\documentclass [11pt,letterpaper]{article}

\usepackage{amsmath,amsthm,amsfonts,amssymb}
\usepackage{graphics}

\usepackage[left=1in,right=1in,top=1in,bottom=1in]{geometry}

\newcommand\im{\textup{Im\,}}
\newcommand\re{\textup{Re\,}}
\newcommand\prsp{(\Omega, \mathcal F, \mathbf P)}
\newcommand\pr{\mathbf P}
\newcommand{\fej}[1]{\noindent\textsc{#1}}
\newcommand{\FF}{\mathcal{F}}
\def\E{\mathbb E}
\def\R{\mathbb R}
\def\eps{\varepsilon}
\begin{document}
\pagestyle{empty}
\begin{center}
\begin{large}
Math 635 -- Spring 2012\\[10pt]{\textbf{Homework 4\\[15pt]}}
\end{large}
\end{center}


\noindent{\textbf {Due:  March 19, 2012}, beginning of the class. Late homework will \textbf{not} be accepted. }\\



\begin{enumerate}
\item Let $B_t$ be a standard Brownian motion.  In this exercise we will prove that with a probability of one the standard Brownian motion started from 0 has infinitely many zeroes in $[0, \delta]$ for any $\delta > 0$.   Let
\[
	B^{-}_t = \min\{B_s, 0\le s \le t\}.
\]
\begin{enumerate}
	\item Noting that $-B_t$ is also a standard Brownian motion, argue that $-B^{-}_t$ has the same distribution as $B_t^* = \max\{B_s, 0\le s \le t\}.$
	\item Show that $P(B_t^* > 0) = P(B_t^{-} < 0) = 1$ for any $t>0$.
	\item Argue that for any $n$, 
	\[
	P(B_t \text{ has no zeros in } (0,1/n)) =0.
	\]
	\item Prove that with probability one,  for any $\delta > 0$ Brownian motion started from zero has infinitely many zeros in $[0,\delta]$.
\end{enumerate}




\item  Suppose that $g$ is globally Lipschitz on $[0,1]$ (Holder continuous with $\alpha = 1$).  Suppose that $H:C[0,1] \to \R$ is continuous.  Then the function $H\circ g: C[0,1] \to \R$, defined via
\[
	(H\circ g)(f) = H(g\circ f)
\] is continuous.

\item Prove that for $f \in \mathcal H^2[0,T]$
\[
	\E I(f) = \E \int_0^T f(s) dB_s = 0.
\]
Hint: show that the result holds for $f_n \in \mathcal H^2_0$, and then make the correct arguments.
\end{enumerate}




\end{document}
