Local Lipschitz constants and a slightly weaker concept

Question

Let $(X, d)$ be a metric space and $f:X \rightarrow \mathbb R$ be a function such that

$$\lim_{r \rightarrow 0} \sup_{0 < y < B(x; r)} \frac{|f(x) - f(y)|}{d(x, y)}$$

exists for some $x \in X$.

Does

$$\lim_{r \rightarrow 0} \sup_{0 < y < B(x; r)} \frac{|f(x) - f(y)|}{r} $$

exists in this case?

Since $d(x, y) < r$, it is clear that $$\lim_{r \rightarrow 0} \sup_{0 < y < B(x; r)} \frac{|f(x) - f(y)|}{d(x, y)} \leq \lim_{r \rightarrow 0} \sup_{0 < y < B(x; r)} \frac{|f(x) - f(y)|}{r} .$$

Then when does the equal sign hold?

Answer 1

To observations:

• The notation $0 < y < B(x; r)$ is unusual.
• Although the first limit exists always (maybe equals $\infty$) as a limit of monotone funcion, the second one not always. We therefore ought to consider upper and lower limits.
• There's a mistake in the question. The condition $d(x,y)<r$ implies that $$\lim_{r \rightarrow 0} \sup_{y\in B(x; r)\setminus\{x\}} \frac{|f(x) - f(y)|}{d(x, y)} \color{red}{\geq} \limsup_{r \rightarrow 0} \sup_{y\in B(x; r)\setminus\{x\}} \frac{|f(x) - f(y)|}{r} .$$

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My example is based on noncontinuous function. If a continuous function is needed one can change a bit the definition of $f$ to be partially linear.

Let $f\colon\Bbb R\to\Bbb R$, $f(2^{-n})=2^{-n}$ and $f(x)=0$ for other points.

Then $\sup_{y\in B(x; r)\setminus\{x\}} \frac{|f(x) - f(y)|}{d(x,y)}=1$ for any $r>0$ and so the first limit is equal to $1$. On the other hand, for $r=2^{-n}$ $$ \sup_{y\in B(x; r)\setminus\{x\}} \frac{|f(x) - f(y)|}{r} = \sup_{y\in B(x; r)\setminus\{x\}} \frac{f(y)}{r} = \sup_{y\in B(x; r)\setminus\{x\}} \frac{2^{-n-1}}{2^{-n}} =\frac 12. $$ Therefore $$ \liminf_{r \rightarrow 0} \sup_{y\in B(x; r)\setminus\{x\}} \frac{|f(x) - f(y)|}{r}\leq \frac 12,$$ which shows that the second limit isn't equal to $1$.

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We can prove that the second limit doesn't exists in our case. Namely, for $r=2^{-n}+4^{-n}$ we have $2^{-n}<r<2^{-n+1}$ and therefore

$$ \sup_{y\in B(x; r)\setminus\{x\}} \frac{|f(x) - f(y)|}{r} = \sup_{y\in B(x; r)\setminus\{x\}} \frac{f(y)}{r} = \sup_{y\in B(x; r)\setminus\{x\}} \frac{2^{-n}}{2^{-n}+4^{-n}} \to 1 \ \ (n\to\infty), $$ so $$\limsup_{r \rightarrow 0} \sup_{y\in B(x; r)\setminus\{x\}} \frac{|f(x) - f(y)|}{r}\geq 1.$$

Final remark. It can be shown that $$\lim_{r \rightarrow 0} \sup_{y\in B(x; r)\setminus\{x\}} \frac{|f(x) - f(y)|}{d(x, y)} \color{red} = \limsup_{r \rightarrow 0} \sup_{y\in B(x; r)\setminus\{x\}} \frac{|f(x) - f(y)|}{r} .$$