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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.

I found this little gem in Rogers and Williams.

The optional stopping theorem (OST) gives that if $X$ is a martingale and $T$ is a bounded stopping time, then $\mathbb{E}[X_T]=\mathbb{E}[X_0]$.

Now take a Brownian motion $B=(B_t:t \geq 0)$, which is a martingale, and the stopping time $T=\inf\{t>0:B_t=1\}$. Obviously $B_T=1$ and OST says that $1=\mathbb{E}[B_T]=\mathbb{E}[B_0]=0$.

So what went wrong? Well quite simply, I did not check that $T$ was actually bounded. Why is it not bounded you say? No? Ok, well I’m going to tell you anyway; it is not bounded because there are many paths that trail below 0 which never hit 1. To properly show this, one must first calculate the supremum of the Brownian motion, then work out the probability that it is less than 1.

Rather nice reminder to always check that the stopping times are bounded before applying the OST. This was provided in a lecture by Nathanael Berestycki.