Let $\kappa$ be an infinite cardinal. Does there always exist a function $f: \kappa \to \kappa$ such that
$f$ is weakly increasing ($f(\alpha) \leq f(\beta)$ for $\alpha \leq \beta < \kappa$)
$f$ is cofinal (equivalently, the image of $f$ is unbounded)
$f$ lies strictly below the diagnoal (i.e. $f(\alpha) < \alpha$ for all $\alpha <\kappa$)
?
I'm willing to relax condition (3) by allowing $f(\alpha) \leq \alpha$ whenenever $\alpha$ has uncountable cofinality.
Also, $f(\alpha)$ need only be defined for sufficiently large $\alpha$, though I'm pretty sure that doesn't make a difference.
For example, this is possible when $\kappa = \aleph_0$, by taking $f(n) = max(n-1,0)$.
First, note that such a function does indeed exist if $\kappa$ has cofinality $\omega$, by an easy extension of the $\kappa=\omega$ case.
I now claim that if $cf(\kappa)>\omega$, then no such function exists. To see this, suppose $cf(\kappa)=\lambda>\omega$ and $f$ has the properties above. Fix an increasing cofinal sequence $S=(s_i)_{i\in\lambda}$ in $\kappa$, and chop $\kappa$ into intervals along this sequence: $I_i=[s_i, s_{i+1})$. Now for each $i$ we have $f(s_i)\in I_j$ for some $j<i$ (ignoring a bounded set of values as usual). Let $h:\lambda\rightarrow\lambda$ be defined by $h(i)=j$ iff $f(s_i)\in I_j$. Then $h$ is regressive, hence constant with value $u$ on a stationary (hence cofinal) set for some $u$. But then $f$ itself can't be both weakly increasing and cofinal, since $f$ maps a cofinal set to a bounded set (namely, $I_u$).