Is there an explicit example of a function in the (topological) support of the law of Brownian motion (with respect to the topology of uniform convergence of continuous functions)?


(You can take "explicit" to mean "doesn't invoke the axiom of choice".)



Deterministic function in the support of Brownian motion
As clarified, you're asking about the law of Brownian motion on a bounded interval $[0,T]$, as a probability measure $\mu$ on $C([0,T])$ (Wiener measure).


The zero function is in the topological support. This amounts to showing that $P(\sup\_{t \in [0,T]} |B\_t| < \epsilon) > 0$ for all $\epsilon$ and you can find several proofs of that statement in [this Math.SE question](https://math.stackexchange.com/questions/38642/showing-that-brownian-motion-is-bounded-with-non-zero-probability?rq=1).


In fact, the topological support is equal to the closed hyperplane $E\_0$ consisting of all functions which vanish at time 0. This follows from the previous assertion and the Cameron–Martin theorem, which asserts that for all $h \in H = H^1\_0((0,T])$ (absolutely continuous paths vanishing at time 0 and with derivative in $L^2$), the translated measure $\mu\_h(A) = \mu(A-h)$ and $\mu$ are mutually absolutely continuous. Since we have just shown $\mu(U) > 0$ for every open neighborhood $U$ of $0$, Cameron–Martin now implies $\mu(U-h) > 0$, which is to say that every open neighborhood of $h$ has positive measure, i.e. $h$ is in the topological support. Since $H$ is dense in $E\_0$, we have that the topological support contains $E\_0$. The reverse inclusion is clear because $E\_0$ is closed and has full measure.


(Keep in mind that being in the topological support doesn't really say "Brownian motion could trace out this path", it says "Brownian motion could trace out paths arbitrarily close to this one".)


