Strictly Convex and Strictly Monotonic Preferences On the assumptions for a well behaved preference, I read recently that a preference must be complete, transitive, continuous, strictly monotonous (if the vectors x≥y and x≠y, then x is strictly preferable to y) or sometimes just monotonous (if x>>y, then x is strictly preferable to y) and strictly convex. But my doubt now follows: if you assume strict convexity (any linear combination with a different bundle in the same indifference curve will be strictly preferable) for a preference, it seems to me that any degree of monotonicity will be strict -- right? Am I missing something?
 A: Definitions
To clarify, preferences that are strongly monotonic state that when $x \geq y$ and $x \neq y$, then $x \succ y$. Preferences are strictly monotonic when $x \geq y$ imply $x \succeq y$ and $x>>y$ imply $x \succ y$. The difference is that strong monotonicity requires only one more of any good to be strictly preferred.
Strict convexity (of preferences) means that when $x \neq y$ and $x \succeq y$, then 
$t x + (1-t) y  \succ y$ for all $t \in (0,1)$. 
Example with no (positive) monotonicity
Consider preferences like those depicted below. The top indifference curve (U1) is strictly less preferred than the bottom (U2). Preferences over these goods (which might better be called "bads") are strictly convex. Whenever $x \neq y$ and $x \succeq y$, then 
$t x + (1-t) y  \succ y$ for all $t \in (0,1)$. 
However, these preferences are in no degree monotonic. (In a sense, they are monotonically decreasing, but we usually take "monotonic preferences" to be increasing as in the definition I gave above.) For the right domain, preferences of this kind are indeed complete and transitive.
So, strictly convex preferences do not imply monotonicity (at least in the direction that you want).

Monotonicity and strict convexity imply strong monotonicity
Suppose we use a definition of monotonicity where $x \geq y$ imply $x \succeq y$. Then, it is true that this form of monotonicity together with strict convexity (as defined above) imply strong monotonicity (as defined above).
To show this, let $x$ and $y$ be given such that $x \geq y$ and $x \neq y$. Then by this form of monotonicity, $x \succeq y$. This allows us to use strict convexity to say that
$\forall t \in (0,1)$ we have $ t x + (1-t) y \succ y$. Now, either $x \succ y$ or $x \sim y$. Suppose by way of contradiction that $x \sim y$. Then $\forall t \in (0,1)$ we have
$$
t x + (1-t) y \succ y \sim x.
$$
But it is clear that $ t x + (1-t) y \leq x$, so by monotonicity $ t x + (1-t) y \preceq x$. This is a contradiction. So it must be that $x \succ y$. Thus, the preferences are strongly monotonic.
A: Here is an attempt to make up for my ugly mistake on the strict convexity of Leontieff preferences ;). 
If I read you right, the guess you want to check is whether for complete, transitive and continuous preferences, strict convexity implies that any monotonic preference relation is also strictly monotonic. I believe your guess is right and this can be shown by contradiction (the construction of the different bundle in the proof are illustrated in the picture).


*

*Assume that there exists $x,y$ such that $x \geq y, x\neq y$ and $y$ is not strictly preferred to $x$.

*By monotonicity, $x_l = y_l$ for some good $l$ (otherwise $x \succ y$).





*

*By completeness we must have $ y \succeq x$. 

*As $x \neq y$ and preferences are continuous, there must exist a bundle $k$, a convex combination of $x$ and $y$, which leaves the agent indifferent with $x$, that is for some $\alpha \in [0,1)$ we have $k = \alpha x + (1-\alpha) y$ and $x \sim k$. (This step is not really needed but helps sticking to your definition of strict convexity).

*By strict convexity, there exists a convex combination $z$ of $k$ and $x$ such that $z$ is strictly preferred to $x$, that is for some $\lambda \in (0,1)$ we have $z = \lambda x + (1-\lambda) y$ and $z \succ x$.

*By continuity, strict upper contour sets are open, so there exists a neighborhood of $z$, $N_\epsilon(z)$ such that for every $m \in N_\epsilon(z)$, $m \succ x$. 

*Take any $w \in N_\epsilon(z)$ such that $w << z$. By construction $w << x$, so by monotonicity we must have $x \succ w$. But by $w \in N_\epsilon(z)$, we must have $w \succ x$, a contradiction.

