Consider a nonlinear dynamical system \begin{equation} \dot{\mathbf{x}}(t) = f(\mathbf{x}(t)), \quad \mathbf{x}(0) = \mathbf{x}_0, \quad t \geq 0 \end{equation} and Lyapunov function $V: D \rightarrow \mathbb{R}$. Since \begin{equation} V'(\mathbf{x}) f(\mathbf{x}) \leq 0, \quad \forall~\mathbf{x} \in D, \end{equation} show $\forall~\mathbf{x}(0) \in D$, $V(\mathbf{x}(t)) \leq V(\mathbf{x}(0))$ for all $t \geq 0$.
This statement holds true to me intuitively, but I think this statement requires some rigorous justification.
Given $V'(\mathbf{x})f(\mathbf{x}) \leq 0$ for all $\mathbf{x} \in D$, then we know that $\dot{V} (\mathbf{x}(0)) = V'(\mathbf{x}(0)) f(\mathbf{x}(0)) \leq 0$ for $\mathbf{x}_0 \in D$. However, to show the statement, I need $\forall~t > 0$, $\dot{V}(\mathbf{x}(t)) \leq 0$. Just knowing $\dot{V} (\mathbf{x}(0)) \leq 0$ is not enough, I think. Any help would be appreciated.