Consider a finite dimensional complex Hilbert space $H$ of dimension $d$ equipped with an inner product denoted by $(\cdot,\cdot)$ and let $\rho$ be a generic density operator, i.e. a positive semi-definite operator with unit trace. This enables us to write
$$\rho = (\rho^{1/2})^\dagger \,\rho^{1/2} \quad . $$
Define $\rho_{nm} \equiv (v_n,\rho\, v_m)$, where $V=\{v_n\}$ is a complete orthonormal basis in $H$. Using the Cauchy-Schwarz inequality, we can show that the off-diagonal elements of the density matrix are bounded by the diagonal elements:
\begin{align}
|\rho_{nm}| &= |(v_n,\,(\rho^{1/2})^\dagger \,\rho^{1/2}\, v_m) |
=|(\rho^{1/2}\,v_n,\,\rho^{1/2}\, v_m) |\\
&\leq \sqrt{(\rho^{1/2}\, v_n,\,\rho^{1/2}\, v_n)\, (\rho^{1/2}\, v_m,\,\rho^{1/2}\, v_m)} \\
&= \sqrt{(v_n,\,\rho \,v_n) \, (v_m,\,\rho \,v_m)} \quad ,
\end{align}
which eventually yields
$$|\rho_{nm}|^2 \leq \rho_{nn}\, \rho_{mm} \quad. \tag{$*$}$$
The above inequality holds for all elements $v_n,v_m \in V$. Note that in particular the off-diagonal elements vanish if one of the corresponding diagonal elements is zero.
Regarding the purity, which we will define
as $$\pi(\rho) \equiv \mathrm{Tr}\rho^2 \quad . $$
The purity shows how 'mixed' / 'pure' a density operator is
and it is bounded by $d^{-1}\leq \pi(\rho) \leq 1$ for all density operators $\rho$. The lower bound is reached for maximally mixed states and the upper bound for pure states. Most importantly, the value of $\pi(\rho)$ is independent of the basis in which you express the density operator as a matrix.
But since every density operator is diagonalizable, there always exists a basis in which there are no off-diagonal elements. For example, a pure state as well as a maximally mixed state each have a basis in which all off-diagonal elements are zero, while their corresponding purities differ. So just from the fact that the off-diagonal elements (in a given basis) are zero or non-zero you don't know, in general, anything about the purity of a density operator.
In other words, the purity of a density operator is basis-independent, in contrast to the appearance of off-diagonal elements.
That being said, we still can conclude two things from equation $(*)$, namely:
$$\mathrm{Tr} \rho^2 \leq \left(\mathrm{Tr} \rho\right)^2 = 1 $$ and further that the equality in $(*)$ holds for all $v_n, v_m \in V$ if only if $\rho$ is a pure state.
To show the latter statement, we first suppose that $|\rho_{nm}|^2 = \rho_{nn}\, \rho_{mm}$ holds for all $v_n,v_m \in V$. Then
$$\mathrm{Tr}\rho^2 = \sum\limits_{nm} |\rho_{nm}|^2 = \sum\limits_{nm} \rho_{nn} \, \rho_{mm} = \left(\mathrm{Tr}\rho\right)^2 = 1 \quad ,$$
which shows that $\rho$ is pure. For the reverse direction, we recall that for a pure density operator $\rho^\psi$ there exists by definition a unit vector $ \psi \in H$ s.t. $\rho^\psi = \psi \psi^\dagger$. Expanding $\psi$ in the orthonormal basis $V$:
$$ \psi = \sum\limits_n c_n \, v_n$$
with $c_n \equiv (v_n,\psi)$ and $\sum\limits_n |c_n|^2 = 1$ shows that $\rho^\psi_{nm} = c_n\, \bar{c}_m $ and thus
$$|\rho^\psi_{nm}|^2 = |c_n|^2\, |c_m|^2 = \rho^\psi_{nn}\, \rho^\psi_{mm} \quad , $$
which completes the proof.
Consequently, for a mixed density operator $\sigma$ we find that there exists at least one pair $v_i,v_j \in V$ such that:
$$ |\sigma_{ij}|^2 < \sigma_{ii}\, \sigma_{jj} \quad ,$$
while for a pure state $\rho$
$$|\rho_{nm}|^2 = \rho_{nn}\, \rho_{mm} \quad \forall\, v_n,v_m \in V $$
holds.
It is stressed that we chose an arbitrary orthonormal basis and hence the above (in)equalities hold for any such basis (specifically the eigenbasis). All in all, we see that equation $(*)$ also serves as a criterion to decide whether a given density operator expressed in an arbitrary complete orthonormal basis is pure or mixed.
