Pathwise stochastic integral as a linear operator on continuous functions

Question

Let $B$ be a Brownian motion. Definining a pathwise stochastic integral $I(f):=\int f~dB$ for certain classes of deterministic functions is straightforward: For instance if $f=\sum_ic_i1\{[t_i,t_{i+1})\}$ is a compactly supported simple function, then $$I(f):=\sum_ic_i(B(t_{i+1})-B(t_i)).$$ More generally if $AC_0$ denotes the set of compactly supported absolutely continuous functions, continuity of $B$ allows to define the stochastic integral via integration by parts $$I(f):=-\int f'(x)B(x)~dx.\tag{1}$$

Two properties of this "stochastic integral operator" on $AC_0$ that are of interest to me are:

1. The operator defines a pathwise stochastic integral, i.e., for almost every outcome $\omega\in\Omega$ in our probability space, the brownian path $B_\omega\in C[0,\infty)$ yields a linear functional $I_\omega:AC_0\to\mathbb R$, as $(1)$ satisfies $$I_\omega(af+bg)=aI_\omega(f)+bI_\omega(g).$$
2. The random functional $f\mapsto I(f)$ is a centered Gaussian process indexed by $AC_0$ with covariance given by the $L^2$ inner product.

Question. Is there a way to extend this operator to deterministic functions with much less regularity (namely, continuous with compact support) while preserving the above two properties simultaneously?

Using Itô's theory, we can of course define $I(f)\in L^2(\Omega)$ for every $f\in L^2[0,\infty)$, but my understanding is that this construction fails the above question. That is, for any fixed $f,g\in L^2$, we have linearity in the sense that $$I(af+bg)=aI(f)+bI(g)\tag{2}$$ as elements of $L^2(\Omega)$, but this is a weaker linearity: the probability-one event on which $(2)$ holds depends on $a,b,f,g$, and thus from this alone we can't define $I$ as an almost sure linear operator $I_\omega:L^2[0,\infty)\to\mathbb R$.

Answer 1

You can do the same for less regular functions, for example for functions with finite $p$-variation for some fixed $p \in [1,2)$ or functions in $\mathcal{C}^\alpha$ for some fixed $\alpha > 1/2$. You cannot go much below that in the sense that there exists no Banach space $\mathcal{B}$ with the following properties:

1. Trigonometric polynomials are dense in $\mathcal{B}$.
2. For almost every $\omega$, $f \mapsto I_\omega(f)$ extends to a continuous linear map on $\mathcal{B}$.
3. $\mathcal{B}$ is large enough so that Brownian paths belong to $\mathcal{B}$ almost surely.

See for example this paper by Terry Lyons. The argument is pretty straightforward: basically one constructs two Brownian paths that are correlated in such a way that the integral of one against the other diverges and derives a contradiction from there.