Suppose we have a closed (compact without boundry) Riemannian manifold $(M,g)$, do we have uniqueness (assuming solutions exist on $[0,T]$ ) for the backward heat equation
$$\frac{\partial f }{\partial t}+\Delta_gf=0$$ $$f(x,0)=f_0(x),$$ where $f_0$ is some Borel function on $M$.
I'm assuming you are supposing that $f\in C^\infty(M)$ for each $t \in [0,T]$, or something along this line. I'm not sure what happens if you allow $f$ to have some singularities. Then, picking any $0 < T_1 < T$, we can consider $f$ as solving a forwards parabolic PDE from $T_1$ to $0$. This puts parabolic regularity at our disposal and will allow you to make the following argument rigorous.
Write $f(t) = \sum_{j=1}^\infty c_j(t) \varphi_j$ where $\Delta_g \varphi_j + \lambda_j\varphi_j=0$ are the eigenfunctions for the Laplacian on $(M,g)$. Then, the backwards heat equation becomes, at the level of the constants $c_j(t)$, $$ c_j'(t) = \lambda_j c_j(t), $$ so $c_j(t) = c_j(0) e^{\lambda_j t}$. Thus, we see that $$ f = \sum_{j=1}^\infty c_j(0) e^{\lambda_j t} \varphi_j. $$ The parabolic regularity argument above shows that for $t \in [0,T)$, this series converges in $L^2(M)$ (along with all derivatives, but this doesn't matter). Thus, we see that $f$ is unique given $f(0)$ (which determines $c_j(0)$.
This method is probably the easiest way to prove what you ask, but certainly is not the most robust. One can prove much more with techniques like Carleman estimates: see, e.g. http://arxiv.org/pdf/0704.1349v2.pdf.
