Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers.
To wit, for $\varepsilon>0$ small (to be determined later) set $X = \varepsilon + Z$ and $Y = -Z$. As $Z$ and $_Z$ have the same law, $X$ dominates $Y$ for every strictly increasing utility function. However, adding now $Z$ on both sides leaves us with $\tilde{X} = 2Z + \varepsilon$ and $\tilde{Y} = 0$ and we have that $\tilde{Y}$ dominates $\tilde{X}$ for each strictly increasing, strictly concave utility function as long as $\varepsilon$ was chosen small enough. [In particular your claim for exponential $U$ seems wrong.]
This clarifies some things, but I still do not understand how you are going from the economic to the mathematic formulation. It seems that the economic formulation depends heavily on the joint distribution of the random variables $X$ and $Y$ and the standard normal perturbation (let's call it $Z$). It seems to be that the desired relation is never true for strictly concave utility functions, save you make some additional assumptions such as independence of the perturbation of $X$ and $Y$. The beauty of utility theory is that it is law invariant - adding a perturbation destroys this.
Your question utterly confuses me. 1) Do you consider the initial wealth $X$ and $Y$ to be random variables or deterministic. You are also switching from lower case to capital letters which makes it even less clear. 2) Do you want to perturb them with the same standard normal random variable or with random variables with the same (standard normal) distribution? 3) Are you asking about linear transformation (as in the title) or for monotonicity under transformation (as in the first paragraph)?
For the case $a=1$ there are infinitely many solutions, at least as long as you do not require any continuity properties. The easiest one $F(-3)=F(1)=1$ and $F(x) = c$ for all other $x \in \mathbb{R}$ for an arbitrary constant $c$.
Finally, the example from (2) is not only a local martingale, but a true martingale which can be checked easily just using the definition of a martingale and the fact the Brownian motion is a martingale. What you think to remember about the quadratic variation being always absolutely continuous is most likely true, but I assume you forgot to remember that somewhere above was a qualification that one assumes that the filtration is Brownian (which is done quite often).
I never spoke about (local) martingales with absolutely continuous paths. They would be quite boring, as the only ones are the constants. I spoke only about absolutely continuous quadratic variation processes.
Indeed $a$ is just doing some reallocation, but for that only $\int_0^T a(s) \, ds$ matters. However this might not be chosen arbitrarily, but might depend on the constraints you have. To understand the problem better, could you provide some economic motivation? In particular, are $X_1$ and $X_2$ to be assumed always nonnegative? And more general it would be interesting to understand why you have two different utility functions for the different processes which are nevertheless additive.
Localization is a standard practice written down in nearly every stochastic analysis textbook. As limits in the definition of stochastic integrals are taken in probability, the generic setting for Ito's formula is neither $L^2$ nor $L^1$ but $L^0$.
