Proof that convexity implies pseudoconvexity

Question

I have two questions.

1. By definition, pseudoconvexity is for functions whose domain is 'convex open' set while convexity is for functions with just 'convex' (not necessarily open) set. Then how does convexity imply pseudoconvexity always? Is it like convexity on a set implies pseudoconvexity on interior of the set, just like continuity on $[a, b]$ implies differentiability on $(a,b)$?
2. Can someone refer me to a book or resource containing the proof for convexity implies pseudoconvexity? I just encountered these definitions in what I am working on and I am not explicitly learning about convex sets and functions, so I do not have a resource to refer to right now, where I can find this proof.

Edit: I am following this definition for pseudoconvexity: Let $S$ be a nonempty open set in $\mathbb{R}^n$, and let $f:S\to \mathbb{R}$ be differentiable on $S$. The function $f$ is pseudoconvex if, for each $x_1$,$x_2\in S$ with $\nabla{f}(x_1)^T(x_2−x_1)\geq 0$, then $f(x_2)\geq f(x_1)$.

Answer 1

1. Not all sources define the "pseudoconvex function" in the same way.

   • Nonlinear Programming: Theory and Algorithms by Bazaraa, Sherali and Shetty, and Invexity and Optimization by Shashi, Mishra and Giorgi give the same definition you do, which requires the function to be differentiable and its domain to be open, but not necessarily convex. To eliminate ambiguity I shall refer to functions that are pseudoconvex in this sense as BS-pseudoconvex.

   • Iterative Solution of Nonlinear Equations in Several Variables by Ortega and Rheinboldt require the domain of pseudoconvexity to be convex, but not necessarily open, and give a definition which does not require the function to be differentiable. They say it is "easy to see" that their condition reduces to to the one in your definition whenever the function is Gateaux differentiable (a weaker condition than full differentiability). The degree to which it is "easy to see" this may well depend on how familiar you are with the relevant concepts. I shall refer to functions that are pseudoconvex in this sense as OR-pseudoconvex.

   $\ $
   It is not true that convexity always implies BS-pseudoconvexity (i.e. pseudoconvexity in the sense in which you have defined it), because a convex function is not necessarily differentiable. What is true in this case is that if a convex function is differentiable on the interior of its domain then it must be BS-pseudoconvex on that interior.

   It is true, however, that convexity always implies OR-pseudoconvexity.

2. I couldn't find a proof in any of the above references that convexity implies pseudoconvexity, merely assertions to that effect. Starting with Ortega and Rheinboldt's definition, however:

   A function $\ g:D\subseteq\mathbb{R}^n\rightarrow\mathbb{R}\ $ is OR-pseudoconvex on a convex set $\ D_0\subseteq D\ $ if, whenever $\ g(x) > g(y)\ $, with $\ x,y\in D_0\ $, there exist $\ \alpha>0\ $ and $\ \tau\in(0,1]\ $ (both depending, in general, on $\ x\ $ and $\ y\ $) such that $$ g((1-t)x+ty)\le g(x)-t\alpha\ \ \text{ for all }\ t\in[0,\tau]\ , $$ the proof is not all that difficult.

   Let $\ g:D\rightarrow\mathbb{R}\ $ be a convex function on the convex set $\ D\subseteq\mathbb{R}\ $ and $\ x,y\in D\ $ such that $\ g(x)>g(y)\ $. Let $\ \tau=1\ $ and $\ \alpha=g(x)-g(y)\ $. Then $\ \alpha>0\ $, and if $\ t\in[0,\tau]\ $ then the convexity of $\ g\ $ implies that \begin{align} g((1-t)x+ty)&\le (1-t)g(x)+tg(y)\\ &=g(x)-t(g(x)-g(y))\\ &= g(x)-t\alpha\ , \end{align} so $\ g\ $ is OR-pseudoconvex on $\ D\ $.

   Now suppose, in addition, that $\ g\ $ is differentiable on the interior of $\ D\ $. Then there exists $\ \delta\in(0,1)\ $ such that $$ \left|\frac{g(x+t(y-x)) - g(x)}{t}-\nabla f(x)^T(y-x)\right|<\frac{\alpha}{2} $$ for all $\ t\in(0,\delta)\ $. Therefore, \begin{align} \nabla f(x)^T(y-x)&\le \frac{g(x+t(y-x)) - g(x)}{t}+ \frac{\alpha}{2}\\ &\le-\frac{\alpha}{2}\\ &<0 \end{align} for any such $\ t\ $. By the equivalent contrapositive, if $\ x,y\ $ lie in the interior of $\ D\ $ with $\ \nabla f(x)^T(y-x)\ge0\ $, then $\ g(x)\le g(y)\ $, so $\ g\ $ is BS-pseudoconvex on the interior of $\ D\ $.