Lévy's characterization of Brownian motion: right-continuous processes Let $(X_u)_{u}$ be right-continuous martingale such that $X_0=0$ such that $(X^2_u-u)_u,(X_u^3-3uX_u)_u,(X_u^4-6uX_u^2+3u^2)_u$ are martingales (with respect to the canonical filtration).
Prove that $X_u$ has continuous paths and deduce that $(X_u)_u$ is a Brownian motion.
Supposing that we proved that $X_u$ has continuous paths then the Brownian motion property could be deduced from Lévy's characterization of BM.
How to prove the path-continuity of $X_u$? (It doesn't seem trivial).
 A: Cool question.
Proposition
Let $(X_u)_{u}$ be right-continuous martingale with $X_0=0$, such that $(X^2_u-u)_u,(X_u^3-3uX_u)_u,(X_u^4-6uX_u^2+3u^2)_u$ are martingales. Then for every integer $M \ge 1$, the path $(X_u)_{u}$ is a.s. continuous in $[0,M]$.
Proof: Let $\{\mathcal F_t\}$ be the canonical filtration of  $\{X_t\}$. Fix $u \ge 0$ and write
$E^u[\,\cdot\,]:=E[\,\cdot \,| \mathcal F_u]$. Denote $\mathcal G_t:=\mathcal F_{u+t}$  for $t \ge 0$. Observe that the process $\{Y_t\}_{t \ge 0}$ defined by $$Y_t:=X_{u+t}-X_u \tag{1}$$ is a
$\{\mathcal G_t\}$-martingale.
Claim: For any bounded $\{\mathcal G_t\}$-stopping time $\tau$, we have
$$E^u[Y_\tau^4]=3E^u[\tau^2]\,.$$
The claim is proved below. Now we will use it to complete the proof of the proposition.
Fix $\epsilon>0$  and $\delta>0$. Let $P^u[\,\cdot\,]:=P[\,\cdot \, | \mathcal F_u]$. Applying the claim to
$$\tau:=\delta \wedge \min\{t\ge 0: |Y_t| \ge \epsilon\}$$
gives
$$P^u[|Y_\tau| \ge \epsilon]\cdot\epsilon^4 \le E^u[Y_\tau^4] \le 3\delta^2\,,$$
so
$$P[|Y_\tau| \ge \epsilon] \le 3\delta^2 \epsilon^{-4}\,. \tag{*}$$
For $k \ge 1$, we will use $(*)$ for $\delta_k={32}^{-k}$ and $\epsilon_k=2^{-k}$ to bound the probability of
$$A_k:=\bigcup_{j=0}^{{32}^k M-1} \Big\{\max_{0 \le t \le \delta_k} |X_{j\delta_k+t}-X_{j\delta_k}| \ge \epsilon_k\Big\} \,.$$
We obtain
$$P(A_k) \le 32^k M \cdot 3\delta_k^2 \epsilon_k^{-4}=3M\epsilon_k\,.$$
By  the Borel-Cantelli lemma, almost surely only finitely many of the events $A_k$ occur. This implies that $\{X_t\}$ is a.s. continuous in $[0,M]$.
$\hspace{6.6in} \Box$
Proof of Claim
$$X_u-u^2= E^u[(X_u+Y_\tau)^2-(u+\tau)]\,, \quad \text{so} \quad E^u[Y_\tau^2-\tau]=0 \,. \tag{2}$$
Similarly,
$$X_u^3-3uX_u=E^u[(X_u+Y_\tau)^3-3(u+\tau)(X_u+Y_\tau)]\,, \quad \text{so by} \; (2),$$
$$E^u[Y_\tau^3-3\tau Y_\tau]=0 \,. \tag{3}$$
Also,
$$X_u^4-6uX_u^2+3u^2=E^u[(X_u+Y_\tau)^4-6(u+\tau)(X_u+Y_\tau)^2+3(u+\tau)^2]\,, $$
so
$$0=E^u[Y_\tau^4+4X_u Y_\tau^3+6X_u^2 Y_\tau^2-6u Y_\tau^2-12\tau X_u Y_\tau-6\tau(X_u^2+Y_\tau^2)+6u\tau+3\tau^2]  \,.  $$
Therefore,
$$E^u[Y_\tau^4]=4X_u E^u[Y_\tau^3-3\tau Y_\tau]+6(X_u^2-u) E^u[Y_\tau^2-\tau] +3E^u[\tau^2]=3E^u[\tau^2]\,. \tag{4}$$
$\hspace{6.6in} \Box$
