Difference between revisions of "0708-1300/Class notes for Tuesday, October 16"

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''Proof''
 
''Proof''
  
W.L.O.G (without loss of generality) we can assume n=1. Intuitively this is reasonable as in lower dimensions the theorem is harder to prove; indeed, a set of size <math>\epsilon</math> in 1D becomes a smaller set of size <math>\epsilon^2</math> in 2D etc. More precisely, for <math>f = (f_1,\ldots,f_n)</math>, <math>f(D_n)\subset f_1(D_m)\times{R}^{n-1}</math>. Applying the proposition that if A is of measure zero in <math>\mathbb{R}</math> then <math>A\times\mathbb{R}^{n-1}</math> is measure zero in <math>\mathbb{R}^n</math> we now see that assuming n=1 is justified.  
+
W.L.O.G (without loss of generality) we can assume n=1. Intuitively this is reasonable as in lower dimensions the theorem is harder to prove; indeed, a set of size <math>\epsilon</math> in 1D becomes a smaller set of size <math>\epsilon^2</math> in 2D etc. More precisely, for <math>f = (f_1,\ldots,f_n)</math>, <math>f(D_m)\subset f_1(D_m)\times{R}^{n-1}</math>. Applying the proposition that if A is of measure zero in <math>\mathbb{R}</math> then <math>A\times\mathbb{R}^{n-1}</math> is measure zero in <math>\mathbb{R}^n</math> we now see that assuming n=1 is justified.  
  
  

Latest revision as of 17:06, 7 November 2007

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Contents

Dror's Computer Program for C+\alpha C

Our handout today is a printout of a Mathematica notebook that computes the measure of the projection of C\times C in a direction t\in[0,\pi/2] (where C is the standard Cantor set). Here's the notebook, and here's a PDF version. Also, here's the main picture on that notebook:

0708-1300-CCShadow.png

Typed Class Notes

The notes below are by the students and for the students. Hopefully they are useful, but they come with no guarantee of any kind.

First Hour

Today's Agenda:

Proof of Sard's Theorem. That is, for f:M^m\rightarrow N^n being smooth, the measure \mu(critical values of f) = 0.


Comments regarding last class

1) In our counterexample to Sard's Theorem for the case of C^1 functions it was emphasized that there are functions f from a Cantor set C' to the Cantor set C. We then let g(x,y) = f(x) + f(y) and hence the critical values are C+C = [0,2] as was shown last time. The sketch of such an f was the same as last class.

Furthermore, in general we can find a C^n such function where we make the "bumps" in f smoother as needed and so f(C') = C'' where C'' is a "very thin" Cantor set. But now let g(x,y,z,\ldots) = f(x) + f(y) + f(z) +\ldots which will have an image of C'' + C'' + C'' + \ldots = an interval.


2) The code, and what the program does, for Dror's program (above) was described. It is impractical to describe it here in detail and so I will only comment that it computes the measure of C+\alpha C for various alpha and that the methodology relied on the 2nd method of proof regarding C+C done last class.


Proof of Sard's Theorem


Firstly, it is enough to argue locally, since a manifold is second countable (that is, a manifold has a countable basis) and a countable union of sets of measure zero has measure zero.

Further, the technical assumption about manifolds that until now has been largely ignored is that our M must be second countable. Recall that this means that there is a countable basis for the topology on M.

As a counterexample to Sard's Theorem when M is NOT second countable consider the real line with the discrete topology, a zero dimensional manifold, mapping via the identity onto the real line with the normal topology. Every point in the real line is thus a critical point and the real line has non zero measure.


We can restrict our neighborhoods so that we can assume M=\mathbb{R}^m and N=\mathbb{R}^n

The general idea here is that if we consider a function g=f' that is nonzero at p but that f is zero at p, the inverse image is (in some chart) a straight line (a manifold). As such, we will inductively reduce the dimension from m down to zero. For m=0 there is nothing to prove. Hence we assume true for m-1.


Now, set D_k = \{p | all partial derivities of f of order \leq k vanish} for k\geq 1

Set D_0 = \{p\ |\ df_p is not onto }. This is just the critical points.


Clearly D_0\supset D_1\supset\ldots\supset D_i\supset\ldots\supset D_m


We will show by backwards induction that \mu(F(D_k)) = 0


Comment:

We have not actually defined the measure \mu. We use it merely to denote that F(D_k)) has measure zero, a concept that we DID define.

Second Hour

Claim 1

F(D_m) has measure 0.

Proof

W.L.O.G (without loss of generality) we can assume n=1. Intuitively this is reasonable as in lower dimensions the theorem is harder to prove; indeed, a set of size \epsilon in 1D becomes a smaller set of size \epsilon^2 in 2D etc. More precisely, for f = (f_1,\ldots,f_n), f(D_m)\subset f_1(D_m)\times{R}^{n-1}. Applying the proposition that if A is of measure zero in \mathbb{R} then A\times\mathbb{R}^{n-1} is measure zero in \mathbb{R}^n we now see that assuming n=1 is justified.


Reminder

Taylor's Theorem: for smooth enough g:\mathbb{R}\rightarrow\mathbb{R} and some x_0 then g(x) = \sum_{j=0}^{m} \frac{g^{(j)}(x_0)}{j!}(x-x_0)^j + \frac{g^{(m+1)}(t)}{(m+1)!}(x-x_0)^{m+1} for some t between x and x_0.


For x_0\in D_m all but the last term vanishes and so we can conclude that f(x) is bounded by a constant times (x-x_0)^{m+1}.


Now let us consider a box B in \mathbb{R}^m containing a section of D_m. We divide B into C_1\lambda^m boxes of side 1/\lambda.

By Taylor's Theorem, f(B_i)\subset of an interval of length C_2\frac{1}{\lambda^{m+1}} where the constant C_2 is determined by Taylor's Theorem. Call this interval I_i

Hence,

f(D_m)\subset\bigcup_{i:B_i\cap D_m \neq 0} f(B_i)\subset \bigcup_{i:B_i\cap D_m \neq 0}I_i

But \sum_{i} length(I_i)\leq C_1\lambda^m C_2\frac{1}{\lambda^{m+1}} which tends to zero as \lambda tends to infinity.

Q.E.D for Claim 1


Claim 2

f(D_k) has measure zero for k\geq 1. We just proved this for k=m.


Now, W.L.O.G. D_{k+1} is the empty set. If not, just consider M^m - D_{k+1} which is still a manifold as D_{k+1} is closed (as it is determined by the "closed" condition that a determinant equals zero)


So, there is some kth derivative g of f such that dg\neq 0.


So D_k\subset g^{-1}(0) but g^{-1}(0) is at least locally a manifold of dimension 1 less. So, f(D_k)\subset f(D_k\cap g^{-1}(0)) which has measure zero due to our induction hypothesis.

Q.E.D for Claim 2


Claim 3

f(D_0) is of measure zero.

Recall D_0 is defined differently from the D_k and so requires a different technique to prove.

W.L.O.G. assume that D_1 is the empty set. So, some derivative of f is not zero. W.L.O.G. \frac{\partial f_1}{\partial x_1} is non zero near some point p. We can simply move to a coordinate system where this is true.

The idea here is to prove that the intersection with any "slice" has measure zero where we will then invoke a theorem that will claim everything has measure zero.

So, let U be an open neighborhood of a point p\in M. Consider f_1:U\rightarrow\mathbb{R} and let df_1 be onto. Using our previous theorem for the local structure of such a submersion W.L.O.G. let us assume f_1:\mathbb{R}^m\rightarrow\mathbb{R} via (x_1,\ldots,x_m)\mapsto x_1. That is, f_1 = x_1.

Our differential df then is just the matrix whose first row consists of (1,0,\ldots,0). Then df is onto if the submatrix consisting of all but the first row and first column is invertible.

For notational convenience let us say f =(f_1,f_{rest}).

Now define U_t = \{t\} \times\mathbb{R}^{m-1}. Also, let us denote "critical points of f" by CP(f) and "critical values of f" by CV(f).

Claim 4

The CP(f) = \bigcup_{t\in\mathbb{R}}\{t\}\times CP(f_{rest}|_{U_t})

\Rightarrow

CV(f) = \bigcup_{t\in\mathbb{R}}\{t\}\times CV(f_{rest}|_{U_t}).

But CV(f_{rest}|_{U_t}) has measure zero by our induction.


Lemma 1

If A\subset I^2 is closed and has \mu(A\cap(\{t\}\times I)) = 0\ \forall t then \mu(A)=0.


Proof

Note: We prove this significantly differently then in Bredon

Sublemma

If \{t\}\times U for an open U cover \{t\}\times I\cap A for a closed A then \exists\epsilon>0 such that (t-\epsilon,t+\epsilon)\times U\supset (t-\epsilon,t+\epsilon)\times I\cap A

Indeed, let d:A-(I\times U)\rightarrow\mathbb{R} be d(x) = |x_1-t| then d is a continuous function of a compact set and so obtains a minimum and since d>0 then min(d)>0 \rightarrow d>\epsilon>0. But this \epsilon works for the claim. Q.E.D


The rest of the proof of Lemma 1, and of Sard's Theorem will be left until next class