Proposition 1.3 (Alternative form of the first-order convexity condition)
The first-order convexity conditions $$ f() \geq f() + ( - ) \cdot \nabla f( ) $$ and $$ ( - ) \left( \nabla f( ) - \nabla f( ) \right) \geq 0 $$ are equivalent, where $, \in D (f)$ .
However, it comes without proof. The first implying the second is easy enough. However, I do not see how the second implies the first. I tried many things but I can't get away without assuming too much. I am at the point where I am not sure if they are equivalent. Question: Is the given equivalence true, and if so, then how does one go about proving the backward direction?
asked Feb 18, 2023 at 16:02 Maths Wizzard Maths Wizzard 2,312 1 1 gold badge 10 10 silver badges 25 25 bronze badges $\begingroup$ Where does this proposition come from? $\endgroup$ Commented Feb 18, 2023 at 17:46 $\begingroup$ This is from lecture notes on a course on variational principles. $\endgroup$ Commented Feb 18, 2023 at 18:04 $\begingroup$ Is there a link to lecture slides? $\endgroup$ Commented Feb 18, 2023 at 18:42$\begingroup$ These are adaptions from notes that I took. Are you asking by any chance because of the formating? $\endgroup$
Commented Feb 18, 2023 at 20:45 $\begingroup$ Not really. Just curious. $\endgroup$ Commented Feb 18, 2023 at 20:54Questions about convex functions of multiple variables can often be reduced to a question about convex functions of a single variable by considering that function on a line or segment between two points.
The two conditions are indeed equivalent for a differentiable function $f:D \to \Bbb R$ on a convex domain $D \subset \Bbb R^n$ .
To prove that the second condition implies the first, fix two points $x, y \in D$ and define $$ l:[0, 1] \to D \, , \, l(t) = x + t (y-x) \, , \\ g:[0, 1] \to \Bbb R\, , \, g(t) = f(l(t)) \, . $$
Note that $$ g'(t) = (y-x) \nabla f(l(t)) \, . $$
For $0 < t < 1$ is $$ g'(t)- g'(0) = (y-x)\bigl(\nabla f(l(t)) - \nabla f(l(0) \bigr) \\ = \frac \bigl(l(t)-l(0)\bigr) \bigl(\nabla f(l(t)) - \nabla f(l(0) \bigr) \ge 0 \, , $$ and the mean-value theorem gives, with some $\xi \in (0, 1)$ , $$ f(x) - f(y) = g(1) -g(0) = g'(\xi) \ge g'(0) = (y-x) \nabla f(x) \, . $$
(Actually, $g'$ is increasing so that $g$ is convex.)