The title is a general statement, and I am trying to apply it in the following problem.
Suppose $$ X_n\stackrel{\mathcal{D}}{\rightarrow} N\left(0, E\left\{\left(\varphi-\varphi^*\right)\left(\varphi-\varphi^*\right)^T\right\}\right) $$
In the textbook Semiparametric theory by Tsiatis Chapter 3, the author mentioned in order for this limiting normal distribution to be $o_p(1)$, we would require that the covariance matrix $$E\left\{\left(\varphi-\varphi^*\right)\left(\varphi-\varphi^*\right)^T\right\}=0^{q \times q}$$.
Intuitively it makes sense, but I don't know how the formal proof goes. Is it just simply because - if we want derive converges to 0 in probability, we need the variance to be zero?
The short title of the question is poorly related to the main question.
We have $X_n \to \xi \sim N(0, A)$ where $A = A^{\mathbf{T}}$. We want to show that $X_n = o_P(1)$ iff $A = 0$. But $$X_n = o_P(1) \Leftrightarrow X_n \overset{P}{\to} 0 \Leftrightarrow \xi = 0 \Leftrightarrow N(0, A) = 0 \Leftrightarrow A = 0$$ q.e.d.
In details: $ \xi = 0 \Leftrightarrow (\xi, c) = 0$ for every nonrandom vector $c$. As $(\xi, c) = c^{\mathbf{T}} \xi \sim N(0, c^{\mathbf{T}} A c)$ we have $$ \xi = 0 \Leftrightarrow (\xi, c) = 0 \ \ \forall c \Leftrightarrow N(0, c^{\mathbf{T}} A c) =0 \ \ \forall c \Leftrightarrow c^{\mathbf{T}} A c \ \forall c \Leftrightarrow A = 0$$