# G. Leoni, Sobolev spaces, Lemma 3.13

1. Jul 4, 2014

### jostpuur

I don't have the whole book A First Course in Sobolev Spaces by G. Leoni myself, but I have obtained a pdf file of the third chapter of it, and I have got stuck in trying to understand the Lemma 3.13. Is there anyone here feeling like understanding it?

The claim is this: $I\subset\mathbb{R}$ is some interval, and $u:I\to\mathbb{R}$ is some function. We assume that $E\subset I$ is some set such that $u$ is differentiable at all points $x\in E$. We also assume that a real constant $M$ exists such that $|u'(x)|\leq M$ for all $x\in E$. It is emphasized that $E$ is not assumed measurable. The claim is that $m^*(u(E))\leq Mm^*(E)$ holds, where $m^*$ is the Lebesgue outer measure.

The proof is supposed to start like this: We fix some $\epsilon > 0$ until the end of the proof. For all $N=1,2,3,\ldots$ we define sets $E_N$ as

$$E_N = \big\{x\in E\;\big|\; \big(x\in [\alpha,\beta]\subset [a,b]\quad\textrm{and}\quad \beta-\alpha < \frac{1}{N}\big)\implies m^*(u([\alpha,\beta])) \leq (M+\epsilon)(\beta-\alpha)\big\}$$

The next intermediate claim is

$$E = \bigcup_{N=1}^{\infty} E_N$$

Since "$\supset$" direction is obvious, the "$\subset$" is the real task. We fix arbitrary $x\in E$. Since we know $|u'(x)|\leq M$, we also know that there exists $\delta > 0$ such that

$$|y-x|<\delta\implies |u(y)-u(x)|\leq (M+\epsilon) |y-x|$$

Therefore:

$$\big(y<x<y'\quad\textrm{and}\quad y'-y < \delta\big)$$
$$\implies\quad |u(y')-u(y)| \leq |u(y')-u(x)| + |u(x) - u(y)| \leq (M+\epsilon)(y'-y)$$

At this point Leoni states, that we now know $x\in E_N$ for all $N$ such that $\frac{1}{N}<\delta$. At a quick glance it seems that there is a mistake. It seems that quantities $|u(y')-u(y)|$ and $m^*(u([y,y']))$ have been confused. Of course it could be that there is no mistake, and the proof of the intermediate claim can be completed with some additional information that was not explicitly mentioned in the book. However, I have been unable to figure out the additional information on my own, and that's why I wrote this post. I have made some changes to the notation, to improve clarity in my opinion. It could also be that I have made some mistakes in this, and I would very much appreciate if that turned out to be the case. I'm not aware of any mistakes myself though.

2. Jul 5, 2014

### jostpuur

No answers during the first 24h. I've been thinking about this, and now to me it seems that the claim by G. Leoni is correct, but I'm still not convinced that his proof is, since to me it seems to be so far from some critical remarks that one cannot reasonably insist that he would have only left "some details" to the reader.

So arbitrary $x\in E$ is fixed. This is how I would proceed: By using the definitions of the derivative and the limit, we see that there must exist $\delta >0$ with the following property:

$$|h|<\delta\quad\implies\quad u(x)+hu'(x)-\frac{\epsilon}{2}|h|\leq u(x+h)\leq u(x) + hu'(x) + \frac{\epsilon}{2}|h|$$

Once this $\delta$ has been fixed, we claim that it also has the property:

$$\big(y<x<y'\quad\textrm{and}\quad y'-y<\delta\big)\quad\implies\quad m^*(u([y,y']))\leq (M+\epsilon)(y'-y)$$

This is a nice claim in the sense that if it is true, we get what we wanted. That was that $x\in E_N$ holds for all $N$ such that $\frac{1}{N}<\delta$. The claim can be proven through an antithesis. According to an antithesis, there exists $y$ and $y'$ with the following properties:

$$y<x<y'\quad\textrm{and}\quad y'-y<\delta\quad\textrm{and}\quad (M+\epsilon)(y'-y) < m^*(u([y,y']))$$

The last condition implies that the set $u([y,y'])$ cannot be contained within a closed interval whose length would be $(M+\epsilon)(y'-y)$. Therefore two points can be found from the set $u([y,y'])$ such that their distance is greater than the mentioned length. Since this set is an image, the two points can be written in the forms $u(z)$ and $u(z')$ for some $z,z'\in [y,y']$. Then we denote $h=z-x$ and $h'=z'-x$, and we have found numbers $h,h'$ with the following properties:

$$|h|<y'-y,\quad |h'|<y'-y\quad\textrm{and}\quad (M+\epsilon)(y'-y) < u(x+h') - u(x+h)$$

We use the approximation obtaind from the definition of the derivative to approximate further

$$(M+\epsilon)(y'-y) < \big(u(x)+h'u'(x)+\frac{\epsilon}{2}|h'|\big) - \big(u(x)+hu'(x) - \frac{\epsilon}{2}|h|\big)$$
$$= (h'-h)u'(x) + \frac{\epsilon}{2}(|h'|+|h|)$$
$$< (h'-h)u'(x) + \epsilon(y'-y)$$

Epsilon terms cancel, and we get

$$M(y'-y) < (h'-h)u'(x)$$

This is in contradiction with the inequalities $|h'-h|<|y'-y|$ and $|u'(x)|\leq M$.

3. Jul 5, 2014

### micromass

Staff Emeritus
I think it's false what he's trying to prove (perhaps not the theorem but the intermediate claim). Consider $\varepsilon = 1$ and $f:\mathbb{R}\rightarrow \mathbb{R}$ defined by $f(x) = 0$ everywhere except at $f(0.9) = 0.9$ and $f(0.85) = - 0.85$. Take $E=\{0\}$ and thus we can take $M = 0$.

Then we know indeed that there is a $\delta>0$ such that
$$|u(y)| \leq 2|y|$$
Indeed, taking $\delta = 2$ works. Now take $N=1$, then certainly $1/N<\delta$. So now the author claims that $0\in E_1$. But take $\alpha = -0.94$, $\beta = 0.05$. Then $u([\alpha,\beta]) = [-0.85,0.9]$. Thus
$$m^*(u([\alpha,\beta])) = 1.75$$
and
$$(M+\varepsilon)m^*([\alpha,\beta]) = 0.99$$
So we certainly have a violation of the inequality.

Of course, the theorem as well as the claim $E = \bigcup E_N$ is true. But his proof seems to fail.

4. Jul 6, 2014

### jostpuur

That was a confusing post. What is the relation of $f$ and $u$?