Problem
Consider the backwards heat equation of the form $$ \left\{ \begin{aligned} u_{t} & = \lambda^2 u_{xx}, & x\in[0,L], \quad t\in[0,T]\\ u(0,t) &= u(L,t) = 0 \\ u(x,T) &= f(x), \end{aligned} \right.\tag{*}\label{*}$$ Establish whether solution is unique and analyze its stability.
Attempt of (dis)proving stability
My attempt to answer the stability question is provided in this post.
Attempt of proving uniqueness
I know that it is possible to use the energy functional method for proving uniqueness of the solution of backwards heat equation in a way we do that for a regular heat equation, with some additional tweaks.
Assume there are two different solutions $u_1$ and $u_2$ of $\eqref{*}$ and define the discrepancy as $w(x,t) : = u_1(x,t) - u_2(x,t)$. By superposition principle $w(x,t)$ is also a solution of \eqref{*}. Define the energy $$ E (t):=\int_0^Lw^2(x,t)\,dx \ge 0, $$ then $$\dot{E}(t) := \frac{dE}{dt} = 2\int_0^L w w_t\,dx, = 2 \int_0^L w w_{xx}\,dx = 2ww_x\big|_{0}^L - 2\int_0^L w_x^2\,dx=-2\int_0^L w_x^2\,dx,$$ $$ \ddot{E}(t) = \frac{d }{dt}\Big(\dot{E}(t) \Big) = -2\frac{d }{dt}\Bigg(\int_0^L w_x^2\,dx \Bigg) = -4\int_0^L w_x w_{xt}\,dx = 4\int_0^L w_{xx} w_{t}\,dx = 4\int_0^L w_{xx}^2 \,dx $$ By Cauchy-Schwarz we have $$ \dot{E}^2 = 4\Bigg(\int_0^L w w_{xx}\Bigg)^2\le 4\bigg(\int_0^Lw^2\,dx \bigg)\cdot \bigg(\int_0^Lw_{xx}^2\,dx\bigg) = E\cdot \ddot{E}$$
What should be the next step in the proof?
EDIT: Thanks to this answer, I was able to get the following:
$$\dot{E}^2\le E \ddot{E} $$
Define $F(t) := \ln\big(E(t)\big)$, then $$ \dot{F} = \frac{\dot{E}}{E}, \quad \ddot{F} = \frac{\ddot{E}E - \dot{E}^2}{E^2} > 0, $$ so that $F(t)$ is convex. By definition of convexity, $$ \forall t_1, t_2 \in [0,T], \ \forall \theta\in [0,1] \quad F\big(\theta t_1 + (1-\theta) t_2 \big) \le \theta F(t_1) + (1-\theta) F(t_2 ) \implies \\ \ln\Big(E\big(\theta t_1 + (1-\theta) t_2 \big) \Big) \le \theta \ln\big(E(t_1)\big) + (1-\theta)\ln\big(E(t_2)\big) = \ln \Big(E^\theta\left( t_1 \right) E^{(1-\theta)}\left( t_2 \right) \Big) \\ E\big(\theta t_1 + (1-\theta) t_2 \big) \le E^\theta\left( t_1 \right) E^{(1-\theta)}\left( t_2 \right) $$ Choosing $t_2 = T$ and assuming arbitrary $t_1 = t$, we get $$E\big(\theta t + (1-\theta) T \big) \le E^\theta\left( t \right) \underbrace{E^{(1-\theta)}\left( T \right)}_{=0} =0 \quad \forall \theta \in [0,1]$$ Since $w(T) = u_1(x,T) - u_2(x,T) = 0$, we know that $E(T) =0 $. But then $$ 0 \le E\big(\theta t_1 + (1-\theta) t_2 \big) \le E^\theta (t)\! \cdot\!0 \implies \\ \forall t\in (0,T) \quad E(t) \equiv 0 \ \implies w(x,t) \equiv 0 \iff \\ u_1(x,t) \equiv u_2(x,t). $$ Q.E.D.
The relation $\dot E^2\le E\ddot E$ brings to mind logarithmic differentiation. (I.e., $\dot E/E$ is the derivative of $\log E$.) So, let $\Phi(t)=\log E(t)$ and calculate $$ \ddot \Phi = \frac{\ddot E}{E}-\frac{\dot E^2}{E^2} $$ Conclude that $\Phi$ is a convex function of $t$. Therefore, it is impossible for it to drop from a finite value to $-\infty$ in finite time. In terms of $E$, this means if the energy was positive at some moment, it will always be positive.