CONVEX FUNCTIONS WITH UNBOUNDED GRADIENT

OLIVER C. SCHNÜRER

Abstract. We show that domains, that allow for convex functions with unbounded gradient at their boundary, are convex.

In this paper, we prove the following

Theorem 1. Let Ω ⊂ ℝ^{n} be any domain. If there exists a convex function u : Ω → ℝ, such that ∇u becomes unbounded near ∂Ω, then Ω is convex.

This problem arises in the context of affine hypersurfaces [1], where a similar statement is proven.

Acknowledgement: We want to thank Neil Trudinger, who brought this problem to our attention and encouraged us to provide a geometric proof. We also want to thank Klaus Ecker and Free University Berlin for support during the preparation of this paper and Ann Björner for converting our sketch into a beautiful illustration.

We will use the following notions of convexity.

Definition and Remark 2. Let Ω ⊂ ℝ^{n} be a given domain.

A function u : Ω → ℝ is called convex, if

u(τx + (1 − τ)y) ≤ τu(x) + (1 − τ)u(y)

for all x, y ∈ Ω and all τ ∈ (0, 1) such that τx + (1 − τ)y ∈ Ω.

A convex function is locally Lipschitz continuous and thus differentiable almost everywhere. We obtain

u(y) ≥ u(x) + ⟨∇u(x), y − x⟩

for all y ∈ Ω and all x ∈ Ω, where u is differentiable.

A function u is called locally convex, if

u(τx + (1 − τ)y) ≤ τu(x) + (1 − τ)u(y)

for all x, y ∈ Ω and τ ∈ (0, 1) such that tx + (1 − t)y ∈ Ω for all t ∈ (0, 1).

For u ∈ C^{2}(Ω), local convexity is equivalent to positive semi-definiteness of its Hessian in Ω.

Note especially that locally convex functions do not need to be convex, unless their domain of definition is a convex set, i. e. x, y ∈ Ω, τ ∈ (0, 1) ⟹ τx+(1−τ)y ∈ Ω.

In order to strengthen Theorem 1, we fix x_{0} ∈ Ω and define Ω_{1} = Ω_{1}(x_{0}) by

Ω_{1} := {x ∈ Ω : tx + (1 − t)x_{0} ∈ Ω for 0 ≤ t ≤ 1}

Date: April 2005, revised September 2005.

2000 Mathematics Subject Classification. 52A99.

Key words and phrases. Convex function, convex domain.

The author is a member of SFB 647/3B “Raum – Zeit – Materie”.

as the largest subdomain of Ω that is star shaped with respect to x_{0}. Note that Ω is convex if and only if Ω = Ω_{1}(x) for all x ∈ Ω.

For convenience, we will assume from now on that x_{0} is the origin. So the domain Ω_{1} = Ω_{1}(0) is star shaped with respect to the origin.

We define

∂_{2}Ω_{1} := {x ∈ ∂Ω_{1} : ∃ε > 0 : tx ∈ ∂Ω_{1}∀t ∈ (1 − ε, 1 + ε)}

and ∂_{1}Ω_{1} := ∂Ω_{1} \ ∂_{2}Ω_{1}.

For Ω as in Theorem 1, we deduce that ∇u becomes unbounded near ∂_{1}Ω_{1}. Thus the following result generalizes Theorem 1 and [1, Lemma 2.4].

Theorem 3. Let Ω ⊂ ℝ^{n} be open and star shaped with respect to 0 ∈ Ω. If there exists a convex function u : Ω → ℝ, such that

∥∇u∥_{L^{∞}(B_{δ}(x)∩Ω)} = ∞

for any x ∈ ∂_{1}Ω and any δ > 0, then Ω is convex.

Here, the set ∂_{1}Ω is defined as above. We consider ∇u only at those points, where u is differentiable.

Proof. We argue by contradiction. Assume that there exist ỹ_{0} and ỹ_{1} in Ω and τ̃ ∈ (0, 1) such that

τ̃ỹ_{0} + (1 − τ̃)ỹ_{1} ∉ Ω.

Let λ > 0 be the supremum over all positive numbers such that

tλỹ_{0} + (1 − t)λỹ_{1} ∈ Ω for all 0 ≤ t ≤ 1.

As 0 is an interior point of Ω, λ is positive. Define y_{0} := λỹ_{0} and y_{1} := λỹ_{1}. Let τ ∈ (0, 1) be such that τy_{0} + (1 − τ)y_{1} ∈ ∂Ω. According to the definition of λ,

μ(τy_{0} + (1 − τ)y_{1}) ∈ Ω for all 0 ≤ μ < 1.

So we deduce that τy_{0} + (1 − τ)y_{1} ∈ ∂_{1}Ω. Thus, there exist x_{i} ∈ Ω, i ∈ ℕ, such that x_{i} → τy_{0} + (1 − τ)y_{1} for i → ∞, ∇u exists at x_{i} for all i, and |∇u(x_{i})| → ∞ as i → ∞. This contradicts the gradient bounds that we will prove in the following.

As Ω is open and star shaped, there exists ε > 0 such that B_{3ε}(y_{k}) ⊂ Ω, k = 0, 1, and B_{3ε}(0) ⊂ Ω. Assume that the balls B_{3ε}(y_{0}), B_{3ε}(y_{1}), B_{3ε}(τy_{0} + (1 − τ)y_{1}), and B_{3ε}(0) are disjoint. By scaling u, we may therefore arrange that |u| ≤ 1 in B_{2ε}(y_{0}) ∪ B_{2ε}(y_{1}) ∪ B_{2ε}(0).

Assume furthermore that, after an appropriate rotation, y_{1} − y_{0} is a positive multiple of e_{1} = (1, 0, …, 0).

We will now use the convexity of u and boundedness of u in balls around y_{0} and y_{1} to prove explicit bounds on ∇u near τy_{0} + (1 − τ)y_{1}. Set u_{k} := \frac{∂u}{∂x^{k}}, 1 ≤ k ≤ n.

The first step is to bound u in B_{ε}(τy_{0} + (1 − τ)y_{1}). Let z ∈ B_{ε}(0) be such that τy_{0} + (1 − τ)y_{1} + z ∈ Ω. Convexity of u and u ≤ 1 in B_{ε}(y_{k}), k = 0, 1, imply that

u(τy_{0} + (1 − τ)y_{1} + z) =u(τ(y_{0} + z) + (1 − τ)(y_{1} + z))
≤τu(y_{0} + z) + (1 − τ)u(y_{1} + z)
≤τ · 1 + (1 − τ) · 1 = 1.

For a lower bound on u, we compare u at τy_{0} + (1 − τ)y_{1} + z =: p, at 2ε\frac{p}{|p|}, and at the origin. We get |p| ≤ |τy_{0} + (1 − τ)y_{1}| + ε,

2ε\frac{p}{|p|} = tp + (1 − t)0

Figure 1. Geometric situation

with t = \frac{2ε}{|p|}, and

|p| ≥ |τy_{0} + (1 − τ)y_{1}| − |z| ≥ 6ε − ε,

as B_{3ε}(τy_{0} + (1 − τ)y_{1}) ∩ B_{3ε}(0) = ∅. Thus

u (2ε\frac{p}{|p|}) ≤ tu(p) + (1 − t)u(0)

and we obtain that

u(p) ≥\frac{1}{t} (u (2ε\frac{p}{|p|}) − (1 − t)u(0))
≥\frac{|τy_{0} + (1 − τ)y_{1}| + ε}{2ε}(−1 − 1)
= − \frac{1}{ε}(|τy_{0} + (1 − τ)y_{1}| + ε).

Thus |u| ≤ 1 + \frac{1}{ε} |τy_{0} + (1 − τ)y_{1}| ≡ C_{0} in B_{ε}(τy_{0} + (1 − τ)y_{1}).

The next step is to bound |u_{1}| in B_{ε}(τy_{0} + (1 − τ)y_{1}). Remember that y_{1} − y_{0} is a positive multiple of e_{1}. As balls of radius 3ε as chosen above are disjoint, we get

|(1 − τ)(y_{0} − y_{1})| = |y_{0} − (τy_{0} + (1 − τ)y_{1})| ≥6ε

and

|τ(y_{0} − y_{1})| = |(τy_{0} + (1 − τ)y_{1}) − y_{1}| ≥6ε.

Thus we obtain

(1)  1 − τ ≥ \frac{6ε}{|y_{0} − y_{1}|} and τ ≥ \frac{6ε}{|y_{0} − y_{1}|}.

Let z ∈ B_{ε}(0) and τy_{0}+(1−τ)y_{1}+z =: p be an arbitrary point in B_{ε}(τy_{0}+(1−τ)y_{1}), where u is differentiable. We deduce that

p + (1 − τ)(y_{0} − y_{1}) ∈B_{ε}(y_{0})

and

p + τ(y_{1} − y_{0}) ∈B_{ε}(y_{1}).

Convexity and boundedness of u in B_{ε}(y_{0}) ∪ B_{ε}(y_{1}) ∪ B_{ε}(τy_{0} + (1 − τ)y_{1}) imply that

τ |y_{1} − y_{0}| ⟨∇u(p), e_{1}⟩ =⟨∇u(p), τ(y_{1} − y_{0})⟩
≤u(p + τ(y_{1} − y_{0})) − u(p)
≤1 + C_{0}

and

(1 − τ) |y_{1} − y_{0}| ⟨∇u(p), −e_{1}⟩ =⟨∇u(p), (1 − τ)(y_{0} − y_{1})⟩
≤u(p + (1 − τ)(y_{0} − y_{1})) − u(p)
≤1 + C_{0}.

We insert the bounds (1) and deduce that

|⟨∇u(p), e_{1}⟩| = |u_{1}| (p) ≤ \frac{1}{6ε}(1 + C_{0}).

The last step is to bound u_{k} in B_{ε}(τy_{0} + (1 − τ)y_{1}) for 2 ≤ k ≤ n. Consider an arbitrary vector ξ ∈ ℝ^{n} such that |ξ| = 1 and ⟨ξ, e_{1}⟩ = 0. As above, let p ∈ B_{ε}(τy_{0} + (1 − τ)y_{1}) ∩ Ω be chosen arbitrarily such that u is differentiable in p. Then we have

p + (1 − τ)(y_{0} − y_{1}) ± εξ ∈ B_{2ε}(y_{0}).

We remark that considering p + τ(y_{1} − y_{0}) ± εξ ∈ B_{2ε}(y_{1}) in the following yields a similar estimate. As u is convex, we get

⟨∇u(p), (1 − τ)(y_{0} − y_{1}) ± εξ⟩ ≤ u(p + (1 − τ)(y_{0} − y_{1}) ± εξ) − u(p).

So we deduce

ε⟨∇u(p), ±ξ⟩ ≤u(p + (1 − τ)(y_{0} − y_{1}) ± εξ) − u(p) + ⟨∇u(p), (1 − τ)(y_{1} − y_{0})⟩
≤ (1 + \frac{|y_{0} − y_{1}|}{6ε}) (1 + C_{0}).

Thus |∇u| is bounded in B_{ε}(τy_{0} + (1 − τ)y_{1}), wherever u is differentiable. This contradicts the assumption that τy_{0} + (1 − τ)y_{1} ∈ ∂_{1}Ω. The theorem follows. □

Remark 4. In Theorem 1, we don’t have to assume Ω being connected. If y_{0} and y_{1} lie in different components of Ω, we find 0 < τ < 1 such that τy_{1} + (1 − τ)y_{0} ∈ ∂Ω. Choose 0 < t < 1, t ≠ τ, such that ty_{1} + (1 − t)y_{0} ∈ Ω. Assume that t > τ. There exists 0 < σ < 1 such that

ty_{1} + (1 − t)y_{0} = σ(τy_{1} + (1 − τ)y_{0}) + (1 − σ)y_{1}.

Thus

u(ty_{1} + (1 − t)y_{0}) ≤ σu(τy_{1} + (1 − τ)y_{0}) + (1 − σ)u(y_{1})

and u(τy_{1} + (1 − τ)y_{0}) is bounded below. A similar argument bounds u from below in a neighborhood (relative to Ω) of τy_{1} +(1−τ)y_{0}. Following the lines of the proof of Theorem 3, we obtain an upper bound on u near ty_{1} + (1 − t)y_{0} and gradient bounds. As ∇u is bounded near τy_{1} + (1 − τ)y_{0}, we obtain a contradiction. We conclude that Ω has to be connected.

Remark 5. Let Ω be any convex domain in ℝ^{n} and d := dist(·, ∂Ω). As d is concave in Ω, −\sqrt{d} is a convex function in Ω with unbounded gradient along ∂Ω. Thus, according to Theorem 1, such functions exist precisely on convex domains.

Reference

[1] Neil S. Trudinger and Xu-Jia Wang, Affine complete locally convex hypersurfaces, Invent. Math. 150 (2002), no. 1, 45–60.

Oliver Schnürer, FU Berlin, Arnimallee 2-6, 14195 Berlin, Germany

E-mail address: Oliver.Schnuerer@math.fu-berlin.de