 Solution

Let $W$ be a Brownian motion, set $X=W$, and defineThis is a continuous local martingale with quadratic variation $t$, hence is Brownian by the Lévy characterization of Brownian motion. Since ${\mathrm{sign}}^2=1$,which gives a weak solution.

Suppose a strong solution existed. It has quadratic variation $t$, so $X$ itself is Brownian. The supplied Tanaka formula gives $|X_t|=B_t+L_t$, and $L$ is adapted to the completed filtration of $|X|$. Hence $B$ and $|X|$ generate the same completed filtration. Strongness would make $X$, and therefore $\mathrm{sign}(X_t)$, measurable with respect to the history of $|X|$. But a Brownian excursion has an independent symmetric sign; in particular, conditionally on the reflected Brownian path, the sign at a fixed nonzero time is not measurable. This contradiction proves that no strong solution exists.