Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Hi,

On page 120, chapter 4, proposition 4.2.7 in Hubbard's Teichmuller Theory book, volume 1, he proves :

Let $U,V$ be open in $C, f:U \to V $ be a homeomorphism and the restriction of $f$ on $U \backslash l$ is $K$ quasiconformal, where $l$ is a line in $U$. Then $f$ is $K$-q.c. in $U$.

FYI: his definition 4.1.1 ( page 112 ) of a $K$-q.c. map $f:U\to V$ is :

1) $f$ is a homeomorphism.

2) Distributional/weak derivatives $f_z,f_\bar{z}$ of $f$ exist almost everywhere and these derivatives are in $L^2_{loc}(U)$.

3)$|\frac{f_\bar{z}}{f_z}|\leq k = \frac {K-1}{K+1} $ for some $K\geq1$.

Now for the proof of Proposition 4.2.7, isn't it $ENOUGH$ to prove/check that the derivatives $f_z, f_\bar{z}$ are in $L^2(K) \forall K$ compact subset of $U$,i.e. condition (2), which does not readily follow from that they are in $L^2_{loc}(K') \forall K' $ compact subsets oi $U \backslash l$. This he proves by using condition (3) stated above .

But what else is there to prove, since we only care about existence of partial derivatives on a full measure set, we do NOT need to prove anything else apart from condition (2), right ? So, what does he do after proving condition (2), in page 120-121 ?

Thanks and Happy Memorial Day!

share|improve this question
add comment

1 Answer

up vote 7 down vote accepted

All that Hubbard has shown up to that point is that the functions $f_z$ and $f_{\bar{z}}$ are locally in $L^2(R)$, i.e. they are defined almost everywhere and are square integrable in $R$ as functions. But all we know at this point is that they are the distributional derivatives of $f$ away from $l$. What he shows after that is that they are actually the distributional derivatives of $f$ in all of $R$ rather than just $R-l$. I can understand the confusion, and I am surprised that Hubbard does not make this more explicit (he is generally very careful about these kinds of issues, both in writing and in conversations).

Near the top of the page, Hubbard remarks that $[\widetilde{Df}]\in L^1(R)$ rather than just $R-l$. This is a key point. He writes that this allows him to justify applying Fubini's theorem, but he also needs this fact to apply the Lebesgue dominated convergence theorem. I'll point out exactly where.

On the series of equalities in 4.2.14, the second equality follows from Lebesgue dominated convergence since $f$ and $\phi$ are continuous and hence bounded on $R$; similarly for the third equality. The fourth equality follows from the fact that away from $l$ our function $f$ has a weak derivative locally in $L^2$ and hence $L^1$. There are now two limits to compute in this sum. The second one is zero since $f$ is continuous and the Lebesgue dominated convergence theorem applies. One may now write the remaining double integral as a single integral over $R_\epsilon$ by Fubini and then push the limit into the integral by Lebesgue dominated convergence (these two steps are where Hubbard's remark is used).

These equalities guarantee that $f_x$ and $f_y$ are indeed the distributional derivatives in $R$. And now, exactly as you've said, this, along with the computations he has already made, give you exactly what you need. I hope this explains everything for you.

share|improve this answer
1  
I'm glad to see your getting you use of my copy, since I'm taking it back when I visit in three days! –  Greg Muller May 31 '11 at 4:48
    
Yep, got a surprising little more mileage out of that book. The OP and I should thank you for asking for the book back, otherwise the book wouldn't have been right next to me when I saw this question and this answer would never have happened. –  Peter Luthy May 31 '11 at 4:59
    
@ Peter Luthy : Thanks very much for your detailed answer, but I guess I am still missing the point : why do we even need to show that $f_z,f_\bar{z}$ are the distributional partial derivatives of $f$ on the whole $U$,not just $U- l$ since the distributional derivatives are defined almost everywhere on U , and l is a measure zero set in U, and we already know that f has distributional derivatives on almost everywhere on $U$,i.e. on $U - l $ ? Isn't the statement "f has distributional derivatives on U" the same as "f has distributional derivatives on $U-l$" ? – Plus 0 secs ago –  Analysis Now May 31 '11 at 14:01
1  
Suppose you are on $R^1$ and $l$ is just the point $\{0\}$. Then the function $f$ which is 0 when $x<0$ and 1 when $x\ge 0$ has distributional derivative 0 away from 0. So $f'$ is defined almost everywhere and $f'$ is the distributional derivative for any function compactly supported in $R^1-\{0\}$. But if $\phi$ is any $C_c^\infty$ supported function which is 1 in a neighborhood of $0$, then $-\int f\phi'=\phi(0)=1$ rather than 0 if $f'$ were actually the distributional derivative on all of $R$. You can come up with clever ones so that $f$ is continuous: see the Cantor-Lebesgue function. –  Peter Luthy May 31 '11 at 14:43
1  
The Cantor-Lebesgue function has derivative 0 almost everywhere but is continuous and strictly increasing. The issue is that for a function on $R$ to have distributional derivatives it must be absolutely continuous; simply having a well-defined almost everywhere derivative is not enough. –  Peter Luthy May 31 '11 at 14:50
show 2 more comments

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.