Let's say I'm interested in two player games.
I'm aware of the complexities of finding mixed Nash equilibria. To really be sure we've found all we have to enumerate all possible combinations of row and column strategies and check for indifference (ignoring other simple tricks).
Although, let's assume some pure Nash equilibria exist. Does this say anything about mixed ones?
If pure strategies do exist can my search for mixed strategies include only those that contain at least one of the strategies from the pure Nash equilibrium in their support?
The answer is no, that is, if you have a $2$-player game with NE in pure strategies, not necessarily you can rely on the fact that the actions that are in the pure NE are also in the support of the mixed NE.
Here there is an example,
$$\begin{bmatrix} 1,1 & 0,0 & 0,0 \\ 0,0 & 1,-1 & -1,1 \\ 0,0 & -1,1 & 1,-1 \\ \end{bmatrix}$$
[Notice that this is really like matching pennies, with the addition of the first row/column.]
There is one pure strategy NE, given by $(T, L)$, and a mixed strategy NE where players equally randomize between Middle/Down and Center/Right, much in the same spirit of matching pennies, and there is neither $T$ nor $L$ in the support of this mixed strategy NE equilibrium.