Recall Chebyshev's Inequality:
Let $f$ be a nonnegative measurable function on $E$. Then for any $\lambda > 0$, $$m \{x \in E \mid f(x) \ge \lambda \} \le \frac{1}{\lambda } \int_{E} f$$
Here is a theorem in my book:
Let $f$ be an increasing function on the closed, bounded interval $[...
Definition. For a real valued function $f$ and an interior point $x$ of its domain, the uppper derivative of $f$ at $x$ denoted by $\overline{D}f(x)$ is defined as follows: $$\overline{D}f(x)=\lim_{h\rightarrow0}\left[ \sup \left \{\frac{f(x+t)-f(x)}{t}: 0<|t|\leq h \right \} \right]$$
I am ...
Eh dunno, it was mostly because the first part was measure theoretic probability and then it became very computational and applied with Markov processes
So basically the story goes as follows. If you have an open subset $U \subset \Bbb R^n$ with Lipschitz boundary (basically having a cone with sufficient solid angle at any point on the boundary which does not intersect $U$ anywhere except at the vertex point suffices), say, then the Dirichlet problem $\Delta F = 0$ with $F|\partial U = f$ has the solution $F(x) := \Bbb E[f(B_\tau)|B_0 = x]$
where $\{B_t\}$ is a Brownian motion on $\Bbb R^n$ with initial distribution $B_0$ and hitting time at $\partial U$ being $\tau$. The expectation is taken with respect to the measure $\nu_x$ called the "hitting distribution at $\partial U$" or the "harmonic measure" where for any Borel subset $A \subset \partial U$, $\nu_x(A) = \Bbb P[B_\tau \in A]$, i.e., probability that the Brownian motion escapes $U$ through $A$.
For any $y \in \partial U$ you can try to solve the Dirichlet problem distributionally as well; take $f = \delta_y$ to be the "delta function at $y$". Then $F(x) = f(B_\tau = y|B_0 = x)$ is the probability density that the Brownian motion starting at $x$ escapes $\partial U$ from $y$.
This is the "Green's function" that keeps appearing everywhere in physics
Should have used a different notation for pdf since I was already using $f$ as boundary condition for the Dirichlet problem
I think people use Green's function in a way general context, where if they have some sufficiently nice linear differential operator $\mathcal{L}$, to solve the equation $\mathcal{L} f = g$ they solve for $\mathcal{L} f = \delta_y$ first, the solution to which they call the Green's function and denote by $f(x) = G(x, y)$, and then the general solution to the original thing is obtained by convolving $g$ with this $G$.
I think that is more complicated @KarlKronenfeld. I did it by that method just now and found that using product rule is much simpler provided you know the differentiation of $[x]$ is 0 at non-integer points and undefined at integer points.
You have to remember that as a result as I said earlier.
For integer values of x, differentiation of [x] is undefined while for non-integer values of x, differentiation of [x] is 0.
BTW, can you help me with this:
For four proper fractions $a, b, c, d$ X writes $a+ b + c >3(abc)^{1/3}$. Y also added that $a + b + c> 3(abcd)^{1/3}$. Z says that the above inequalities hold only if a, b,c are positive. (a) Both X and Y are right but not Z. (b) Only Z is right (c) Only X is right (d) Neither of them is absolutely right.
> The view that machines cannot give rise to surprises is due, I believe, to a fallacy to which philosophers and mathematicians are particularly subject. This is the assumption that as soon as a fact is presented to a mind all consequences of that fact spring into the mind simultaneously with it. It is a very useful assumption under many circumstances, but one too easily forgets that it is false.
Does this equality $m^*(I) + m^*(J) - 2\varepsilon = m^*(I\cap J) + m^*(I \cup J) - 2\varepsilon$ only holds when $I\cap J=\emptyset$? or can still hold even when the intersection is nonempty ?
Recall Chebyshev's Inequality:
Let $f$ be a nonnegative measurable function on $E$. Then for any $\lambda > 0$, $$m \{x \in E \mid f(x) \ge \lambda \} \le \frac{1}{\lambda } \int_{E} f$$
Here is a theorem in my book:
Let $f$ be an increasing function on the closed, bounded interval $[...
