It's not freely available if they are not free to copy... and I really wonder who could possibly be harmed by copying these videos around. I can't imagine that Milnor would object to having his work widely disseminated. Do you think if we copy something he made 50 years ago (which really should be in the public domain by now), he'll be less motivated to generate new lectures at his 84 years of age?
Your usage of the term "pirated" seems to indicate more indignation than is in order.
It's not even free beer. If I go to an event and get a free beer, I'm not forbidden from giving my free beer to anyone else. It is ridiculous that we are forbidden from sharing a video that was created 50 years ago just because Disney wants to make sure Mickey Mouse never falls out of copyright.
For the set of real numbers
R, he starts with a function
f: R --> R
such that f(x) is zero for
x <= 0, strictly positive
otherwise, and infinitely differentiable. There is
such a function, the same or
similar, in an exercise in
Rudin, Principles of Mathematical
Analysis.
At one time I used that function
in Rudin
to show that for positive
integer n and closed subset
C of R^n there exists function
f: R^n --> R
so that f(x) = 0 for all x in
C, strictly positive otherwise,
and infinitely differentiable.
That result is comparable with
the classic Whitney extension
theorem -- Whitney assumed
a little more and got a little
more.
I discovered this result
for and used it in some work in the constraint
qualifications of the Kuhn-Tucker
conditions: I constructed
a counterexample that showed
that the Zangwill and Kuhn-Tucker
constraint qualifications
are independent.
So, back to Milnor's lecture!
I want to see if he drags out
the inverse and implicit
function theorems (essentially
the local nonlinear version
of what is standard from
Gauss elimination in
systems of linear equations!).
Neat. Did you do it by making such a function for the open ball, and then, given an open set defined as a countable union of open balls, define a sequence of such functions scaled appropriately so that their derivatives all converge uniformly?
I took a countable dense set in R^n - C, x_j, j = 1, 2, ...,
and for each j took the closed circle with center x_j and radius
d(x_j, C) (distance from x_j to
C which is well defined due to
compactness). Then I defined a
function g_j: R^n --> R 0 outside the
circle and on the circle with
values on the radii much as in
the function Milnor, Rudin, and
I mentioned. So, g_j is
infinitely differentiable,
strictly positive on the interior
of the circle at x_j, and
0 otherwise. Then I added
the functions g_j in a
convergent way, again using
the exponential trick.
The resulting function
was as desired, 0 on C,
strictly positive otherwise,
and infinitely differentiable.
No. The domain of f is all of
R^n so that f(x) has to
be defined for all x in R^n.
Then there is the requirement
that f be strictly positive otherwise, that is, in the open set outside C.
The result is a bit amazing:
It says that any closed set
can be the level set for
an infinitely differentiable
function.
Can add some interest when
consider how bizarre some
closed set are:
So, in the plane
can take C as a sample path of
Brownian motion. So, there's
an infinitely differentiable
function zero on that sample path
and positive otherwise.
Next, the Mandelbrot set is
closed. So, there's an
infinitely differentiable
function positive everywhere
but zero on the Mandelbrot set.
Next consider the Cantor set
or Cantor sets of positive
measure. Same story.
So, a really smooth,
infinitely differentiable, function can have for
a level set
anything at all considering
that any level set of a
continuous function is closed.
There's a famous paper in
mathematical economics by
Arrow, Hurwicz, and Uzawa
with a question with no
answer, and my work answers
the question. Yes, Arrow
and Hurwicz won prizes in
economics -- IIRC so far
poor Uzawa has not won
such a prize!
For something intuitive,
consider an infinitely
differentiable mountain
range, and pour in some water
to make a lake or many
lakes (assume a porous
mountain) all with the
same altitude. Then, the lake
boundary is a level set of
an infinitely differentiable
function. As we know from
Mandelbrot, commonly, roughly
lake boundaries are fractals.
So, we're fine; everything
holds together: My result
shows that for Mandelbrot's
fractal lakes, the ground
can be smooth, infinitely
differentiable. Did
Mandelbrot know that?
For rational p/q for positive
whole numbers p and q with p/q
in lowest terms, consider in
R^2 a ray from the origin of
length 1/q at angle p radians.
Do this for all such countably
infinitely many rationals. Then
the set of all these rays is
closed but is a spiny urchin,
really bizarre. That's the
closed set I used for my
counterexample.
Can also apply this to
zero day detection of
anomalies in server farms
and networks: So, suppose
collect 20 times a second
numerical data on 10 variables.
Then that vector of 10
variables has a probability
distribution. Suppose it also
has a continuous density.
Then pour in water to make
lakes -- all the same altitude.
Get a point in a lake, then
declare an anomaly. The volume
of land under the lakes is
essentially the false alarm rate.
Pour in more water and get
a higher false alarm rate but
also a higher detection rate.
Here the set where we declare
an anomaly has the highest possible area for the given
false alarm rate, and there's
another sense in which the
detector has optimal combinations
for false alarm rate and
detection rate.
A function f is differentiable if it has a derivative just as
in calculus or, as in high school,
a tangent line. If the derivative, another function, is
differentiable in the same sense,
then the function f is twice
differentiable. For any positive integer n, in the same way, can say what it means
for function f to be n-times
differentiable. Finally, if
for each positive integer n,
function f is n-times
differentiable, then function
f is infinitely differentiable.
Why do we care about function
f being infinitely differentiable? Because that's
necessary for function f
to have a Taylor series
expansion. Why care? It says
we have a start on a ready made way to
approximate f as accurately
as we wish with, also some
good, first-cut information
on the function from if only
from the first few terms.
With more, as Milnor mentioned,
the function f might be
analytic which means that
it is equal to its Taylor
series expansion.
In R^n, a set is open if
it contains a little ball
around each of its points.
So, in R the interval (0,1)
is open. A set C a subset of
R^n is closed if and only
if R^n - C is open. Open
sets have some nice properties;
so do closed sets.
E.g., we know from calculus
from uses of deltas and epsilons
what a continuous function is.
Well we can generalize to topoloty, Milnor's subject:
Quite generally, function f
is continuous if
the inverse image of
each open set in the range
set of f is open in the
domain set of f --
and this definition works with
astounding generality.
Topology is heavily interested
in continuity.
In R^n, a each bounded, closed
set is compact which is
really a generalization of
it being finite and finger
lick'n good: E.g., in calculus,
each f: R --> R continuous
on a compact set has a
Riemann integral on that set.
Biggie stuff.
Can take open and closed off to,
say, Banach space (complete
normed vector space)
and the famous open mapping
and closed graph theorems.
For the result I discussed,
open and closed were
central concepts.
The Kuhn-Tucker conditions
are in optimization. So,
imagine you are in a
dark cave with
walls and a floor that is
not flat. If you put down
a ball and it rolls, then
you are not at the bottom.
So, a necessary condition
for being at the bottom
is that the ball not roll.
For that it may be that
the floor would let
the ball roll but the walls
stop it. So, when do the
walls do this? Roughly the
normal vectors of the walls
in contact with the marble
block the motion for the
marble to roll lower.
But for the simple Kuhn-Tucker
equations, in terms of
gradients, work, the
walls must be sufficiently
nice, that is, satisfy
some constraint qualifications.
There are lots of constraint
qualifications, some easier
to verify than others.
In nearly all routine engineering
problems, the constraints
do satisfy some, typically
several, constraint qualifications.
But if want to dig into
lots of details, look
at pathological cases,
etc. then can wonder about
constraint qualifications.
Kuhn and Tucker have a
statement of some constraint
qualifications (CQ), and
so does Zangwill. Are the
two CQs equivalent? One
direction was no. I showed
that the other direction was
also no and, thus, that the
two CQs are independent,
with neither implying the other.
My proof was by counterexample,
from my result, and was
bizarre and tricky, which
is likely why it was new.
For my counterexample, I
used my result about
infinite differentiability,
but that was much more
than I needed. I showed
infinite differentiability
only out of curiosity --
just wondered if it was true.
The result is curious
because intuitively an
infinitely differentiable
function is supposed to
be very smooth (Milnor
gave a precise definition
but I was just being intuitive)
but some of the level sets
can be wildly irregular,
e.g., the Mandelbrot
set and Brownian motion.
E.g., Brownian motion
never has a tangent line,
not even a first derivative,
but it is the level set
of an infinitely differentiable
function. So, the juxtaposition
is curious.
Yet again
a math guy stands in awe
of reality that math
discovers one little theorem
at a time.
So, now you know more about
what the heck we are talking
about!
Long ago I read a quote from a post about mathematicians who are good communicators. I'd love to come across it again.
Of John Milnor, this post proclaimed, "When he speaks, you understand."
I try to notice rare compliments, like that one. Feynman received one like that:
A female engineer says about him: "Yes, [Feynman's sexism] really annoys me," she said. "On the other hand, he is the only one who ever explained quantum mechanics to me as if I could understand it."
What causes the halos in this film? I notice whenever there is an unusually light or dark spot, it gets a halo around it of the opposite lightness. See his hand as he starts writing shortly after this time stop to see what I mean: https://youtu.be/1LwkljjLBns?list=PLelIK3uylPMFHC6Xny11XFXgw...
It's not a filter, as others have suggested, but rather a standard artifact of vacuum-tube-based television cameras that were in use at the time. (I don't know why this lecture wasn't filmed, but the extensive dark halo effect makes it clear this was shot with a TV camera. This was the early era for magnetic video tape, but I assume that's how the recording was preserved.)
Anyway, the point is that these TV cameras are based on the fact that incoming light will dislodge electrons from a thin plate in a vaccum tube in an amount proportional to brightness. A very bright spot in the image produces a shower of electrons that is more powerful than the rest of the tube (the part that detects the electrons) can deal with. The net result of this "splash" of electrons is a mild desensitization of the detection apparatus around the bright spot. This makes the nearby stuff appear darker.
You mentioned that dark spots also seem to have a bright halo, but I don't see that in the video, and it isn't consistent with the usual artifacts of these cameras. Are you sure?
Could also be a high dynamic range operator: they do the same kind of thing to preserve contrast across large ranges of intensity. See Larson's paper http://gaia.lbl.gov/btech/papers/39882.pdf etc.
I really appreciate how he writes and draws the slides as he speaks. With math lectures, this always encourages me to think deeply about the details and implications of each line. I find it better than pre-written slides, now so easily created and presented using Power Point.
I think this style of presentation is still very common in maths departments. Mathematicians tend to favour low-tech presentations. Blackboards are still in vogue!
https://www.simonsfoundation.org/science_lives_video/profess...
Published sequel here: http://www.ams.org/notices/201106/rtx110600804p.pdf