I'm trying to follow the following statement on pg 325 Durrett's book on probability where he states that the following holds by symmetry:
$$E_{y}(B_{u})=y $$
The subscript $y$ implies that we are assuming that almost surely our base point is $y$, that is $\omega(0)=y$. In which case, I think of $B_{u}$ as sending $\omega \mapsto \omega(u)$, why is the best guess for this $y$?
Under the measure $P_y$ the process $(B_u -y)_{u\geq 0}$ is a Brownian motion starting from $0$. Now the symmetry: The process $(-(B_u -y))_{u\geq 0} = (-B_u + y)_{u\geq 0}$ is also a Brownian motion starting from $0$. But this is overkill.
$B_u -y$ has centered normal distribution with variance $u$, thus this distribution is symmetric, this means $B_u -y$ has the same distribution as $-(B_u -y)$.
For a symmetric random variable $Z$ we have $E(Z) =E(-Z) = -E(Z)$, hence $E(Z) = 0$.
This means in this case $E_y( B_u -y ) = 0$, which is your statement.