*Preamble:*
This question is similar to the one in [total variation distance between two solutions of SDE](https://mathoverflow.net/questions/115090/total-variation-distance-between-two-solutions-of-sde) . The difference is that in my case the drift is the same but there are different diffusion coefficients.


*Question:*
Consider the two SDE:
\begin{align}
dX\_t &= f(X\_t) dt + \frac{\sigma(X\_t)}{\sqrt{n}} dW\_t\\
dY\_t &= f(Y\_t) dt + \frac{\bar\sigma\_n(Y\_t)}{\sqrt{n}} dW\_t,
\end{align}
where $t \in [0,T]$, $X\_0 = Y\_0 = 0$, $W\_t$ is the standard 1-dimensional Brownian motion. The hypotheses are that the drift $f$ and the diffusion coefficient $\sigma > 0$ are bounded and Lipschitz, and the $\bar\sigma\_n$ are defined as the piecewise constant approximations of $\sigma$, constant on the intervals of the form $\big[\frac{j}{n}T,\frac{j}{n}T\big)$.


Can we conclude that the total variation distance between the two solution processes tends to zero as $n$ tends to infinity?


*Considerations*
There should be no problem proving that there is $L^2$ convergence, thanks to Theorem 10, chapter V of Protter's "Stochastic Integration and Differential Equations", but I think that this is not enough to obtain the convergence in total variation. In the answer to the question linked above, there is a mention to a possible solution of a similar problem via Malliavin calculus, but since I am not at ease with this theory I would gladly accept any suggestion or reference in that direction.



Total variation distance between diffusion processes with different volatility coefficient
No. the total variation distance is always one, since the quadratic variation of the processes is different, and so the measures are mutually absolutely singular.


