Recent content by economicsnerd

Related question: If everything above were ##C^2##, then ##v## would satisfy the DE, ##r(x)v(x) = f(x) + \mu(x)v'(x) + \tfrac12 v''(x)##. I'd love to know if, when ##\mu, f, r## are discontinuous, there's still any meaningful sense in which the DE is satisfied.

Let ##\mu: \mathbb{R}\to \mathbb{R}##, ##f: \mathbb{R}\to \mathbb{R}##, and ##r: \mathbb{R}\to [1, \infty)## be bounded measurable functions (which may be discontinuous). I'm interested in the function ##v:\mathbb{R}\to\mathbb{R}## given by ##v(x) = \mathbb E \left[ \int_0^\infty e^{-\int_0^t...

I did mean ##\tfrac{\partial}{\partial y} f(y, x) = -\tfrac{\partial}{\partial x} f(x,y)##, not ##\tfrac{\partial}{\partial y} f(x, y) = -\tfrac{\partial}{\partial x} f(x,y) ##. So I'm after the set of functions whose first partial is an antisymmetric function. I'm wondering if there's a way...

Sorry for the terribly vague title; I just can't think of a better name for the thread. I'm interested in functions ##f:[0,1]^2\to\mathbb{R}## which solve the DE, ##\tfrac{\partial}{\partial y} f(y, x) = -\tfrac{\partial}{\partial x} f(x,y) ##. I know this is a huge collection of functions...

And notice that stevendaryl's argument that ##cf(\aleph_a) \leq |a|## can be slightly modified to show ##cf(\aleph_a) \leq cf(a)##. Indeed, letting ##C## be some cofinal subset of ##a##, one can verify that ##\{\aleph_c: \ c\in C\}## is cofinal in ##\aleph_a##.

Let ##u,v## be vectors in the same Euclidean space, and define the symmetric matrix ##M = uv'+vu'##, the sum of their two outer products. I'm interested in whether or not ##M## is positive (semi)definite. Does anybody know of any equivalent conditions that I might phrase "directly" in terms of...

Hi Krylov, 1) As it turns out, what you've provided is the definition I'm aware of for measurability of a map whose codomain is the space of probability measures on ##(B, \mathcal{B})##. In fact, I'm used to seeing the "standard" ##\sigma##-algebra on ##\mathcal P(B, \mathcal{B})## as the one...

I think I found one that works okay. For some fixed ##d>0## and ##\lambda\in(0,1)##, let ##g(x)## denote the "gap" between the graph of ##f## and its secant between ##x## and ##x+d##, measured proportion ##\lambda## of the way along. That is, let ##g(x):= [\lambda f(x+d) + (1-\lambda) f(x)] -...

Any others?

[I asked this question over a year ago, but I thought I'd try again.] Let ##I\subseteq \mathbb R## be an interval and ##f:I\to\mathbb R## be a ##C^\infty## function. I have the following characterizations: 1) ##f'\geq 0## everywhere iff ##f## is increasing. 2) ##f''\geq 0## everywhere iff...

1.5 years later, I'm still curious about this!

I don't buy that argument. I don't think it follows from ##(a,c)\in A\times C## and ##c\in C## that ##a\in A##. Let ##C= \mathbb R^\infty## and ##A=\mathbb R##. Let ##c=(1,1,1,...)\in C## and ##a = (0,0)\in \mathbb R^2.## Then ##a\notin A##, but ##(a,c)=(0,d)##, where ##d=(0,1,1,1,...)\in...

Assume we're only dealing with real numbers. Let ##\Phi_0(z)\equiv##"##\exists a,b\in\mathbb R \text{ such that } z = a^2, \enspace 1-z = b^2 \text{ and } b\neq 0##." It's straightforward to check that ##\Phi_0(z)## is true if and only if ##z\in[0,1)##. For any ##n\in \mathbb N##, let...

The usual definition of a set-valued map being continuous is that it's both upper and lower semicontinuous. So of course, it can't be continuous without being lower semicontinuous.

It appears to be lower semicontinuous but not upper semicontinuous. To see it's lower semicontinuous, fix an open set ##V\subseteq \mathbb R^2## which intersects ##F(\alpha)## for some given ##\alpha\in[0,2\pi]##. That is, ##\lambda(\cos\alpha,\sin\alpha) \in V## for some ##\lambda\geq0##...