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498 J. Van Schaftingen / Journal of Functional Analysis 236 (2006) 490–516<br />

For every s>0, the set {x ∈ Rn : ϕ(x) > s} is open and boun<strong>de</strong>d. Moreover, by Sard’s <strong>le</strong>mma,<br />

for almost every s>0, for every y ∈ ϕ−1 ({s}), ∇ϕ(y) = 0. Hence ∂{x ∈ Rn : ϕ>s} is smooth<br />

and<br />

<br />

<br />

<br />

<br />

<br />

∇u(x) dx<br />

=<br />

<br />

<br />

<br />

<br />

u(y)ν(y)dH n−1 <br />

<br />

(y) <br />

<br />

{x∈R n : ϕ>s}<br />

∂{x∈R n : ϕ(x)>s}<br />

AH n−1 ∂ x ∈ R n : ϕ(x) > s .<br />

A simi<strong>la</strong>r reasoning for s s + H n−1 ∂ x ∈ R n : ϕ(x) < −s ds<br />

<br />

= A |∇ϕ| dx. ✷<br />

2.5. Geometric characterization of Vn−1(R n )<br />

Proposition 2.11. If n 2, for every u ∈ C(R n ),<br />

uDn−1(R n ) = sup<br />

γ ∈C 1 (S 1 ;R n )<br />

<br />

1 <br />

<br />

˙γ u<br />

L1 (S1 )<br />

γ(t) <br />

<br />

˙γ(t)dt<br />

.<br />

The proof repeats the argument of Bourgain and Brezis for the equiva<strong>le</strong>nce between the inequality<br />

(1.7) and<br />

<br />

u γ(t) ˙γ(t)dtCs,p˙γ L1 (S1 ) uWs,p (Rn ),<br />

for every u ∈ W s,p (R n ) with sp = n [3].<br />

Proof. First note that<br />

S 1<br />

uDn−1(R n ) = sup<br />

f ∈D(R n ;R n )<br />

div f =0<br />

f L 1 (R n ;R n ) 1<br />

S 1<br />

<br />

<br />

<br />

uf d x<br />

.<br />

Let ρ ∈ D(B(0, 1)) be such that ρ 0 and <br />

R n ρdx= 1, and <strong>le</strong>t ρε(x) = ρ(x/ε)/ε n . Define<br />

R n<br />

<br />

1<br />

fε(x) =<br />

ρ<br />

˙γ L1 (S1 )<br />

γ(t)−x ˙γ(t)dt.<br />

S 1

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