Let $a\in \mathbb{R}$ and consider the IVP \begin{eqnarray} u_t+au_x&=&0 \quad \quad (x,t) \in \mathbb{R} \times \mathbb{R}^+\\ u(x,0)&=&u_0(x) \quad \quad x \in \mathbb{R}. \end{eqnarray}
For $u_0\in C_c^1,$ there exists at most one solution(differentiable) and this can be proved by enegry estimates.
If $u_0 \in L^1(\mathbb{R})\cap L^{\infty}(\mathbb{R}),$ we can use the fact that any weak solution to transport equation is infact entropy solution (in the sense of Kruzkov) and hence its unique.
Is there a more direct means of proving the uniqueness of the weak solution of the transport equation without relying on Kruzkov's theory? In other words can we prove contraction estimates or some sort of energy estimates only using the definition of weak formulation?
I don't have enough points to comment, so here is a comment:
If $u_0 \in C^{\infty}$, the unique smooth solution is $u(t, x) = u_0(x - at)$. This can be proved by considering $z(t) = u(t_0 + t, x_0 + at)$, and noting that $z'(t) = 0$, $z(0) = u(t_0, x_0)$, to obtain $u(t_0 + t, x_0 + at) = u(t_0, x_0)$, so $$u(t, x) = u(0 + t, x - at + at) = u(0, x - at) = u_0(x - at).$$
If $u_0 \in \mathcal{D}(\mathbb{R})$, then at least, because $f(t, x) = x - at$ is a submersion, the composition $u_0 \circ f \in \mathcal{D}(\mathbb{R}^+ \times \mathbb{R})$ can be defined, and satisfies the chain rule; see page 84 of "Introduction to the Theory of Distributions" by Friedlander and Joshi. So $u(t, x) = u_0(x - at)$ is a distributional solution. Moreover, $u \in C([0, T], \mathcal{D}(\mathbb{R}))$. It is reasonable to ask whether this $u$ is the only solution in the class $C([0, T], \mathcal{D}(\mathbb{R}))$.