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

Question

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.

Answer 1

I doubt that this identity holds true. Just consider $f:=1$, then the assertion reads $$B_t = 0,$$ which is obviously not correct.

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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 $f \in C^1(\mathbb{R})$.