# Equivalent conditions for local martingales

I am having some trouble solving this task:

Consider the solution $$X$$ of the SDE in $$\mathbb{R}^n$$ $$\mathrm{d} X_t = b (X_t) \mathrm{d} t + \sigma (X_t) \mathrm{d} B_t,$$ where $$B$$ is a $$n$$-dimensional Brownian motion and $$b : \mathbb{R}^n_{} \rightarrow \mathbb{R}^n$$, $$\sigma : \mathbb{R}^n \rightarrow \mathbb{R}^{n\times n}$$ locally bounded coefficients.

Let $$a (x) = \sigma (x) \sigma(x)^T \in \mathbb{R}^{n \times n}$$ and for all $$f \in C^2 (\mathbb{R}^n)$$ let $$\mathcal{L}f (x) = b (x) \cdot \nabla f (x) + \frac{1}{2} \mathrm{Tr} [a(x) \nabla^2 f (x)], \qquad x \in \mathbb{R}^n$$ where $$\nabla^2 f (x)$$ is the $$\mathbb{R}^{n \times n}$$ matrix of second derivatives of $$f$$. Prove that the following conditions are equivalent

i) For any $$f \in C^2 (\mathbb{R}^d)$$, the process $$M^f_t = f (X_t) - f (X_0) - \int_0^t \mathcal{L}f (X_s) \mathrm{d} s$$ is a local martingale.

ii) For any $$v \in \mathbb{R}^d$$, the process $$M^v_t = v \cdot X_t - v \cdot X_0 - \int_0^t v \cdot b (X_s) \mathrm{d} s$$ is a local martingale with quadratic variation $$[M^v]_t = \int_0^t v \cdot a (X_s) v \mathrm{d} s.$$

iii) For any $$v \in \mathbb{R}^d$$ the process $$Z^v_t = \exp \left( M^v_t - \frac{1}{2} \int_0^t v \cdot a (X_s) v \mathrm{d} s \right)$$ is a local martingale.

Hint: use the fact that linear combinations of exponentials are dense in $$C^2$$ w.r.t. uniform convergence on compacts for the functions and its first two derivatives (assumed wihtout proof)

"$$i)\Rightarrow ii)$$": Consider $$f(X_t)=v\cdot X_t$$. Then $$f$$ is $$C^2$$ with $$\nabla f=v$$ and $$\nabla^2 f=0$$, thus we have $$M_t^f=v\cdot X_t-v\cdot X_0-\int_0^t v\cdot b(X_s)\mathrm{d} s=M_t^v$$ Now by assumption $$M_t^f=M_t^v$$ is a local martingale. Furthermore $$M^v_t=v\cdot X_t-v\cdot X_0-\int_0^t v\cdot(\mathrm{d} X_s-\sigma(X_s)\mathrm{d} B_s) =-v\cdot X_0-\int_0^t v\cdot\sigma(X_s)\mathrm{d} B_s.$$ This is an Ito process, so the quadratic variation is given by $$[M^v]_t =\int_0^t(v\cdot\sigma(X_s))(v\cdot\sigma(X_s))^T\mathrm{d} s =\int_0^t v\cdot\sigma(X_s)\sigma(X_s)^T v\mathrm{d} s =\int_0^t v\cdot a(X_s)v\,\mathrm{d} s.$$ "$$ii)\Rightarrow iii)$$": I know that $$Z^v_t=\exp\left( M^v_t-\frac{1}{2}[M^v]_t \right),$$ which looks familiar, but I have to confess that I have only taken an introductory course in probability so far and starting pretty much from scratch in terms of martingales and stochastic processes.

"$$iii)\Rightarrow i)$$": By the hint, $$f$$ can be approximated by linear combinations of exponentials, e.g. there exists a sequence of functions $$f_k(x)=\sum\limits_{i=0}^{n_k-1}\lambda_{i_k}\exp(t_{i_k}|x|)$$ such that $$\lim_{k\rightarrow\infty}\sup_{x\in K} |f(x)-f_k(x)|+|f'(x)-f_k'(x)|+|f''(x)-f_k''(x)|=0$$ for all compact sets $$K\subset\mathbb{R}^n$$...

I don't know how to get from the assumption in iii) and the approximation to the conclusion that $$f$$ is a local martingale. How do I get my $$Z^v_t$$ in there? I suspect I have to choose the $$v$$ accordingly to have $$f_k$$ in the form of $$f_k(X_t)=\sum_i\lambda_i Z^v_t$$?