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enter image description here $\mu_{n-1}$ is the $n-1$ dimensional measure and $\operatorname{meas}$ is the $n$-dimensional one.

$I(\varrho)$ is the ball of radius $\varrho$ around a fixed point $y$ in the domain $\Omega\subset \mathbb{R}^n$ and $A(t,\varrho)$is the subset of $I(\varrho)$ where a certain function $u$ is greater than $t$.

Surely $I(\varrho) \geq 2\tau(t;\varrho)$ infact, $$\operatorname{meas}{A(t;\varrho)}+\operatorname{meas}(I(\varrho)\setminus A(t;\varrho))= \operatorname{meas}I(\varrho)$$

It seems to me that it is claimed $x > y $ implies $ A-x > A-y$

Fortunately, it turns out the inequality still holds but you need a bit of work (EDIT)

Original article: De Giorgi's original article pages 5-6

EDIT:

Let us assume that $\operatorname{meas}(I(\varrho))=1$ then $\tau(t;\varrho)=x\leq \frac{1}{2}$ what we need to prove is that for $\alpha < 1$ $$(1-x)^{\alpha} + x^{\alpha} -1 \geq 2x^{\alpha}-(2x)^{\alpha}$$

but considering $$f(x)=\left( \frac{1-x}{x}\right) ^{\alpha} -\frac{1}{x^{\alpha}}$$

we can prove that it is decreasing in $(0,\frac{1}{2})$ and $f(\frac{1}{2})=1-2^{\alpha}$

From this we obtain the inequality and hence justify the line.

Due to the several mistakes the question has changed a lot since I posted it, just to be clear I am asking the following now:

1) Is my "fixing" of the proof alright or am I oversimplifying things?

2) Do you think it was an actual mistake that fortunately did not lead to a problem or it is something obvious enough to be omitted in an article?

P.S. sorry if the question is not specific enough, I think it was when I first posted it!

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1 Answer 1

up vote 5 down vote accepted

Your explanation looks correct, but there are other ways to see it. For example, it's a special case of Karamata's majorization inequality for the concave function $f(z) := z^\alpha$. When $0 \le x \le 1/2$, the sequence $1,x$ majorizes $1-x,2x$, so $f(1) + f(x) \le f(1-x) + f(2x)$, which is the inequality you want. (And Karamata's inequality is overkill, since the version for sequences of length 2 is easier to prove. All you need to check is that $f(z)+f(y-z)$ is an increasing function of $z$ for $z \le y/2$, which follows immediately from concavity. Then apply this with $y=1+x$.)

I don't see any reason to think De Giorgi made a mistake here, and I would assume he had an argument based on concavity. Maybe there's a one-line version he thought was obvious.

However, I wouldn't say it's obvious the way he has written it, and he could have made it easier to read with a little more explanation. Even a hint like "and it follows from the concavity of $z \mapsto z^{1-1/r}$ that..." could help.

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thank you, I thought it could have been a mistake since the rest of the paper explains quite carefully every step (for istance, the previous inequality $meas(I(\varrho)) \geq 2\tau$ is more trivial than this!) possibly he just overlooked it, or maybe you are right and there is some trivial justification! –  Moritzplatz Jun 30 '13 at 18:40

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