We suppose that $X$ is a continuous local martingale such that $X_0=0$ and $$\forall u \in \mathbb{R}_+,\sup_{r\in[0,u]}E[e^{X_r/2}]<\infty.$$
Prove or disprove $e^{X-[X]/2}$ is a martingale.
This seems to be a version of Kazamaki's condition, I know how to prove the theorem if we supposed that $e^{X/2}$ is a sub-martingale.
How can we prove it when $$\forall u \in \mathbb{R}_+,\sup_{r\in[0,u]}E[e^{X_r/2}]<\infty?$$
This doesn't appear to be true. As an example (adapted from Revuz and Yor's Continous Martingales and Brownian Motion), let $Z_t = (X_t,Y_t)$ be a planar Brownian motion starting from $(1,1)$ and define $M_t := -\alpha \ln |Z_t|$ for some $\alpha \in [\frac{1}{2},2)$. Ito's formula shows that $M$ is a local martingale, and it is continuous because $Z$ almost surely never hits $(0,0)$. Note that we have \begin{align*} \mathbb{E}[e^{\frac 12 M_t}] &= \mathbb{E}[|Z_t|^{-\frac{\alpha}{2}}] \\ &= \frac{1}{2 \pi t} \int_{-\infty}^\infty \int_{-\infty}^\infty (x^2+y^2)^{-\frac \alpha4}e^{-((x-1)^2+(y-1)^2)/(2t)}dxdy \\ &= \frac{1}{2 \pi t} \int_0^\infty r^{1-\frac{\alpha}2} e^{-(r^2+2)/(2t)} \int_0^{2\pi} e^{\frac rt (\cos \theta + \sin \theta)}d\theta dr \\ &\le \frac 1t \int_0^{\infty}r^{1-\frac \alpha2} e^{-\frac{1}{2t}(r^2-2 \sqrt 2 r + 2)}dr \\ &= \frac 1t \int_0^{\infty}r^{1-\frac \alpha2} e^{-\frac{1}{2t}(r-\sqrt 2)^2}dr \\ &= t^{1-\frac \alpha 2} \int_0^\infty u^{1-\frac \alpha2} e^{-\frac{1}{2t}(tu-\sqrt 2)^2}du, \end{align*} after the change of variables $u = \frac rt$. Since $1-\frac \alpha2 > 0$, the integral clearly converges. Thus, we have that $\sup_{t \in [0,u]}\mathbb{E}[e^{\frac 12 M_t}] < \infty$ for all $u > 0$.
However, $\mathcal E(M)$ is not a martingale. To see this, first let $\rho_t := |Z_t|$ and compute \begin{align*} d\rho_t &= \frac{X_t}{\rho_t}dX_t + \frac{Y_t}{\rho_t}dY_t + \frac 12 \rho_t^{-1}dt \\ &= d\beta_t + \frac 12 \rho_t^{-1}dt, \end{align*} where $\beta_t := \int_0^t \frac{X_s}{\rho_s}dX_s + \int_0^t \frac{Y_s}{\rho_s}dY_s$ is a Brownian motion by Levi's characterization. We also compute \begin{align*} d\ln \rho_t &= \rho_t^{-1}d\beta_t, \end{align*} so $\langle \beta,\ln \rho\rangle_t = \int_0^t \rho_s^{-1}ds$.
Assume working towards a contradiction that $\mathcal E(M)$ is a martingale. Let $d\mathbb{Q} := \mathcal E(M)_Td\mathbb{P}$. Then, by Girsanov's theorem, $W_t := \beta_t + \alpha \int_0^t \rho_s^{-1}ds$ is a Brownian motion on $[0,T]$ under $\mathbb{Q}$. However, since $\rho_t = \beta_t + \frac 12 \int_0^t \rho_s^{-1}ds$, this implies $W_t = \rho_t + \left(\alpha - \frac 12\right) \int_0^t \rho_s^{-1}ds$. But since $\rho_t > 0$ a.s. and $\alpha - \frac 12 \ge 0$, this implies $W_t > 0$ a.s., contradicting that $W$ is a Brownian motion. Thus we conclude $\mathcal E(M)$ is not a martingale.