Is there any relation between monotone operators and increasing operators? If $V$ is some Riesz space (vector lattice for a partial order, say a Banach space), and if an operator $A\colon V \to V'$ is monotone, i.e
$$\langle A(u) - A(v), u-v \rangle \geq 0,$$
does it mean that $f \geq g \implies A(f) \geq A(g)$, with the induced order for the dual space? What about the converse?
As far as I can see, there is no relationship.
 A: As you conjectured, there no implication between the two notions holds. In order to find a counterexample for each implication, let us endow $\mathbb{R}^2$ with the componentwise order and identify it with its dual space in the usual way. We can find linear counterexamples for each implication:
The symmetric matrix
$$
  A =
  \begin{pmatrix}
     1 & -1 \\
    -1 &  1
  \end{pmatrix}
$$
has spectrum $\{0,2\}$ and is thus positively semi-definite. Hence, it is monotone in the sense specified in the question. But $f \ge g$ does not imply $Af \ge Ag$; indeed, consider for instance $f = (1,0)^T$ and $g = 0$.
The symmetric matrix
$$
  B =
  \begin{pmatrix}
    0 & 1 \\
    1 & 0
  \end{pmatrix}
$$
satisfies $Bf \ge Bg$ whenever $f \ge g$ (since all entries of $B$ are $\ge 0$). But it is not monotone in the sense specified in the question since, for $u = (1,-1)^T$ and $v = 0$,
$$
  \langle Bu - Bv, u-v \rangle 
  = 
  \langle Bu, u \rangle
  = 
  -2.
$$
(From a more abstract point of view, this is due to the fact that the spectrum of $B$ is $\{-1,1\}$.)
