Some notation:
Let $B(t)$ be a brownian motion, let $\mathcal F^B$ the sigma algebra generated by $\{B(t):a\leq t\leq b\}$.
Let $L_B^2(\Omega)$ be the Hilbert space of square-integrable functions that are measurable with respect to $\mathcal F^B$.
Assume $J_1$ is defined to be the closure of the linear space spanned by constant functions and polynomial chaos of degree $\leq 1$.
Now define $K_1$ in order to satisfy $$J_1=K_1\oplus \mathbb R.$$
Now suppose we have a deterministic function $f\in L^2[a,b]$, $I(f)=\int_a^b f(s) dB(s)$.
I am trying to prove the following assertion.
Then it follows from direct calculations that $I(f)^2-\|f\|_{L^2[a,b]}^2$ is orthogonal to $\mathbb R$ and $K_1$.
To prove that it's orthogonal to $\mathbb R$ lets write $$\langle I(f)^2-\|f\|_{L^2[a,b]}^2, \alpha\rangle=\alpha\cdot\mathbb E\big[I(f)^2-\|f\|_{L^2[a,b]}^2\big] , \alpha\in\mathbb R$$ This equals $0$ since the Wiener integral is an isometry.
Now in order to show that it's orthogonal to $K_1$ I need to show that
$$\int_{\Omega}(I(f)^2-\|f\|_{L^2[a,b]}^2)\cdot g dP =0$$
Since by assumption any $g\in K_1$ is orthogonal to $\mathbb R$, and $\|f\|_{L^2[a,b]}^2\in\mathbb R$ the previous is equivalent to show that
$$\int_{\Omega} I(f)^2\cdot g dP=\mathbb E\big[I(f)^2\cdot g\big]=0$$
Here's where I get stuck, I know that $g$ is some linear combination of Wiener integrals, but I am not able to see how to show the expectations of the product is actually zero.
If for instance $g=I(f)$ then it's easy to show that the expectations equals zero, the same if $g$ is independent from $I(f)$ (I don't think this is possible).
I would be glad to receive some guidance. Thanks!
Since $g$ is in the first homogenous chaos, there is a deterministic $h$ such that $g = I(h)$. This means that from your work it is enough to show that $$\mathbb{E}[I(f)^2 I(h)] = 0$$
This will follow immediately from the Wick Formula for moments of a multivariate Gaussian (see here) as soon as we check that $(I(f),I(f),I(h))$ is a centered Gaussian vector. This is immediate since $$\alpha I(f) + \beta I(h) = I(\alpha f + \beta h)$$ which is centered Gaussian since $\alpha f + \beta h$ is deterministic.