A Nash equilibrium $y = (y_i, y_{-i}) \in \Delta$ is said to be strict if for all players $i$,
$$u(y_i; y_{-i}) \geq u(x_i; y_{-i}), \forall x_i, y_i \neq x_i$$
Is it true that no mixed equilibrium can be strict? (and only pure strategies can be strict?)
Mixed-strategy Nash equilibria are necessarily always weak, while pure-strategy Nash equilibria can be either strict or weak. We can see this because if there are greater than or equal to two pure strategies that are best responses to the other players strategy then any mixture of them is also a best response. For example take the game,
\begin{array}{|c|c|} \hline & H & T \\ \hline H & (2, 1) & (0, 0) \\ \hline T & (0, 0) & (1, 2) \\ \hline \end{array}
the mixed strategy for the row player is $(\frac{2}{3}, \frac{1}{3})$ and $(\frac{1}{3}, \frac{2}{3})$ for the column player, with each player having an expected payoff of $\frac{2}{3}$. Now assuming the column player does not deviate from their equilibrium mixed strategy, every strategy, $(p, 1 - p)$, is a best response for the row player, so they can deviate if they want.