# Does the domain of integration have to be convex?

The following theorem is straightforward to prove if we ignore the convexity of the domain :

Suppose that $\text{R}$ is a convex region in the plane and that the function $g : \text{R} \to \mathbb{R}$ has continuous bounded partial derivatives. Show that the surface $S = {(x, y, g(x, y)) \ | \ (x, y) \ \text{in} \ \text{R}}\}$ has area equal to that of $\text{R}$ if and only if the function $g : \text{R} \to \mathbb{R}$ is constant.

For the proof I used the fact that $\iint_D \Big(\sqrt{1+f_x^2+f_y^2} -1\Big)\,dA = 0 \iff f_x=0 =f_y \iff f = \text{const.}$

But why it is supposed that the region to be convex? I can't see why it is necessary to use convexity of the region in the proof. Is there a counterexample i.e. a non-convex region that the mentioned theorem fails to hold?

• Can't one approximate any region by a union of convex sets and apply the theorem to each set to prove it for general regions? – Marco Sep 4 '18 at 0:02
• @Marco, I think so, that's what I thought too! However I don't understand why the author has put convexity condition on the theorem ... – Emma Sep 4 '18 at 0:16
• Is the proof easier if R is assumed to be convex in a technical way? – Marco Sep 4 '18 at 0:17

The key property you need is that $R$ is connected. Indeed, you use this to conclude your proof when you write:

$$f_x = f_y = 0 \Leftrightarrow f = \mathit{const}$$

The general fact that we are using here is:

Theorem A. Let $f \colon \Omega \to \mathbb{R}$ be a differentiable function, where $\Omega$ is an open connected set in $\mathbb{R}^n$. Then $\nabla f = 0$ in $\Omega$ if and only if $f$ is constant.

It is easy to give a counter-example to this when $\Omega$ is not connected: take a function which is piecewise constant, with a different value for the constant on each connected component of $\Omega$. For instance take $\Omega = D_1 \cup D_2$ where $D_1$ and $D_2$ are two disjoint open disks, and define $f \colon \Omega \to \mathbb{R}$ by $f = 1$ on $D_1$ and $f = 2$ on $D_2$.

Your problem probably assumes that $R$ is convex instead of merely connected because it is easier to prove Theorem A. In fact, the outline of the proof of the theorem is as follows:

1. Prove the theorem in the case where $R$ is convex. Hint: use the mean value theorem.
2. Prove the theorem in the general case where $R$ is connected (and nonempty): let $x_0 \in \Omega$, show that $\{x\in \Omega ~\colon~ f(x) = f(x_0)\}$ is nonempty, open and closed in $\Omega$. Hint: to show openness, use the fact that $\Omega$ is locally convex, and use 1.

Remark: Note that you can also relax the hypothesis that $f$ has continuous and bounded partial derivatives: it is enough to assume that $f$ is differentiable. The reason that your problem makes this extra assumption is probably to easily guarantee that the area of the surface is well-defined and finite. In fact they probably forgot to also say that $R$ is bounded.

• Questions! : 1. Same as your answer, the link (wiki) also assumes R to be connected; isn't it path-connected instead? Because as I understood, it uses path-connectedness as the possibility of connecting two arbitrary points by a number of joint (straight) lines and use of MVT on each line (?). + 2. If I understood correctly, in the proof, a convex set is a special case of a (path-?)connected set such that it replaces many joined-lines with one line; that implies, there should be a theorem telling that "for any point of a path-connected set there is some neighborhood which is convex"; if so ... – Emma Sep 6 '18 at 9:29
• could you let me know a source for the proof? I can't find any on internet. Thanks! – Emma Sep 6 '18 at 9:30
• @Emma For an open set $R$, being connected is equivalent to being path-connected. As you noted, it is also equivalent to being path-connected with paths that are piecewise linear, and if you are aware of this (or can prove it, it's not very difficult), then you can prove my theorem A directly (without following the two-step proof I suggested), but I personally prefer my proof (it's similar, but matter of taste). – Seub Sep 6 '18 at 14:22
• As for a reference, I'm not going to try to look for one, although I'm sure it's in plenty of places. But try to prove it yourself: it's really not very hard. Step 1 is an immediate consequence of the MVT. For step 2, nonemptyness is clear, closedness is just by continuity of f. In order to prove openness, start by observing that any point in $\Omega$ has a convex neighborhood in $\Omega$ (a small ball works), and then you're almost done: just use 1. – Seub Sep 6 '18 at 14:25

The theorem does not remain true if you remove the word "convex", because then that allows for a disconnected region $R$.

Counter-example:

$$R = ([-2,-1] \cup [1,2]) \times [0,1]$$

$$g(x,y) = \begin{cases} 0 & \mathrm{if\ } x < 0 \\ 1 & \mathrm{if\ } x > 0 \end{cases}$$ Then both $R$ and $S$ have area 2, but $g$ is not constant.

Okay, then what if we replace the word "convex" with the word "connected"? Then the theorem is still not true, because you can have line-like pieces with no area connecting components with different values of $g$.

Counter-example:

\begin{align*} R_1 &= [-1,0] \times [0,1] \\ R_2 &= (0,1) \times \{0\} \\ R_3 &= [1,2] \times [0,1] \\ R &= R_1 \cup R_2 \cup R_3 \end{align*} $$g(x,y) = \begin{cases} 0 & \mathrm{if\ } x \leq 0 \\ 3x^2-2x^3 & \mathrm{if\ } 0 < x < 1 \\ 1 & \mathrm{if\ } x \geq 1 \end{cases}$$

Then $g$ has continuous bounded partial derivatives, and again $R$ and $S$ both have area 2, but $g$ is not constant.

But if we know $R$ is convex (and has a well-defined area per your favorite set measure theory), we can prove the theorem something like this:

Form a set $R'$ by subtracting from $R$ all points $(x,y)$ that can be enclosed in some small box $B_{xy}$ such that the set $R \cap B_{xy}$ has measure zero. Then $R'$ is also convex, and the area of $R'$ is the same as the area of $R$, and the area of the curve $S'$ with support $R'$ is the same as the area of $S$. A slight change to your argument shows the partial derivatives $g_x$ and $g_y$ are always zero on $R'$ (just not necessarily all of $R$). If we choose any two points $a, b \in R'$, then there must be a point $c$ arbitrarily close to $a$ and a point $d$ arbitrarily close to $b$, neither of them on the line through $a$ and $b$ (by construction of $R'$). The convex hull of these four points is in $R'$, and contains a path of horizontal and vertical segments joined end-to-end, with endpoints within the small box containing $a$ and $c$ and the small box containing $b$ and $d$. So the absolute difference $|g(a)-g(b)|$ can be shown to be as small as desired, so it must always be zero.

• The two counter-examples are crystal-clear! But I couldn't understand the last paragraph. This theorem has been asked in an undergraduate's 1st-year exam to b proved; and there should be some proof not using advanced analysis; if so, I would appreciate it if you please provide some clarifications. I read the last paragraph several times but still have difficulties understanding it. – Emma Sep 4 '18 at 2:42
• @Emma Hmm, I think I did overcomplicate it: for one thing, I think for any convex set with non-zero area, the $R'$ I used is actually identical to $R$. And there's probably some theorem I'm forgetting that would go from $g$ having partial derivatives to $g$ being differentiable, in which case we could just use the one straight segment between two points, instead of needing just horizontal and vertical segments. – aschepler Sep 4 '18 at 3:04
• I will try to find a rudimentary proof. Thank you :) – Emma Sep 4 '18 at 3:19
• Can you please introduce me some book that will cover the technicalities in the last paragraph of your answer? I've already started Real Analysis for Graduate Students by R F Bass but if there is a better/broader book I can switch to your recommended one. – Emma Sep 4 '18 at 17:21
• I don't really agree with this: "If we replace the word "convex" with the word "connected"? The theorem is still not true, because you can have line-like pieces connecting components with different values of g". The problem is that your domain is not open, and talking about the differentiability (or existence of partial derivatives) of $f$ only really makes sense when $\Omega$ is open. I think the most sensible way to define differentiability of a function on a non-open $\Omega$ is to say that it extends to a differentiable function on an open neighborhood. In this case, the theorem is true. – Seub Sep 6 '18 at 3:18