Paley-Wiener type integral: Conditions for the integrand $f$

Question

Let $(B_t)_{t \geq 0}$ be a Brownian Motion with $B_0=0$. Then define the stochastic integral for $f \in C^1([0,1], \mathbb{R})$ ($f$ does not have to be in $BV$) via an integration by parts formula, involving a "normal" Riemann-type integral:
$$ \int_0^1 f(s)dB_s=f(1)B_1-\int_0^1f'(s)B_sds $$ For this type of stochastic integral, there is supposed to hold the isometry: $$ \mathbb{E}\ [\big(\int_0^1 f(s)dB_s)^2 ]=\int_0^1 f(s)^2ds $$ Using lenghty calculations, involving the identites $\mathbb{E}[B_r B_s]=\min(r,s)$, Fubinis Theorem, splitting up integrals, finding a closed "form" for $\min(s,r)$ under the integral as well as $\frac{d}{ds}f(s)^2=2f'(s)f(s)$ and the integration by parts formula $\int_a^b f'(s)s=f(s)s|_a^b-\int_a^b f(s)$, I could not get the desired result. Instead, I got stuck with terms involving $f(1)$, not involving a square (the squares could have canceled out with $\mathbb{E}[B_1^2f(1)^2]=f(1)^2$).
My question is: Is there any type of (boundary) conditions required for the functions that can be integrated via this stochastic integral? A condition like e.g. $f(1)=0$ would be the simplest way to provide a solution. I am aware that my condition mentioned above does not seem to be a necessary condition, as the constant case $f(s)=c \neq 0$ shows.
Edit: For clarification, the leftover term had the form $f(1)\int_0^1 f′(s)sds$, see the answer by zhoraster for more details.

Answer 1

Using lenghty calculations ... I could not get the desired result

Let us try: $$ \mathbb{E}\left[\left(f(1)B_1 -\int_0^1 f'(s) B_s ds\right)^2 \right] = f(1)^2 - 2f(1)\int_0^1 f'(s) s\,ds\\ + 2\int_0^1 \int_0^1 f'(s)f'(u) \min(s,u)ds\,du \\ = f(1)^2 - 2f(1) \int_0^1 f'(s) s\,ds + 2\int_0^1 f'(s) s \int_s^1 f'(u) du \, ds \\ = f(1)^2 - 2f(1) \int_0^1 f'(s) s\,ds + 2\int_0^1 f'(s) s\big(f(1)-f(s)\big)ds \\ = f(s)^2s\bigg|_0^1 - \int_0^1 2f'(s)f(s) s\,ds = \int_0^1f(s)^2 ds. $$ What am I doing wrong?