Coupling of Wiener processes Is it possible to find a coupling of two Wiener processes $W^0, W^x$ (i.e. two Wiener processes defined on a common probability space). One starting from $0$ and the other from $x$ such that 
$W_t^0 - W_t^x \rightarrow_t 0$ almost surely and in $L^1$.
Using some random-walks considerations I suspect that it is not possible to have convergence in $L^1$ but I do not know how to prove it.

The answer is: $W_t^0 - W_t^x$ cannot converge to $0$ in $L^1$
The proof is as follow (it is a slightly extended version of the proof by smalldeviations). Obviously
$|W_t^0 - W_t^x|_1\geq  \inf _{\gamma\in \Gamma } \{\int _{\mathbf{R}^d\times \mathbf{R}^d} |x-y| \text{d}\gamma(x,y)\},$
where $\Gamma$ is the set of all couplings of $\mathcal{L} (W_t^0)$ and $\mathcal{L} (W_t^x)$ ($\mathcal{L}$ is the law of given variable). By the duality formula (see http://en.wikipedia.org/wiki/Transportation_theory) the right hand side is equal to
$\sup \{ \int_{\mathbf{R}^d} \phi (x)  \mathrm{d} \mu (x) + \int_{\mathbf{R}^d} \psi (y)  \mathrm{d} \nu (y) \},$
where the supremum runs over all pairs of bounded and continuous functions such that $\phi(x)+\psi(y)\leq |x-y|$ and $\mu =\mathcal{L}(W_t^x)$ and $\nu = \mathcal{L} (W_t^0)$. In our case it is sufficent to take $\phi(x) = x, \psi(y)=y$. Then the expression under the sup is equal to:
$\mathbf{E}(W^x_t) - \mathbf{E}(W^0_t) = x-0 =x$. Therefore $|W_t^0 - W_t^x|_1\geq x$.
 A: The a.s.-convergence is easy to achieve.For example:
Let W0 be a standard Wiener process on any probability space. Define X(t)=x-W0(t). This is a Wiener process emitted from x. Define τ to be the first time when W0 hits x/2. This is an a.s.-finite stopping time and W0(τ)=X(τ)=x/2. Let Wx coincide with X up to τ and coincide with W0 after τ.
The L1 convergence is impossible since it would have implied that E Wx(t)- E W0(t)  converges to 0.
A: If there is no coupling s.t. the distance goes to 0 in $L^{1}$ (which I agree seems likely), you might want to look up an introduction to the Wasserstein-1 distance (which is exactly expected $L^{1}$ distance after an optimal coupling). This is the language that I've seen this type of problem most often discussed in. The field of finding optimal couplings for given metrics is 'optimal transport'.
I vaguely recall that there are theorems about optimal couplings (in only certain $L^{p}$ only, of course!) never giving rise to crossing lines. In a 1-dimensional problem, such as the one you have, this would tell you what the optimal coupling is explicitly (in this case, if you construct your Brownian motion via Donsker's theorem, it says: whenever $W^{0}$ takes a move in the $\alpha$ percentile, make $W^{x}$ also move in the $\alpha$ percentile... in other words, my probably-misremembered theorem would imply coupling doesn't help your $L^{1}$ distance at all in this case).
Cedric Villani has two excellent books on the subject, at least one of which was available for free download the last time I checked, and you should be able to find 'this sort' of theorem. Please don't take my word for the statements.
Edit: Here is (I believe) a proof... though it might fit in the category of so-simple-its-wrong. First of all, we have starting points x,0 and add two normal (0,a) random variables X and Y to them. Plugging in the obvious cost functions, the "Kantorovitch Duality" formula tells us that the $L^{1}$ distance between x+X and 0+Y is at least x (while plugging in the independent coupling to the standard way of writing this metric tells us it is at most x). So, at time a, the $L^{1}$ distance between the brownian motions must be at least x (since at time a they have the same distribution as x+X and 0+Y, and we have found this lower bound for ALL couplings, and in particular all couplings that come from them both being brownian motions). In particular, the $L^{1}$ distance can't go to 0.
A: Here is another solution to this problem: $W^0$ and $W^x$ are both martingales and therefore so is their difference $W^0-W^x$. Then $|W^0-W^x|$ is a submartingale ($|\cdot|$ is convex). It follows that $E[|W^0_t - W^x_t|] \ge E[|W^0_0 -W^x_0|] =|x|$ for all $t$. 
A: I do not know how to prove it either, but I would be quite surprised if this was possible. The heuristic is that the fastest coupling of two Brownian motions (in the sense that the coupling inequality is in this case an equality) is the mirror coupling (actually their are other ones, but with a slightly more stringent definition of coupling -Markovian coupling- the mirror coupling is the fastest one), and this coupling does not converge to $0$ in $L^1$. This is a good article about coupling of Brownian motions.
A: this is indeed a very nice proof. I just wanted to mention that there is no need of Kantorovich duality here (which is a highly non-trivial result) since you merely use the fact that:
$$\inf \iint |x-y| \, \gamma(dx,dy) \geq \int \phi(x) \, \mu(dx) + \int \psi(y) \, \nu(dy)$$
for any functions $\phi(x)+\psi(y) \leq |x-y|$, which is obvious.
