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Let a convex $ U \subset_{op} \mathbb{R^n} , n \geq 2$, with the usual inner product. A function $F: U \rightarrow \mathbb{R^n} $ is monotone if $ \langle F(x) - F(y), x-y \rangle \geq 0, \forall x,y \in \mathbb{R^n}.$

Let $f:U \rightarrow \mathbb{R}$ differentiable. Show that $f$ is convex $\iff \nabla f:U \rightarrow \mathbb{R^n}$ is monotone.

My attempt on the right implication: I already proved that if $f$ is convex and 2-differentiable then $f''(x) \geq 0$. But this exercise only says f is 1-differentiable. Then I tried the following: $f$ is convex $\iff \forall x,y \in U $ the function $\varphi:[0,1] \rightarrow \mathbb{R}$, defined by $ \varphi(t) = f((1-t)x+ty)$ is convex. Then $\varphi'$ is non-decreasing, then $\nabla \varphi(x) \geq 0$... but I'm stucked here.

My attempt on the left implication:

$ |\nabla \varphi (x) - \nabla \varphi (y)|| x-y| \geq | \langle \nabla \varphi (x) - \nabla \varphi (y), x-y \rangle | \geq 0$

And so $ |\nabla \varphi (x) - \nabla \varphi (y)| \geq 0 $ then $\nabla \varphi $ is non-increasing and then (By an already proved Theorem) it is convex.

Can someone please verify what I did and give me a hint?

Thanks.

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  • $\begingroup$ If you define $F$ only on $U$, then it is difficult to verify $\langle F(x) - F(y), x-y \rangle\ge 0$ for every $x,y\in\mathbb R^n$. $\endgroup$ – user251257 Mar 28 '16 at 19:34
  • $\begingroup$ Did you mean isn't difficult? $\endgroup$ – user286485 Mar 28 '16 at 20:19
  • $\begingroup$ no. How do you compute $F(x)$ for $x\notin U$? $\endgroup$ – user251257 Mar 28 '16 at 20:20
  • $\begingroup$ As much as I love the humor in the error, the term is monotone or monotonic, not monotonous :-) Convex functions are anything but monotonous! math.stackexchange.com/q/365717/52878 $\endgroup$ – Michael Grant Jun 15 '18 at 13:22
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1) If $f$ is convex, then $$ f(y)\geq f(x) + \nabla f(x)\cdot (y-x) $$

and $$ f(x)\geq f(y) + \nabla f(y)\cdot (x-y) $$

so that by adding $$ (y-x)\cdot( \nabla f(x) - \nabla f(y)) \leq 0 $$

2) Assume that $\nabla f$ is monotone : Define $A =\{ x| f(x)\leq a\}$. If $A$ is not convex, then there are $x,\ y\in \partial A$ s.t. $$ \nabla f(x)\cdot (y-x),\ \nabla f(y)\cdot (x-y) >0 $$ Hence $$ (\nabla f(x) -\nabla f(y))\cdot (y-x) >0 $$

It is a contradiction.

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