I am learning the quadratic variation of stochastic process, and I am working on an exercise stating that 



> 
> If for all $t$, we have $$0=A\_{0}(t)+A\_{1}(t)W(t),$$ where $(A\_{0}(t),\mathcal{F}\_{t})$ and $(A\_{1}(t),\mathcal{F}\_{t})$ are processes with $C^{1}$ trajectories, and $W\_{t}$ is a brownian motion with respect to $(\mathcal{F}\_{t})$ then $A\_{0}(t)=0$ and $A\_{1}(t)=0$ for all $t$.
> 
> 
> 


My current attempt:


We have $-A\_{0}(t)=A\_{1}(t)W(t)$, and set $g(t)=-A\_{0}(t)$. Since $g(t)$ is $C^{1}$, it is of bounded variation, and thus it has zero quadratic variation.


Hence, $A\_{1}(t)W(t)$ is of zero quadratic variation. I want to start from here to show that $A\_{1}(t)W(t)$ is zero for all $t$, but I don't know what to do..


Please help! Thank you!



Show that if $A_{0}(t)+A_{1}(t)W(t)=0$ for all $t$ with $A_{0}$ and $A_{1}$ differentiable in $t$ and $W(t)$ a Wiener process, then $A_{0}=A_{1}=0$
$\newcommand\De{\Delta}$ $\newcommand\ep{\epsilon}$ $\newcommand\om{\omega}$ $\newcommand\Om{\Omega}$
Take any real $a$ and $b$ such that $a<b$. For each natural $n$ and all $k=0,\dots,2^n$, let $t\_k:=t\_{k;n:a,b}:=a+k(b-a)/2^n$. Then 
$$\sum\_{k=1}^{2^n}(\De\_kW)^2\to b-a \tag{1}$$
(as $n\to\infty$) for almost all paths/realizations of $W$, where $\De\_k B:=B(t\_k)-B(t\_{k-1})$ for any, possibly random function $B$; see e.g. [Theorem 2, formula (5)](https://ocw.mit.edu/courses/sloan-school-of-management/15-070j-advanced-stochastic-processes-fall-2013/lecture-notes/MIT15_070JF13_Lec8.pdf). Without loss of generality, (1) holds for all paths/realizations of $W$. So, in what follows we may assume that the functions $W,A\_0,A\_1$ are non-random. 


Suppose now that the function $A\_1$ is nonzero. Since $A\_1$ is $C^1$, it is continuous, and so, we may assume that that our $a$ and $b$ are such that $|A\_1|\ge\ep$ on $[a,b]$ for some real $\ep>0$. From the condition $A\_0+A\_1W=0$ it follows that 
$$-\De\_k A\_0=A\_1(t\_k)\De\_k W+W(t\_{k-1})\De\_k A\_1,$$
whence
$$\ep^2(\De\_k W)^2\le A\_1(t\_k)^2(\De\_k W)^2
\le2(\De\_k A\_0)^2+2W(t\_{k-1})^2(\De\_k A\_1)^2.$$
Since $A\_0,A\_1$ are in $C^1$ and $W$ is continuous, we have 
$$(\De\_k A\_0)^2+(\De\_k A\_1)^2\le M(\De t\_k)^2\quad\text{and}\quad W(t\_{k-1})^2\le M$$
for some real $M>0$ and all $k=0,\dots,n$, where $\De t\_k:=t\_k-t\_{k-1}=(b-a)/2^n$. So,
$$\ep^2\sum\_{k=1}^{2^n}(\De\_k W)^2
\le2M\sum\_{k=1}^{2^n}(\De t\_k)^2+2M^2\sum\_{k=1}^{2^n}(\De t\_k)^2.$$
Letting now $n\to\infty$, we get $\ep^2(b-a)\le0$, which is a contradiction. 


Thus, $A\_1=0$ and hence $A\_0=-A\_1W=0$. 


