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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 i

by graycat 11y ago

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. How 'bout that!

analognoise 11y ago
See, this is why I am on HN. I'm interested but not as advanced. You included good examples; now I have to figure out wtf you are talking about. :)
- graycat 11y ago
  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!

by graycat 11y ago

analognoise 11y ago

See, this is why I am on HN. I'm interested but not as advanced. You included good examples; now I have to figure out wtf you are talking about. :)

graycat 11y ago
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!