Skip to main content
MathOverflow

Return to Question

Became Hot Network Question
deleted 44 characters in body
Source Link
Nate River
Nate River
  • 6.2k
  • 2
  • 23
  • 99

We say a measurable subset $S$ of $\mathbb R^n$ is measure dense if for every open set $U \subset \mathbb R^n$, $U \cap S$ is of positive Lebesgue measure.

Let $n \geq 2$, and let $f: \mathbb R^n \to \mathbb R$ be a Lipschitz continuous function with strict Lipschitz constant $L > 0$.

That is, $|f(x) - f(y)| < L|x - y|$ for all $x \neq y$ in $\mathbb R^n$.

Question: Let $n \geq 2$. Can there exist a function $f: \mathbb R^n \to \mathbb R$ suchIs it possible that $|Df| = L$ on a measure dense set?

Note: Here $Df$ denotes the total derivative of $f$, and $|\cdot|$ the operator norm of a linear map.

We say a measurable subset $S$ of $\mathbb R^n$ is measure dense if for every open set $U \subset \mathbb R^n$, $U \cap S$ is of positive Lebesgue measure.

Let $f: \mathbb R^n \to \mathbb R$ be a Lipschitz continuous function with strict Lipschitz constant $L > 0$.

That is, $|f(x) - f(y)| < L|x - y|$ for all $x \neq y$ in $\mathbb R^n$.

Question: Let $n \geq 2$. Can there exist a function $f: \mathbb R^n \to \mathbb R$ such that $|Df| = L$ on a measure dense set?

Note: Here $Df$ denotes the total derivative of $f$, and $|\cdot|$ the operator norm of a linear map.

We say a measurable subset $S$ of $\mathbb R^n$ is measure dense if for every open set $U \subset \mathbb R^n$, $U \cap S$ is of positive Lebesgue measure.

Let $n \geq 2$, and let $f: \mathbb R^n \to \mathbb R$ be a Lipschitz continuous function with strict Lipschitz constant $L > 0$.

That is, $|f(x) - f(y)| < L|x - y|$ for all $x \neq y$ in $\mathbb R^n$.

Question: Is it possible that $|Df| = L$ on a measure dense set?

Note: Here $Df$ denotes the total derivative of $f$, and $|\cdot|$ the operator norm of a linear map.

added 47 characters in body
Source Link
Nate River
Nate River
  • 6.2k
  • 2
  • 23
  • 99

We say a measurable subset $S$ of $\mathbb R^n$ is measure dense if for every open set $U \subset \mathbb R^n$, $U \cap S$ is of positive Lebesgue measure.

Let $f: \mathbb R^n \to \mathbb R$ be a Lipschitz continuous function with strict Lipschitz constant $L > 0$.

That is, $|f(x) - f(y)| < L|x - y|$ for all $x \neq y$ in $\mathbb R^n$.

Question: Let $n \geq 2$. Can there exist a function $f: \mathbb R^n \to \mathbb R$ such that $|Df| = L$ on a measure dense set?

Note: Here $Df$ denotes the total derivative of $f$, and $|\cdot|$ the operator norm of a linear map.

We say a measurable subset $S$ of $\mathbb R^n$ is measure dense if for every open set $U \subset \mathbb R^n$, $U \cap S$ is of positive Lebesgue measure.

Let $f: \mathbb R^n \to \mathbb R$ be a Lipschitz continuous function with strict Lipschitz constant $L > 0$.

That is, $|f(x) - f(y)| < L|x - y|$ for all $x \neq y$ in $\mathbb R^n$.

Question: Can there exist a function such that $|Df| = L$ on a measure dense set?

We say a measurable subset $S$ of $\mathbb R^n$ is measure dense if for every open set $U \subset \mathbb R^n$, $U \cap S$ is of positive Lebesgue measure.

Let $f: \mathbb R^n \to \mathbb R$ be a Lipschitz continuous function with strict Lipschitz constant $L > 0$.

That is, $|f(x) - f(y)| < L|x - y|$ for all $x \neq y$ in $\mathbb R^n$.

Question: Let $n \geq 2$. Can there exist a function $f: \mathbb R^n \to \mathbb R$ such that $|Df| = L$ on a measure dense set?

Note: Here $Df$ denotes the total derivative of $f$, and $|\cdot|$ the operator norm of a linear map.

Source Link
Nate River
Nate River
  • 6.2k
  • 2
  • 23
  • 99

On functions with strict Lipschitz constant

We say a measurable subset $S$ of $\mathbb R^n$ is measure dense if for every open set $U \subset \mathbb R^n$, $U \cap S$ is of positive Lebesgue measure.

Let $f: \mathbb R^n \to \mathbb R$ be a Lipschitz continuous function with strict Lipschitz constant $L > 0$.

That is, $|f(x) - f(y)| < L|x - y|$ for all $x \neq y$ in $\mathbb R^n$.

Question: Can there exist a function such that $|Df| = L$ on a measure dense set?