When are the total variation distance and Hellinger distance comparable? The total variation distance between (say discrete) probability distributions, represented as vectors over their support, is defined to be
$$\Delta(\vec p,\vec q) = \frac{1}{2}\lVert \vec p-\vec q\rVert_1.$$
The (squared) Hellinger distance is then defined to be
$$H^2(\vec p, \vec q) = \frac{1}{2}\lVert \sqrt{\vec p}-\sqrt{\vec q}\rVert_2^2,$$
where we use the convention that $\sqrt{\vec p} := \left(\sqrt{p_1},\dots,\sqrt{p_k}\right)$ is the (coordinate-wise) square root.
These quantities satisfy the following bounds
$$H^2(\vec p, \vec q) \leq \Delta(\vec p, \vec q) \leq \sqrt{2} H(\vec p, \vec q).$$
Here, $H(\vec p, \vec q) := \sqrt{H^2(\vec p, \vec q)}$.
Constant factors might be slightly off based on my normalizations, but I don't particularly care about this.
Instead, it would be convenient for me if $\Delta(\vec p, \vec q) = \Theta(H^\alpha(\vec p,\vec q))$ for some $\alpha \in [1,2]$, i.e. if the lower and upper bounds were tight (up to constant factors).
I doubt this holds in general.
I would be interested if it even held for some large subset of all distributions — for example if it held for $\vec p, \vec q$ such that $\forall i: 0<c < \vec p_i/\vec q_i < C$, or some other "simple" condition on $\vec p, \vec q$ (I'm being intentionally somewhat vague by what this means).
 A: $\newcommand{\R}{\mathbb R}\newcommand{\al}{\alpha}\newcommand{\be}{\beta}
\newcommand{\De}{\Delta}$What you want is impossible for any reasonable class of probability distributions, including the class defined by your condition that $0<c<p_i/q_i<C$ for some real $c,C$ and all $i$.
Indeed, for simplicity of writing, let $p=(p_1,\dots,p_k):=\vec p$ and $q=(q_1,\dots,q_k):=\vec q$. Let $P_k$ denote the set of all probability vectors $p=(p_1,\dots,p_k)\in\R^k$. Let us say that a neighborhood $N_q$ of $q\in P_k$ is $c$-good for some $c\in(0,1)$ if the vector $p$ defined by the conditions
\begin{equation}
    p_1:=q_1+h,\quad p_2:=q_2-h,\quad p_i:=q_i\text{ if }i\ge3 \tag{0}\label{0}
\end{equation}
for $h:=c\min(q_1,q_2)$ is in $N_q$.
The first sentence of this answer is formalized as

Claim: Suppose that for some real $\al$ and $\be$
there is a constant $c\in(0,1)$ such that for each natural $k$ there is some $q\in P_k$ such that
\begin{equation}
    c\le kq_i\le1/c \tag{1}\label{1}
\end{equation}
for all $i\in[k]:=\{1,\dots,k\}$ and
\begin{equation}
    cH^\al(p,q)\le\De(p,q)\le H^\be(p,q)/c \tag{2}\label{2}
\end{equation}
for all $p\in P_k$ in some $c$-good neighborhood $N_q$ of $q$. Then $\al\ge2$ and $\be\le1$. So,
it is impossible for a relation of the form $cH^\al(p,q)\le\De(p,q)\le H^\al(p,q)/c$ to hold for some real $\al$ and all such $q,p$.

Proof: For all $i\in[k]$, let
\begin{equation}
    p_i:=(q_i+(k-m)h)\,1(i\le m)+(q_i-mh)\,1(i>m), 
\end{equation}
where $k\to\infty$, $m\sim k/2$, and $h>0$ is small enough so that $mh<q_i$ for $i>m$ and $p\in N_q$. Then $\De(p,q)=(k-m)mh\asymp k^2h$ and $H(p,q)\asymp k^2h$, so that $\De(p,q)\asymp H(p,q)$. Letting now $h\downarrow0$ fast enough so that $k^2h\to0$, we see that the inequality $\De(p,q)\le H^\be(p,q)/c$ in \eqref{2} can hold only if $\be\le1$.
Now let $p$ be as in \eqref{0}. Then $\De(p,q)=h\asymp1/k$ and $H^2(p,q)\asymp1/k$, so that $\De(p,q)\asymp H(p,q)^2$. Letting now $k\to\infty$, we see that the inequality $cH^\al(p,q)\le\De(p,q)$ in \eqref{2} can hold only if $\al\ge2$.
$\Box$
