Bony–Brezis theorem

Let F be closed subset of a C² manifold M and let X be a vector field on M which is Lipschitz continuous. The following conditions are equivalent:

• Every integral curve of X starting in F remains in F.
• (X(m),v) ≤ 0 for every exterior normal vector v at a point m in F.

Following Hörmander (1983), to prove that the first condition implies the second, let c(t) be an integral curve with c(0) = x in F and dc/dt= X(c). Let g have a local maximum on F at x. Then g(c(t)) ≤ g (c(0)) for t small and positive. Differentiating, this implies that g '(x)⋅X(x) ≤ 0.

To prove the reverse implication, since the result is local, it enough to check it in Rⁿ. In that case X locally satisfies a Lipschitz condition

${\displaystyle \displaystyle {|X(a)-X(b)|\leq C|a-b|.}}$

If F is closed, the distance function D(x) = d(x,F)² has the following differentiability property:

${\displaystyle \displaystyle {D(x+h)=D(x)+\min _{z}2h\cdot (x-z)+o(|h|),}}$

where the minimum is taken over the closest points z to x in F.

To check this, let

${\displaystyle \displaystyle {f_{\varepsilon }(h)=\min _{z}2h\cdot (x-z),}}$

where the minimum is taken over z in F such that d(x,z) ≤ d(x,F) + ε.

Since f_ε is homogeneous in h and increases uniformly to f₀ on any sphere,

${\displaystyle \displaystyle {f_{0}(h)\geq f_{\varepsilon }(h)\geq f_{0}(h)-C(\varepsilon )|h|,}}$

with a constant C(ε) tending to 0 as ε tends to 0.

This differentiability property follows from this because

${\displaystyle \displaystyle {D(x+h)\leq |x+h-z|^{2}\leq |z-x|^{2}+2h\cdot (x-z)+|h|^{2}=D(x)+f_{0}(h)+|h|^{2}}}$

and similarly if |h| ≤ ε

${\displaystyle \displaystyle {D(x+h)\geq D(x)+f_{\varepsilon }(h)+|h|^{2}.}}$

The differentiability property implies that

${\displaystyle \displaystyle {\lim {\delta \downarrow 0}{D(c(t+\delta ))-D(c(t)) \over \delta }=2\min _{z}X(c(t))\cdot (c(t)-z),}}$

minimized over closest points z to c(t). For any such z

${\displaystyle \displaystyle {2X(c(t))\cdot (c(t)-z)=2X(z)\cdot (c(t)-z)-2(X(z)-X(c(t)))\cdot (c(t)-z).}}$

Since −|y − c(t)|² has a local maximum on F at y = z, c(t) − z is an exterior normal vector at z. So the first term on the right hand side is non-negative. The Lipschitz condition for X implies the second term is bounded above by 2C⋅D(c(t)). Thus the derivative from the right of

${\displaystyle \displaystyle {e^{-2Ct}D(c(t))}}$

is non-positive, so it is a non-increasing function of t. Thus if c(0) lies in F, D(c(0))=0 and hence D(c(t)) = 0 for t > 0, i.e. c(t) lies in F for t > 0.

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• Barrier certificate