Let $B_t$ be a standard Brownian motion. I am interested in the equation
$$\int_0^1 X_s \, ds = B_1 \quad a.s$$
Besides the obvious solution $X_s=B_1$ for all $s\in [0,1]$, how can we come up with solutions ? Can we characterize in a nice way the set of solutions ? Are all solutions of the form $B_1 f(s)$ with $\int_0^1 f(s) \, ds =1$ ? If so, how can we prove it ?
With each solution $X_s$, we associate the value $V_X= \int_0^1 E[X_s^2]\,ds$. Among all solutions, is it true that the one that minimizes $V_X$ is $X_s = B_1 $ and is unique ?
Well, trivially, every solution is of the form $X_s = B_1 + Y_s$ where $\int_0^1 Y_s\,ds =0$ a.s., so this reduces to characterizing processes whose time integral is zero. I can't think of any "nicer" way to characterize the latter set.
In particular, it is certainly not true that every solution is of the form $X_s = B_1 f(s)$; for example take $X_s = B_1 + \sin(2\pi s)$.
It is true that $X_s = B_1$ is the unique minimizer of $V_X$. Note that $V_X$ is just the squared norm in the Hilbert space $L^2(\Omega \times [0,1])$. If $Y_s$ satisfies $\int Y_s\,ds = 0$, then by Fubini's theorem $$\int_0^1 E[B_1 Y_s]\,ds = E\left[B_1 \int_0^1 Y_s\,ds\right] = 0$$ so that $B_1$ is orthogonal to $Y_s$. Hence by the Pythagorean theorem, if $X_s = B_1 + Y_s$ we have $\|X\|^2 = \|B_1\|^2 + \|Y\|^2$, or in your notation, $V_X = 1 + V_Y$. So this is clearly minimized iff $Y=0$.