I am trying to understand (rigorously) why the existence of the Brownian local time implies that the sample paths $t \to W_t$ cannot be differentiable $\mathbb{P}$-a.s. anywhere on $(0, \infty)$. I suspect that the local time must be zero at a (supposed) differentiable point, thus leading to a contradiction, but I struggle to prove this.
Any hints would be appreciated.
Fix $\omega \in \Omega^* \subset \Omega$ such that the Brownian local time $(t, x) \mapsto L_t(x, \omega)$ is continuous in its two arguments. Suppose $W(\omega)$ is differentiable in $t_0 \in (0, \infty)$. Set $x := W_{t_0}(\omega)$. The differentiability at $t_0$ tells us that \begin{equation*} \lim_{t \rightarrow t_0} \frac{W_t(\omega) - x}{t - t_0} \end{equation*} exists in $\mathbb{R}$ and, in particular, that this quotient is bounded in a neighborhood of $t_0$, that is, there exists a $C > 0$ and a $\delta > 0$ such that for all $t \in (t_0 - \delta, t_0 + \delta)$ \begin{equation*} \left| W_t(\omega) - x \right| \leq C |t-t_0|. \end{equation*} Therefore \begin{equation*} \lambda^1\left( \left\{ t \in (t_0 - \delta, t_0 + \delta) \, | \, \left| W_t(\omega) - x \right| \leq C |t-t_0| \right\} \right) = \lambda^1\left( (t_0 - \delta, t_0 + \delta) \right) = 2 \delta, \end{equation*} but, using the definition of the occupation time $\Gamma_t$, the fact that $L_t$ is a Lebesgue-density for $\Gamma_t$, and the continuity of the local time, we obtain \begin{align*} 1 &= \frac{\lambda^1\left( \left\{ t \in (t_0 - \delta, t_0 + \delta) \, | \, \left| W_t(\omega) - x \right| \leq C |t-t_0| \right\} \right)}{2 \delta} \\[0.25cm] &\leq \frac{\lambda^1\left( \left\{ t \in (t_0 - \delta, t_0 + \delta) \, | \, \left| W_t(\omega) - x \right| \leq C \delta \right\} \right)}{2 \delta} \\[0.25cm] &\leq \frac{\lambda^1\left( \left\{ t \in [0, t_0 + \delta] \, | \, \left| W_t(\omega) - x \right| \leq C \delta \right\} \right) - \lambda^1\left( \left\{ t \in [0, t_0 - \delta] \, | \, \left| W_t(\omega) - x \right| \leq C \delta \right\} \right)}{2 \delta} \\[0.25cm] &= \frac{\Gamma_{t_0+\delta}(B_{C\delta}(x), \omega) - \Gamma_{t_0-\delta}(B_{C\delta}(x), \omega)}{2 \delta} \\[0.25cm] &= \frac{1}{2 \delta} \int_{B_{C\delta}(x)} L_{t_0+\delta}(y, \omega) - L_{t_0-\delta}(y, \omega) \, dy \rightarrow 0, \quad \text{as } \delta \downarrow 0, \end{align*} a contradiction.
