# Can we show $\int_0^tf(s){\rm d}B_s=-\int_0^tf'(s)B_s{\rm d}s$ for $f\in C^1(\mathbb R)$?

Let $(\Omega,\mathcal A,\operatorname P)$ be a probability space, $(B_t)_{t\ge 0}$ be a Brownian motion on $(\Omega,\mathcal A,\operatorname P)$ and $f\in C^1(\mathbb R_{\ge 0})$. Can we show that $$\int_0^tf(s){\rm d}B_s=-\int_0^tf'(s)B_s{\rm d}s\;\;\;\operatorname P\text{-almost surely}\tag 1$$ for all $t\ge 0$? I've tried to obtain $(1)$ by the Itō formula in the following way: Let $$g(t,x):=f(t)x\;\;\;\text{for }t\ge 0\text{ and }x\in\mathbb R\;.$$ Then, $g\in C^{1,\:2}(\mathbb R_{\ge 0}\times\mathbb R)$ with $$\frac{\partial g}{\partial t}(t,x)=f'(t)x\;\;\;\text{and}\;\;\;\frac{\partial g}{\partial x}(t,x)=f(t)\tag 2$$ for all $t\ge 0$ and $x\in\mathbb R$. Thus, $$f(t)B_t=\int_0^tf(s)\;{\rm d}B_s+\int_0^tf'(s)B_s\;{\rm d}s\;\;\;\operatorname P\text{-almost surely}\tag 3$$ for all $t\ge 0$. Since we would need $f(t)B_t=0$ (i.e. $f(t)=0$) to obtain $(1)$ and since this needs to be the case for any $t\ge 0$, I start to think that $(1)$ is wrong. However, since Did wrote that $(1)$ is trivial for $f\in C^2(\mathbb R_{\ge 0})$, I probably made a mistake.

• In my opinion the answer is no because Brownian Motion is almost everywhere not differentiable. But I'm no expert on the subject. – JonesY Jul 24 '16 at 18:52
• Is your beef with the fact that $f$ is only $C^1$ and not $C^2$? This should be explained in the question... – Did Jul 24 '16 at 18:55
• Why don't you address my specific request? Note that when $f$ is $C^2$, the identity is direct. – Did Jul 25 '16 at 10:07
• Yeah it is: Itô. (As always...) – Did Jul 25 '16 at 10:13
• @Did I've updated the question and deleted my obsolete comments. – 0xbadf00d Jul 25 '16 at 13:23

I doubt that this identity holds true. Just consider $f:=1$, then the assertion reads $$B_t = 0,$$ which is obviously not correct.
As @Did pointed out in a comment, Itô's formula shows that the identity $$\int_0^t f(s) \, dB_s = f(t) B_t - \int_0^t f'(s) B_s \, ds$$ holds for $f \in C^1(\mathbb{R})$.
• I don't understand your answer. Why do you think we need $f\in C^2$? $(3)$ from the question holds for any $f\in C^1$ by the Itō formula. – 0xbadf00d Jul 26 '16 at 9:57
• @0xbadf00d So, $f$ $C^1$ here yields $g:(t,x)\to f(t)x$ of class $C^{1,2}$ in your book, right? No mystery remaining, it seems. – Did Jul 26 '16 at 12:44
• @0xbadf00d As Did pointed out, there are several versions of Itô's formula... and if you assume $f \in C^2$, you are, somehow, on the safe side. Anyway, you are right, it works fine for $f \in C^1$. – saz Jul 26 '16 at 13:32