$a_n$ consistency and other consistency From Jun Shao's Mathematical Statistics

Definition 2.10 (Consistency of point estimators). Let $X = (X_1 , ..., X_n)$
  be a sample from $P ∈ \mathcal P$ and $T_n(X)$ be a point estimator of $θ$ for every $n$.
Let $\{a_n\}$ be a sequence of positive constants diverging to $∞$. $T_n(X)$ is called $a_n$-consistent for $θ$ if $a_n[T_n(X) − θ] = O_p(1)$ w.r.t. any $P∈\mathcal P$. 

$O_p(\cdot)$ is defined here. 
When $a_n := \sqrt{n}$, the consistency is called root $n$ consistency.
Do one of $a_n$ consistency and weak consistency imply the other? if not, how about $a_n := \sqrt{n}$?
Do one of $a_n$ consistency and strong consistency imply the other? if not, how about $a_n := \sqrt{n}$?
Do one of $a_n$ consistency and $L^2$ consistency imply the other? if not, how about $a_n := \sqrt{n}$?
Thanks and regards!
 A: $a_n$-consistency implies weak consistency
This is only a partial answer to your full query, but I hope it gets you started by showing how you can connect the consistency property in question with the standard consistency property.  Since $a_n |T_n - \theta| = O_p(1)$ there exists a function $c: \mathbb{R}_+ \rightarrow \mathbb{R}_+$ such that:
$$\sup_{n \in \mathbb{N}} \mathbb{P}(a_n |T_n - \theta| \geqslant c(\varepsilon)) < \varepsilon
\quad \quad \quad \text{for all } \varepsilon > 0.$$
Consequently, for all $\varepsilon>0$ we have:
$$\sup_{n \geqslant N} \mathbb{P} \Big( |T_n - \theta| \geqslant \frac{c(\varepsilon)}{a_n} \Big) 
\leqslant \sup_{n \in \mathbb{N}} \mathbb{P} \Big( |T_n - \theta| \geqslant \frac{c(\varepsilon)}{a_n} \Big) < \varepsilon.$$
Now, since $\{ a_n \}$ diverges to infinity, for any $\phi>0$ and $\epsilon>0$ you can choose $N$ sufficiently large to ensure that $a_n \geqslant c(\varepsilon)/\phi$ for all $n \geqslant N$.  Consequently, for any $\phi>0$ and $\epsilon>0$ you have:
$$\begin{align}
\lim_{n \rightarrow \infty} \mathbb{P}(|T_n - \theta| > \phi)
&\leqslant \lim_{N \rightarrow \infty} \sup_{n \geqslant N} \mathbb{P}(|T_n - \theta| > \phi) \\[6pt]
&= \lim_{N \rightarrow \infty} \sup_{n \geqslant N} \mathbb{P} \Big( |T_n - \theta| > \frac{\phi a_n}{a_n} \Big) \\[6pt]
&\leqslant \lim_{N \rightarrow \infty} \sup_{n \geqslant N} \mathbb{P} \Big( |T_n - \theta| \geqslant \frac{c(\varepsilon)}{a_n} \Big) \\[8pt]
&< \lim_{N \rightarrow \infty} \varepsilon \\[8pt]
&= \varepsilon. \\[6pt]
\end{align}$$
Since this inequality holds for all $\varepsilon>0$, we can say that for all $\phi>0$ we have:
$$\lim_{n \rightarrow \infty} \mathbb{P}(|T_n - \theta| > \phi) = 0,$$
which demonstrates weak consistency of $T_n$ in estimating $\theta$.
A: We can see that if $a_n[T_n(X_n) - \theta] = O_p(1)$, for any $b_n \to \infty$:
$$\begin{align*}\frac{a_n}{b_n}[T_n(X_n) - \theta] &= \frac{1}{b_n}  O_p(1) \\
&= o_p(1)O_p(1) \\
&= o_p(1)\end{align*}$$
This shows $\frac{a_n}{b_n}[T_n(X_n) - \theta] = o_p(1)$
If $a_n \to \infty$, then we can take $b_n = a_n$, and get that
$$T_n(X_n) - \theta = o_p(1)$$
Which proves that $a_n$ consistency implies weak consistency.
