Let $n\in\mathbb{N}$ and consider the Sobolev space $W^{1,\infty}(\mathbb{R}^n)=\lbrace u\in L^{\infty}(\mathbb{R}^n):\partial_iu\in L^{\infty}(\mathbb{R}^n) \rbrace$. A function is in $W^{1,\infty}$ iff it is bounded and Lipschitz continuous. We know also from Rademacher theorem that any Lipschitz function is differentiable almost everywhere. Thus we may define the following norms on the space:

\begin{align} \vert\vert u \vert\vert_1:&=\vert\vert u\vert\vert_{\infty} + \text{ess sup}_{x\in\mathbb{R}^n}\vert\vert Du(x)\vert\vert,\\ \\ \vert\vert u \vert\vert_2:&=\sup_{x\neq y}\bigg\lbrace \vert u(x)\vert + \frac{\vert u(x)-u(y)\vert}{\vert x-y\vert} \bigg\rbrace, \end{align}

where $\vert\vert Du(x)\vert\vert$ denotes the usual operator norm. I was told that both norms are equal, i.e. $\vert\vert u \vert\vert_1 = \vert\vert u \vert\vert_2$ for all $u\in W^{1,\infty}(\mathbb{R}^n)$. Unfortunately, I could not find any reference and couldn't prove it either.

For example, let my explain how I tried to prove $\vert\vert u \vert\vert_2 \leq \vert\vert u \vert\vert_1$: For any $x,y\in \mathbb{R}^n$ and any bounded $L$- Lipschitz function $u$ we have: $u(x)+\frac{\vert u(x)-u(y)\vert}{\vert x-y\vert}\leq \sup_{x\in\mathbb{R}^n}\lbrace u(x)\rbrace + L$. Now it would suffice to prove $$L= \sup_{x\in\mathbb{R}^n \backslash\mathcal{N}} \vert\vert Du(x)\vert\vert$$

for all null sets $\mathcal{N}\subset \mathbb{R}^n$ such that $u_{\vert \mathbb{R}^n \backslash\mathcal{N}}$ is differentiable. I think this should be correct, but I was only able to prove $L \geq \sup_{x\in\mathbb{R}^n \backslash\mathcal{N}} \vert\vert Du(x)\vert\vert$. The other inequality seems to need a generalization of the mean value theorem for Lipschitz-functions in the following sense: $$ \vert u(x)-u(y)\vert \leq \sup_{z\in\mathbb{R}^n\backslash \mathcal{N}} \vert\vert Du(z)\vert\vert \cdot \vert x-y \vert, \forall x,y\in \mathbb{R}^n$$

I have found only the following mean value theorem in a book by Dieudonne: Let $\mathcal{B}$ be a Banach space and $T:\mathbb{R}\supset [a,b]\rightarrow \mathcal{B}$ be continuous. If there is a countable set $\mathcal{N}\subset [a,b]$ such that $T$ is differentiable on $[a,b]\backslash \mathcal{N}$ and $\vert T'(x)\vert \leq M$ then $\vert T(a)-T(b)\vert \leq M\cdot \vert a-b \vert$.

My questions:

  1. Is the assertion $\vert\vert u \vert\vert_1 = \vert\vert u \vert\vert_2$ $\forall u\in W^{1,\infty}(\mathbb{R}^n)$ indeed correct? If so, what would be a good reference or method to prove it?
  2. How can I prove (if possible) the inequality $L \leq \sup_{x\in\mathbb{R}^n \backslash\mathcal{N}} \vert\vert Du(x)\vert\vert$?
  3. Is it possible to generalize the mean value theorem to Lipschitz continuous functions?

Thank you for you patience and help!

  • 3
    $\begingroup$ I don't think the two norms are equal. Define $v(x)= \sup_{y\neq x} \frac{\vert u(x)-u(y)\vert}{\vert x-y\vert}$. Then $\Vert u\Vert_1 = \Vert u\Vert_\infty + \Vert v \Vert_\infty$ while $\Vert u\Vert_2= \Vert \vert u \vert + v \Vert_\infty$. $\endgroup$ May 26, 2021 at 12:09
  • 1
    $\begingroup$ The equality is between the two seminorms, that is the supremum of the Euclidean norms of incremental ratios and the ess supremum of the operator norms of the differential. The operator norm is meant to be w.r.to Euclidean norms too. $\endgroup$ May 26, 2021 at 13:24
  • $\begingroup$ @Pietro Majer: I think I do not understand your comment. Could you please explain what you mean in more detail? Do you mean that I have misunderstood something basically regarding the norms? $\endgroup$ May 26, 2021 at 13:36
  • 2
    $\begingroup$ Just a side remark: I'm saying that to make assertion 1 true you may modify the definition of $\|u\|_2$ into $\|u\|_\infty$ plus the supremum of the absolute value of the incremental ratios $\endgroup$ May 26, 2021 at 13:46

1 Answer 1


Write, using the Fundamental Theorem of Calculus for a $C^1$ function, $$\frac{u(x)-u(y)}{x-y} = \frac1{x-y} \int_C Du d\sigma,$$ where $C$ is the line segment from $x$ to $y$. This gives $$\left|\frac{u(x)-u(y)}{x-y}\right|\leq \|Du\|_{L^\infty}.$$ Then by density you obtain $$\| \cdot \|_2 \leq \| \cdot \|_1.$$

The other inequality, as Nicolas Tholozan pointed out, is not true. Take $f:[0,2]\to[0,1]$ given by $$ f=\begin{cases} 1- \frac12 x & \text{ if } x<1\\ \frac32 - x & \text{ if } 1\leq x\leq\frac32\\ 0 & \text{ if } x>\frac32 \end{cases} $$ Then, $\|u\|_1=2$, while $\|u\|_2=\frac32$.

  • $\begingroup$ Thank you a lot! This answer is very helpful $\endgroup$ May 26, 2021 at 13:18
  • 2
    $\begingroup$ In dimension n≥2 the first identity is only true for a.e. x,y, since u may fail to be differentiable on the whole segment from x to y. But it is enough to conclude. $\endgroup$ May 26, 2021 at 13:30

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