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This post will be essentially about functions of bounded variation of one variable. The main source is the book “Functions of Bounded variation and Free Discontinuity Problems” by Ambrosio, Fusco and Pallara. Before we give the definition of a bounded variation function let us recall what exactly does is mean for a function to belong in $W^{1,1}(0,1)$. Recall that any function $f\in L^{1}(0,1)$ can be seen as a distribution $T_{f}$ i.e.  as a bounded linear functional on $C_{c}^{\infty}(0,1)$$T_{f}:C_{c}^{\infty}(0,1)\to\mathbb{R}$  with

$T_{f}(\phi)=\int_{0}^{1}f(x)\phi(x)dx,\quad \forall \phi\in C_{c}^{\infty}(0,1).$

In that case we say that the distribution $T_{f}$ is representable by the function $f$. Given any distribution $T$ we can define its distributional derivative  $DT$  to be the distribution defined as

$DT(\phi)=-D(\phi'),\quad \forall \phi\in C_{c}^{\infty}(0,1)$.

In the special case where the distribution $T$ can be represented by a function in the way we show above the distributional derivative $DT_{f}$ will be

$DT_{f}(\phi)=-\int_{0}^{1}f(x)\phi'(x)dx,\quad \forall \phi\in C_{c}^{\infty}(0,1).$

There is a remarkably nice proof of the Lebesgue decomposition theorem (described below) by von Neumann. This leads immediately to the Radon-Nikodym theorem.

Theorem:

If $\mu$ and $\nu$ are two finite measures on $(\Omega,\mathcal{F})$ then there exists a non-negative (w.r.t. both measures) measurable function $f$ and a $\mu$-null set $B$ such that

$\nu(A)=\int_A f \, d\mu+ \nu(A \cap B)$

for each $A \in \mathcal{F}$.

Proof:

Let $\pi:=\mu+\nu$ and consider the operator

$T(f):=\int f\, d\nu$. Read the rest of this entry »

The fundamental theorem of algebra states that $\mathbb{C}$ is algebraically closed, that is;

Theorem:

For any non-constant polynomial $p$ in $\mathbb{C}$, there exists a $z\in \mathbb{C}$ such that $p(z)=0$.

Proof:

Let $B=(B_t: t \geq 0)$ be a Brownian motion on $\mathbb{C}$ and suppose for a contradiction that a non-constant polynomial $p$ does not have any zero’s. Let $f:=1/p$, then $f$ is analytic and tends to 0 at infinity. Pick such that $\alpha < \beta$ and note that $\{Re f \leq \alpha\}$ and $\{Re f \geq \beta\}$ contain an open set, which can be done due to the fact that $f$ is continuous and non-constant.

Now $f(B_t)$ is a continuous local martingale (by using Ito’s formula) and moreover it is bounded. Hence by the Martingale convergence we have that $f(B_t) \rightarrow f(B)_\infty$ a.s. and in $L^1$.

This last statement is contradicted by the fact that Brownian motion is recurrent on the complex plane, in particular, it visits $\{Re f \leq \alpha\}$ and $\{Re f \geq \beta\}$ infinitely many times which gives that

$\lim\inf f(B_t) \leq \alpha < \beta \leq \lim \sup f(B_t)$ a.s.