# Brownian motion, Feynman integral-1

Brownian motion first appeared in biology. Later Einstein, Smoluchowsk and other people used heat equation to derive some equations for Brownian motions. They viewed Brownian motion as some diffusion process. Later Wienner gave a rigorous foundation of Brownian motions, he constructed the Wienner measure on the space of the paths of the Brownian motions(in general situation is the space $\Gamma=C_{(0)}([0,T],\mathbb{R}^d)$, consisting of continuous functions with value $0$ at time $t=0$). In economics, some economists proposed some models which turned out to be Brownian motion, too.

The story did not stop here. Another aspect came from quantum mechanics. Feynman developed his own formulation of quantum mechanics using the Feynman integral. Later Kac found that removing the factor $\sqrt{-1}$ in the Feynman integral gives a solution to the heat equations.

So Feynman integrals, Brownian motions are related by the heat equations, or diffusion processes.

In this post I will try to give several formulations of the Brownian motions from different aspects.

The first approche relies on Gaussian measures on Hilbert spaces. So, we need first say something on this.

On the real line $\mathbb{R}$, we have the standard Lebesgue measure $dx$ with $dx([0,1])=1$. Then we define the Gaussian measure to be $d\mu=\frac{1}{\sqrt{2\pi\sigma^2}}e^{-\frac{|x-\alpha|^2}{2\sigma^2}}dx$ where $\alpha, \sigma$ are two constants with $\alpha\in\mathbb{R},\sigma\in\mathbb{R}_{>0}$.

For superior dimensions, we say that a probability measure $d\mu$ on $\mathbb{R}^d$ is Gaussian if there is a linear map $h:\mathbb{R}^d\rightarrow \mathbb{R}$ such that the induced measure on $\mathbb{R}$, $d\mu\circ h^{-1}$ is Gaussian. We can show that the Fourier transform $\mu(y)=\int_{\mathbb{R}^d}e^{i}d\mu(x)$ of a Gaussian measure $d\mu$ is $\hat{\mu}(y)=e^{i-\frac{1}{2}}$ where $\alpha$ is a vector in $\mathbb{R}^d$ and $Q$ is a positive operator on $\mathbb{R}^d$. In fact, there is a one-to-one correspondence between the set of Gaussian measures and the set of the pairs $(\alpha,Q)$, so we can write $d\mu=N(\alpha,Q)$.

We can continue this process to any dimension, even infinite dimension. That is to say, if $(H,<,>)$ is a real Hilbert space, then a probability measure $d\mu$ on $H$ is said to be Gaussian if there is a continuous linear map, $h:H\rightarrow \mathbb{R}$ such that $d\mu\circ h^{-1}$ is a Gaussian measure on the real line. Just like the above statements, we can show that there exists a mean vector $m_{\mu}$ and a covariance operator $Q_{\mu}$(a vector $m\in H$ is a mean vector if $=\int_Hd\mu(z)$ for all $x\in H$, assuming that given $x$, the function $H\rightarrow \mathbb{R},z\mapsto $ is integrable. And a covariance operator is a symmetric, positive definite operator $Q:H\rightarrow H$ such that $=\int_Hd\mu(z)$, assuming again some integrability). We can show that $Q$ is a nuclear operator, which, roughly speaking, is so operator for which the trace can be defined and is finite, and is independent of the choice of basis. Conversely, we can show that for any such pair $(m,Q)$ with $Q$ a nuclear opeartor, there is a unique probability measure on $H$ which is Gaussian and $m_{\mu}=m,Q_{\mu}=Q$. In the same way, we write $d\mu=N(m,Q)$.

To define Brownian motion, we set $H=L^2([0,\infty))$(equipped with the usual Lebesgue measure) and $\mu=N(0,Q)$ where $Q$ is a symmetric, positive definite nuclear operator on $H$. Thus we can define the square root of $Q$, which we choose to be positive, too($\sqrt{Q}>0$). For any $t\geq0$, we write $z=\chi([0,t])$ the characteristic function of the interval, $[0,t]$. We can show that $\sqrt{Q}(H)$ is a dense subspace of $H$(what is more, we have that, $d\mu(\sqrt{Q}(H))=0$, this is a remarkable fact). So, there is a sequence $(z_n)\in (\sqrt{Q}(H))^{\mathbb{N}}$ such that $z_n\rightarrow z$. Then we define $B_t(x)=\lim_n$.

Then I say that $B_t$ is a Brownian motion. This means that $B_t$ is a random variable on $H$, and for $s, $B_t-B_s,B_v-B_u$ are independent and $\int_H|B_t(x)-B_s(x)|^2d\mu(x)=|t-s|$. These identities are not difficult to prove.

In the next post, I will try to establish Brownian motion from the point of view of diffusion process.