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Question:

Under what circumstances does local convexity imply global convexity?

Motivation:

Classically, a twice differentiable function $f:\mathbb{R} \rightarrow \mathbb{R}$ is convex if and only if the second derivative is nonnegative everywhere. In this recent question, Derivative of Convex Functional, it's shown that the same result holds for twice Frechet differentiable functionals on Banach spaces, $f:X\rightarrow \mathbb{R}$.

In both these cases, we have a result saying something to the effect of: "local convexity implies global convexity". How far can this idea be generalized?

The following hypothesis, which may or may not be true, expresses the idea in the most general context I can think of.

Conjecture: Let $C$ be a connected subset of a topological vector space, and let $\{ U_\alpha \}_{\alpha \in A}$ be an open cover of the boundary $\partial C$. If $U_\alpha \cap C$ is convex for all $\alpha \in A$, then $C$ is convex.

Informally, "Inspect the boundary of a connected set with a (variable-size) magnifying glass. If, everywhere you look, it looks convex, then the set is globally convex."

Example: $C$ is a square in $\mathbb{R}^2$, and $U_\alpha=B_i$ are balls covering all 4 edges and corners.

enter image description here

Non-example: $C$ is two disjoint disks in $\mathbb{R}^2$.

enter image description here

From this we see that there is a topological element to the question - if the connectedness condition is relaxed, it's easy to come up with counterexamples.

Notes:

  • I've linked the key terms to the wiki pages. Is this good style on M.SE? Most people who could answer the question would already know the definition. On the other hand, when I'm answering a question that's not immediately obvious, a lot of times I'll open up the wiki page and look around even if I already know the definition.

  • A special case I've been considering is where the space is Banach, and the set's boundary is path connected and compact. In this case I think it's true but the proof is elusive so far..

  • In the comments Chris Eagle suggests reducing it to a 2D problem. I'm not sure exactly how this works and if it will generalize to spaces other than $\mathbb{R}^n$.

  • In the comments Cardinal notes that it is trivally false in the discrete topology - the boundary of any set can be covered by open points. Joriki points out that this isn't a problem since all nontrivial sets of interest are not connected in the discrete topology.

  • George Lowther notes that the conjecture is false in $\mathbb{R}^2$ unless a further constraint is added that the set is either closed or open. The open unit square unioned with it's vertices is locally convex, but does not contain it's edges so is not globally convex.

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More or less by definition, a set is convex iff its intersection with any two-dimensional subspace is convex. So it's enough to prove the result for $\mathbb{R}^2$ and prove that your condition ensures the intersection with any two-dimensional subspace is connected. –  Chris Eagle May 16 '12 at 13:12
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@cardinal: In response to your penultimate comment: Because the only connected sets in the discrete topology are the empty set and the one-point sets, and the conjecture is valid for these sets. –  joriki May 20 '12 at 15:37
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@cardinal: Also, $\mathbb{R}^2$ with the discrete topology is not a TVS. The only topology that makes $\mathbb{R}^n$ a TVS is the standard one. –  Chris Eagle May 20 '12 at 16:31
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The conjecture fails in $\mathbb{R}^2$ unless you assume that $C$ is closed (or open). For example, take the interior of the unit square unioned with the four vertices. –  George Lowther May 20 '12 at 21:24
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I would love to see your proof as well, @GeorgeLowther. If you type it up, I'm sure other people following the question and future google searchers would appreciate it as well. –  Nick Alger May 22 '12 at 16:14
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2 Answers

up vote 7 down vote accepted
+500

We suppose that $X$ is a locally convex vector space. Before showing that the conjecture is true (under some conditions), we show a helpful lemma.

Lemma. If a subset $C$ of $X$ has the property that it's boundary can be covered by an open cover $\{U_\alpha\}_{\alpha\in A}$ with the property that each $U_\alpha\cap C$ is convex, then the triangle $$ \Delta(x,y,z):=\{(1-t)((1-s)x+sy)+tz:s,t\in[0,1)\} $$ is contained in $C$ whenever the lines $\ell_{x,y}:=\{(1-s)x+sy:s\in[0,1]\}$ and $\ell_{x,z}:=\{(1-t)x+tz:t\in[0,1]\}$ are contained in $C$.

Here is a picture that contains most of the terms involved in the proof

Proof. Suppose that $\Delta(x,y,z)$ is not contained in $C$. In the following, the boundary and interior are considered relative to the triangle $\Delta(x,y,z)$.

Among $s\in[0,1)$ with the property that the line $$ \{(1-t)((1-s)x+sy)+tz:t\in[0,1)\}\cap \partial(C\cap\Delta(x,y,z))\neq\varnothing $$ there is a minimum $s_0$. Then there is also a minimum $t_0$ among the $t\in[0,1)$ with the property that $$ u:=(1-t)((1-s_0)x+s_0y)+tz:t\in\partial(C\cap\Delta(x,y,z)) $$ Since $u$ is on the boundary of $C$, there exists an $\alpha\in A$ with the property that $u\in U_\alpha$. Then $U_\alpha\cap C\cap\Delta(x,y,z)$ is convex. But note that the set $$ \ell_{x,y}\cup\ell_{x,z}\cup\Delta(x,(1-s_0)x+s_0y,z), $$ where the latter triangle has a similar definition as $\Delta(x,y,z)$, is contained in $C$. The convexity of $U_\alpha\cap C\cap\Delta(x,y,z)$ then implies that there is a $\varepsilon>0$ such that the point $$ (1-(t_0-\varepsilon))((1-s_0)x+s_0y)+(t_0-\varepsilon)z $$ is on the boundary of $C\cap\Delta(x,y,z)$ unless $t_0=0$. The minimality of $t_0$ then gives us that $t_0=0$. A similar argument shows that $s_0=0$ and it follows once again from the convexity of the boundary of $C$ that $x,y,z$ are on a line. But then $\Delta(x,y,z)$ is (contained in) this line. QED.

We have the following corollary (in the open case, an easy adjustment of the above proof has to be given).

Corollary. If $X$ is either open or closed, the triangle $$ \overline{\Delta(x,y,z)}=\{(1-t)((1-s)x+sy)+tz:s,t\in[0,1]\} $$ is contained in $C$.

Now we are ready to answer positive to the conjecture in the case that $C$ is open or closed.

Proposition. Suppose that $X$ is a locally convex vector space and suppose that $C$ is a closed [open] subset of $X$. If there exists a cover $\{U_\alpha\}_{\alpha\in A}$ of $\partial C$ with the property that $U_\alpha\cap C$ is convex for each $\alpha\in A$, then $C$ is convex.

Proof. Suppose $C$ is a subset of a locally convex vector space $X$ and that $\{U_\alpha\}_{\alpha\in A}$ is a cover of the boundary of $C$ with the property that $U_\alpha\cap C$ is convex for each $\alpha\in A$.

First, let $B$ be a maximal among the subsets of $A$ with the property that the convex hull of $\bigcup_{\alpha\in B}U_\alpha$ is contained in $C$. For example, such a maximal $B$ can be found with Zorn's lemma, using the poset $$ P:=\big\{B\subseteq A:\mathrm{Hull}\big(\textstyle\bigcup_{\alpha\in B}U_\alpha\cap C\big)\subseteq C\big\} $$ ordered with inclusion. Here, $\mathrm{Hull}$ denotes the operation of taking the convex hull.

Now let $C^\prime$ be a maximal convex subset of $C$ which contains $U_\alpha$ for each $\alpha\in B$. Here's an outline of the rest of the proof.

  1. First we will show that if for $\alpha\in A$ there is a $\beta\in B$ such that $U_\alpha\cap U_\beta\cap C\neq\varnothing$, then $\alpha$ is in $B$.
  2. Then we will show that no boundary points of $C^\prime$ are interior points of $C$. (Here we will use the assumption that $X$ is locally convex.)

With these two observations the proof is easily finished. Since boundary points of $C^\prime$ are boundary points of $C$, it follows that $C^\prime$ is closed in $C$. The set $$ U:=\textstyle\bigcup_{\alpha\in B} U_\alpha\cup\mathrm{int}\,C^\prime $$ is an open set with the property that $x\in C^\prime$ whenever $x\in U\cap C$, therefore $C^\prime$ is also open in $C$. The connectedness of $C$ together with the non-emptiness of $C^\prime$ (whenever $C$ is non-empty) now implies that $C^\prime= C$.

Proof of 1. Suppose that $x\in U_\alpha\cap U_\beta\cap C$. Note that for any $y\in U_\alpha$, for any $\gamma\in B$ and for any $z\in U_\gamma$, the line $$ \ell_{y,z}:=\{(1-t)y+tz:t\in[0,1]\} $$ is contained in $C$, which is a consequence of the corollary because the we have $\ell_{x,y}\subseteq U_\alpha\cap C\subseteq C$ and $\ell_{x,z}\subseteq C^\prime\subseteq C$. By the maximality of $B$, it follows that $\alpha\in B$.

Proof of 2. Suppose that $x^\prime$ is in $\partial C^\prime\cap\mathrm{int}\,C$. Let $V$ be a convex open neighborhood of $x^\prime$ which is contained in $C$, let $y$ be a point of $V\setminus \bar{C^\prime}$ and let $x$ be a point of $V\cap C^\prime$. It follows by the corollary that for every $z\in C^\prime$ the line $\ell_{y,z}$ is contained in $C$, since the lines $\ell_{x,y}$ and $\ell_{x,z}$ are. Therefore, $C^\prime$ can be extended to include $y$, which contradicts the maximality of $C^\prime$. Thus we may conclude that no boundary points of $C^\prime$ are interior points of $C$. QED.

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Thanks Egbert. I think this works, but I'm going to mull it over for a bit and make sure I really understand before accepting. –  Nick Alger May 22 '12 at 16:16
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That's alright, given that I've already made a flawed attempt :) –  Egbert May 22 '12 at 16:18
    
@Egbert: I don't have time to read through right now, but the statement of the first lemma and the diagram looks very like what I had in mind when I mentioned that I had a proof in an earlier comment. –  George Lowther May 22 '12 at 16:48
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@Egbert: Local convexity is not required. I'm not quite sure how it crept into your proof. One way to prove the result is to show that any two points in C can be joined by a polygonal curve (in C), which uses connectedness. Then use the lemma (which doesn't require local convexity) to inductively reduce the number of edges of the polygonal curve to 1. So, the line joining any two points in C is contained in C, hence C is convex. –  George Lowther May 22 '12 at 21:22
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Ok, this seems to work. Since the time is up I'm accepting this answer and awarding the bounty. I hope that doesn't discourage other people from posting if they have a proof that doesn't depend on local convexity of the underlying space. –  Nick Alger May 24 '12 at 20:15
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There is a classical result known as the Tietze-Nakajima Theorem which more or less answers this question (at least in finite dimensional vector spaces).

Theorem If $X$ is a closed, connected and locally convex subset of $\mathbb{R}^n$ then $X$ is convex.

Here locally convex means exactly what you proposed in your post: for every $x\in X$ there is a ball $B_{\varepsilon}(x)$ centered at $x$ such that $B_{\varepsilon}(x)\cap X $ is convex in the usual sense. As you've noted, the definition is only important for points that are not in the interior of the set.

The proof is given an extremely nice exposition in Section 2 of the paper Revisiting Tietze-Nakajima: Local and Global Convexity of Maps (which also contains a nice generalization of your conjecture, as the title suggests). I recommend reading about it there (if you are still interested): http://www.utm.utoronto.ca/~karshony/papers/convexity-corrected2.pdf

The original papers are:

H. Tietze. “Uber Konvexheit im kleinen und im großen und ¨uber gewisse ¨ den Punkter einer Menge zugeordete Dimensionszahlen." Math. Z. 28 (1928) 697–707.

S. Nakajima, “Uber konvexe Kurven and Flaschen”, Tohoku Mathematical ¨ Journal, vol. 29 (1928), 227–230.

Edit: I've decided to add a sketch of the proof here:

  1. Observe that if the set $X$ is also compact then it is uniformly locally convex (the balls can all be chosen with the same radius).
  2. Define the distance $d_X(x,y)$ between two points $x,y\in X$ to be the infimum of the lengths of paths from $x$ to $y$ in $X$.
  3. For any two points $x_0,x_1\in X$ there exists a midpoint $x_{1/2}$, half the distance between $x_0$ and $x_1$ (this uses closedness of the set).
  4. Now take any two points $x_0,x_1\in X$ and iteratively find midpoints until you have a sequence $x_0, x_{1/2^j},\ldots ,x_{1-1/2^j},x_1$ of points so that the distance between each in $X$ is $d_X(x_0,x_1)/2^j$.
  5. If we intersect $X$ with a large closed ball $\overline{B}$ that contains all these points (give it radius $2d_X(x_0,x_1)$, for example), then by uniform local convexity of $Y = X\cap \overline{B}$, there is some fixed $\varepsilon$ such that $B_{\varepsilon}(y)\cap Y$ is convex, for all $y\in Y$.
  6. Let $j$ be large enough that $$x_{k/2^j-1/2^j},x_{k/2^j+1/2^j}\in B_{\varepsilon}(x_{k/2^j})$$ for all $k$. By local convexity, this says every three consecutive points in the sequence are contained in a line segment in $X$. So the line segment $[x_0,x_1]$ is contained in $X$.
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Thanks for the reference! The finite dimensional condition is much more restrictive than I was looking for though, as it rules out convex analysis in function spaces, convex analysis in pdes, and so on. Also, I think the poster whose answer I accepted is correct, and that applies generally. –  Nick Alger Feb 2 at 20:26
    
You might be interested in some of the other references in the paper I linked. I haven't read them, but they might consider more general spaces. –  AnonymousCoward Feb 2 at 20:32
    
Ok, will do. Much appreciated! –  Nick Alger Feb 2 at 20:35
    
I suppose that this proof breaks down in the `infinite dimensional case' since those spaces won't have Heine-Borel, which is used in step 5. –  AnonymousCoward Feb 2 at 20:39
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