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I started out with the following question:

Say $\Omega$ is a nice bounded domain in $\mathbb{R}^{n-1}$. (One can imagine it being a unit ball in $\mathbb{R}^{n-1}$.) Let $f:\Omega\rightarrow \mathbb{R}^+\cup \{0\}$ be a convex function. Does it follow that the surface area of the graph of $f$ is no less than the area of $\Omega$.

This is intuitively correct, since $\Omega$ is flat while the graph of $f$ might be curved. But in terms of a rigorous proof, I am having trouble.

Now, with the help of multivariable calculus, if $f$ is differentiable everywhere, then it follows that,

$$Area(Graph(f))=\int_{\Omega}\sqrt{1+|\nabla f|^2}dx_1\cdots dx_{n-1}\geq \int_{\Omega}dx_1\cdots dx_n=Area(\Omega).$$

The problem is, my $f$ is not differentiable everywhere. But since $f$ is convex, it is almost everywhere differentiable, suggesting that the first integral in the above equation does make sense. If only the first equality is true for convex functions, then the problem is settled. I have not found such a theorem that says that.

Does anybody know that if the formula in calculus still holds for convex functions? Or is there an easier way to justify the original problem?

henryforever14
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The answer depends on the definition of area. If it's defined by means of the integral formula in the post, then ... the formula holds by definition. One still has to be careful with the differentiability assumption: simply assuming that derivative exists almost everywhere is not enough. (We would not want to say that the graph of the Cantor function has length $1$.)

But if $f$ is Lipschitz (hence differentiable a.e.), the integral indeed gives a geometrically natural value of the area. The proof is not easy (the book by Evans ang Gariepy has it). The integral makes sense because every Lipschitz function is differentiable almost everywhere. And a convex function is locally Lipschitz.

The other approach is to define area as $2$-dimensional Hausdorff measure. From the definition, it is not hard to prove that Hausdorff measure does not increase under $1$-Lipschitz maps. Since the orthogonal projection $(x,f(x))\mapsto (x,0)$ is such a map from the surface onto $\Omega$, the result follows. Convexity is irrelevant with this approach; $f$ could be any function whatsoever.

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  • Of course my definition for "area" is the $n-1$ dimensional Hausdorff measure. I suppose the last part of your answer doesn't restrict itself to 3 dimension. The argument works for any dimension with no modification. Thanks for this viewpoint. I was tempted to ask about the proof when $f$ is not differentiable at all. – henryforever14 Jun 10 '14 at 10:50