y$ approaches $x$. Since $y \to x$ as $|x-y| \to 0$, we can parametrize $y$ by $y=x+r\theta$, where $r$ is the distance $|x-y|$ and $\theta$ is some angle in $\mathbb{R}^n$ (see n-DIMENSIONAL SPHERE)

For any $r$ we have $$\sup_{θ} \frac{|f(x)-f(x+r\theta)|}{|x-(x+r\theta)|}=\sup_{y:\;|x-y|= r} \frac{|f(x)-f(y)|}{|x-y|}\leq \sup_{y:\;|x-y|\leq r} \frac{|f(x)-f(y)|}{|x-y|}$$ Therefore, \begin{align} 0&\leq\lim_{y \to x} \frac{|f(x)-f(y)|}{|x-y|}\\ &=\lim_{|y-x| \to 0} \frac{|f(x)-f(y)|}{|x-y|}\\ &=\lim_{r \to 0}\frac{|f(x)-f(x+r\theta)|}{|x-(x+r\theta)|}\\ &\leq \lim_{r\to 0}\left(\sup_{θ} \frac{|f(x)-f(x+r\theta)|}{|x-(x+r\theta)|}\right)\\ &\leq \lim_{r\to 0}\left(\sup_{y:\;|x-y|\leq r} \frac{|f(x)-f(y)|}{|x-y|}\right)\\ &= 0. \end{align} By the definition of the derivative, this limit is equivalent to $|f'(x)|$, thus we have: $$ \forall x\in B,\quad |f'(x)| = 0. $$ We can conclude that $f$ is constant on the entire ball $B$.

Ricky
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  • Thanks, I understand it's not very rigorous. I'm working to fix it. What I'm thinking is for $y$ to approach $x$ from a sphere and we choose the angle that will make the value largest. – Ricky Aug 30 '23 at 14:04
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    The only thing I'll add (hopefully I'm not too much of a bother...) is that the limit is not $|df(x)|$ per se, or at least it's not obvious. The approach that's more natural to me is to say that it means that $df(x)(h) = 0$ for all $h$ by definition of the differential. Otherwise this is a good proof! – Bruno B Aug 30 '23 at 14:42
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Let $x \in B$ be fixed. Then, for $r > 0$ small enough that $x + h$ belongs in $B$ when $|h| \leq r$, we have: $$0 \leq \frac{|f(x+h) - f(x)|}{|h|} \leq \sup_{h',\, |h'| \leq |h|} \frac{|f(x+h') - f(x)|}{|h'|}$$ Taking the $\limsup_{|h| \to 0}$ on this inequality, we obtain: $$\begin{split} 0 \leq \limsup_{|h| \to 0} \frac{|f(x+h) - f(x)|}{|h|} &\leq \limsup_{|h| \to 0} \sup_{h',\, |h'| \leq |h|} \frac{|f(x+h') - f(x)|}{|h'|}\\ &\leq \limsup_{r \to 0}\sup_{h',\, |h'| \leq r} \frac{|f(x+h') - f(x)|}{|h'|} = 0\end{split}$$ where the last inequality (which is an equality) holds thanks to the fact that the expression only depends on $|h|$ and not $h$.
Hence $$\limsup_{|h| \to 0} \underbrace{\frac{|f(x+h) - f(x)|}{|h|}}_{\geq 0} = 0$$ and thus $\liminf = \limsup = 0$, meaning that the limit exists and is equal to: $$\lim_{|h| \to 0} \frac{|f(x+h) - f(x)|}{|h|}= 0$$ Finally, this means that the differential of $f$ at $x$ is the zero functional, therefore, since $B$ is connected and this holds for all $x \in B$, $f$ is constant.

Bruno B
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The other answers are fine but I would like to point out that this result is also true in much greater generality, and with a non-uniform assumption on the limit, provided we simply satisfy a rectifiability condition.

Namely, let $X$, $Z$ be metric spaces, and let $f\colon X\to Z$ satisfy the infinitesimal Lipschitz condition that for each $x\in X$, we have $$\lim_{r\to 0}\sup_{y\in X \text{, } d(x,y)\leq r} \frac{d(f(x),f(y))}{d(x,y)}=0\text{.} $$

Then $f$ is constant on every rectifiable curve in $X$. In particular, if $X$ is rectifiably connected, then $f$ is constant.

Proof

The idea here is we combine rectifiability and the point-separating nature of Lipschitz functions to reduce to the real case.

That is, on every rectifiable path $\gamma\colon[0,l(\gamma)]\to X$, parametrized by arc-length, and every $1$-Lipschitz function $L$ on $Z$, we have at each $s_0\in[0,l(\gamma)]$ that

\begin{align*} |(L\circ f\circ\gamma)(s)-(L\circ f\circ\gamma)(s_0)| &\leq d((f\circ\gamma)(s),(f\circ\gamma)(s_0))\\ &\leq \omega(d(\gamma(s),\gamma(s_0))) \cdot d(\gamma(s),\gamma(s_0))\\ &\leq \omega(d(\gamma(s),\gamma(s_0))) \cdot |s-s_0|\text{,} \end{align*} where $\omega(\delta)\to 0$ as $\delta\to 0$, whereby $(L\circ f\circ \gamma)'(s_0)=0$. Thus $L\circ f\circ \gamma$ is constant on $[0,l(\gamma)]$, so that if $p,q \in \gamma([0,l(\gamma)])$, $L(f(p))=L(f(q))$. Applying this to the $1$-Lipschitz function $L(z)=d(z,f(p))$, we get that $$0=d(f(p),f(p))=d(f(q),f(p))\text{,}$$ and so $f(p)=f(q)$.

Remark

Without a condition like rectifiable connectedness, there are simple counterexamples to this result, e.g., let $X=[0,1]$ with the "snowflake" metric $d(x,y)=|x-y|^{\frac{1}{2}}$, and let $Z=\mathbb R$ with the usual metric, then let $f$ be the identity.

M W
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