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I'm a little bit confused about different ways of approximating by smooth functions, in particular, quasiconformal mappings.

So if a map $\phi : R\to R'$ is $K$-quasiconformal map on a relatively compact set $R$, it is true that $\phi \in W^{1,2}(R)$. And I know that $\phi$ can be approximated in $W^{1,2}(R)$ by smooth functions in $C^\infty(R)$, meaning that

$ \lim_{n->\infty}||\phi_n - \phi||_{W^{1,2}(R)} = ((|| \phi_n - \phi ||_{2})^2 + || D\phi_n - D\phi ||_{2})^2)^{1/2} = 0 $

But it seems like it's possible to find smooth functions $\phi_n$ that actually

1) $\phi_n\to\phi$ (pointwise, or possibly uniform?)

2)$||D\phi_n -D\phi||_2 \to 0$.

So my question is, how is this possible? (i.e. What theorem?)

The sequence I get from the approximation theorem in $W^{k,p}$ guarantees 2), but not the first one. I think I get a.e. uniform or a.e. convergence at best...

Note: I'm not entirely sure what convergence 1) is. I'm looking at a proof given by Ahlfors that shows that q.c. maps send null sets to null sets in his "Lectures on Quasiconformal Mappings." For the purpose of the proof, the a.e. convergence is enough, but Ahlfors doesn't say "a.e.," so I'm wondering if I'm missing something...

EDIT: Here's a word-by-word quote, as requested by Willie Wong

" THEOREM 3. Under a q.c. mapping the image area is an absolutely continuous set function. This means that null sets are mapped on null sets, and that the image area can always be represented by

$$A(E) = \int\int_E J \, dx \, dy$$

PROOF. $\phi = u+iv$ can be approximated by $C^2$ functions $u_n+iv_n$ in the sense that $u_n \to u$, $v_n \to v$ and

$\int\int |u_x - (u_n)_x|^2 \, dx \, dy \to 0$

$\int\int |v_x - (v_n)_x|^2 \, dx \, dy \to 0$ etc.

Consider rectangles $R$ such that $u$ and $v$ are absolutely continuous on all side.... "

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Retagging because the question itself has not much to do with quasiconformality. Also, (reference-request) is for questions asking about explicit references, like "In which paper was blah first proven?" or "Where can I find a statement of foobar?" –  Willie Wong Aug 1 '11 at 16:45
    
@Willie - So I'm reading a proof by Ahlfors, but he doesn't make any reference to Sobolev spaces. I originally had the q.c., because I thought maybe I was missing something that was true for q.c. maps but not necessarily for all of $W^{1,2}$. –  Braindead Aug 1 '11 at 17:06
    
@Willie - Also, I AM looking for a statement or a theorem or a reference, since it looks like Ahlfor is using a different approximation theorem. Like I wrote before, he makes no reference to Sobolev spaces, and he explicitly writes $\phi_n\to\phi$ and the convergence of the partial derivatives in $L^2$. –  Braindead Aug 1 '11 at 17:08
    
You need to include a lot more material. For example, an exact quote of what Ahlfors wrote would help greatly. At the present, "it seems like it's possible to find smooth functions $\phi_n$ that actually ..." is unjustified. Did Ahlfors actually claim that? Did you see it asserted somewhere? Obviously since you cannot prove it yourself, it does not "seem possible" to you. –  Willie Wong Aug 1 '11 at 17:24
    
I included the original quote. –  Braindead Aug 1 '11 at 17:45
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1 Answer

up vote 5 down vote accepted

(Edited after checking the Google Books preview of the book.)

You should really check the definition that Ahlfors gave in his book. From the snipplet preview I see on Google Books, his definition of a quasiconformal maps requires that $\phi$ is a topological mapping, aka a homeomorphism of $\Omega$ onto its image. This particular implies that $\phi$ is continuous. So the standard mollifier construction will give you a sequence $\phi_n$ that converges to $\phi$ pointwise. And on any compact subset, $\phi$ is uniformly continuous, and so $\phi_n \to \phi$ uniformly pointwise.

In other words, the question you wanted to ask is

If $\phi \in C^0\cap W^{1,2}$, can we find a sequence $\phi_n \in C^2$ s.t. $\phi_n$ converges to $\phi$ pointwise and $\|D\phi_n-D\phi\|_2 \to 0$.

And the answer to that is yes. Let $\eta$ be a standard bump function and define the mollified $\phi_n = \eta_{1/n} * \phi$ via convolution in the usual way, so that $\eta_{1/n}$ is supported in the ball of radius $1/n$. Then by continuity of $\phi$, for every $x$ and for every $\epsilon > 0$, there exists a corresponding $\delta$ such that $|\phi(x) - \phi(y)| < \epsilon$ if $y$ is in the ball of radius $\delta$ around $x$. For $n > \delta^{-1}$, then,

$$ |\phi(x) - \phi_n(x)| \leq \int \eta_{1/n}(y) |\phi(y)-\phi(x)| dy \leq \epsilon $$

using that $\int \eta_{1/n}(y) dy = 1$. It is easy to see how this proof can be modified to use uniform continuity when available.

The convergence in $\cdot{W}^{1,2}$ you know how to prove already, so I omit.

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I feel like you misunderstood my question. I know that it's possible to find $\phi_n$ in $C^\infty \cap W^{1,2}$ that converges to $\phi$ in the $W^{1,2}$ norm. The question is whether there was a different approximation theorem, which allows $\phi_n \to \phi$ maybe pointwise or uniform (without the a.e. part) with the partial derivatives converging in $L^2$. –  Braindead Aug 1 '11 at 17:15
    
If $\phi$ is asserted to be merely a function in $W^{1,2}$, the statement that $\phi_n \to\phi$ pointwise makes no sense, as $\phi$ is an equivalence class of functions differing on sets of measure zero. –  Willie Wong Aug 1 '11 at 17:22
    
Oh no. $\phi$ is a quasiconformal mapping. It is an orientation preserving homeomorphism with some extra properties. One of these extra properties is that it is differentiable almost everywhere, and that the the partial derivatives are locally $L^2$. –  Braindead Aug 1 '11 at 17:29
    
So $\phi$ is not merely a $W^{1,2}$, which is why I had the quasiconformal tag on the post. –  Braindead Aug 1 '11 at 17:30
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The point is that if you view the quasiconformal criterion in the distributional sense, there is no requirement a priori that $\phi$ is continuous. But in the case at hand you are using Ahlfors' definition which assumes that $\phi$ is continuous, and that is a rather important property that you should've stated in the question. –  Willie Wong Aug 1 '11 at 18:14
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