I need to prove the following statement. Let $ f:\mathbb{R}^{n}\to \mathbb{R}$ be a differentiable function. If $\forall x,y\in$ dom$(f)$, $f(y)\le f(x)\Rightarrow \nabla f(x)^{T}(y-x)\le 0$, then $f$ is quasiconvex function
The idea of the proof is the following. For the converse let us suppose that $f$ is not a quasiconvex function. Then there exist $x,y\in$dom$(f),$ $f(y)\le f(x),$ there exists $\lambda\in(0,1)$ such that $f(\lambda x+(1-\lambda)y)>\max\{f(x),f(y)\}=f(x)$. Let us define a function $g$ as follows: $g(\lambda)=f(\lambda x+(1-\lambda)y)$. Then $g(\lambda)>g(1)$.
Now, I do not understand the next steps. By the Mean value theorem there exists a $\tau\in (0,1)$ such that $g`(\tau)<0,$ and thus $\nabla f(z)^{T}(x-z)>0$, where $z=g(\tau)$. Could you explain me the part of that proof? Thank you.
Here is my proof that does not use the mean value theorem but some basic calculus analysis. I hope this can help you a bit about the proof of quasi-convexity that bothers me quite a while. Proof the quasi-convexity of $f$ by contradiction. Firstly, we assume that the set $A =\{\lambda |f(\lambda x + (1 - \lambda)y) > f(x) \ge f(y), \lambda \in (0,1) \}$ is not empty. Then by the assumption $\nabla f(\lambda x + (1 - \lambda)y)^{T}(x - (\lambda x + (1 - \lambda)y))\le 0$ and $\nabla f(\lambda x + (1 - \lambda)y)^{T}(y - (\lambda x + (1 - \lambda)y))\le 0$. $\Rightarrow$ $\nabla f(\lambda x + (1 - \lambda)y)^{T}(x - y)\le 0$ and $\nabla f(\lambda x + (1 - \lambda)y)^{T}(y - x)\le 0$. Which is equivalent to for any $\lambda \in A$, we have $\nabla f(\lambda x + (1 - \lambda)y) = 0$. Next we proof the contradiction part by prooving that the minimum $\lambda \in A$ violate the previous finding. let $\lambda^{*}$ be the minimum element in $A$, we declare that $\nabla f(\lambda^{*} x + (1 - \lambda^{*})y)^{T}(y - x) < 0$. If not, i.e. $\nabla f(\lambda^{*} x + (1 - \lambda^{*})y)^{T}(y - x) \ge 0$. Then direction $y - x$ is an ascending direction for $f$ at the point $\lambda^{*} x + (1 - \lambda^{*})y$, for it is formed an accute angle with the gradient. Then there exist $\epsilon > 0$ such that $f(\lambda^{*} x + (1 - \lambda^{*})y + \epsilon(y-x)) = f((\lambda^{*}-\epsilon) x + (1 - (\lambda^{*} - \epsilon))y) > f(\lambda^{*} x + (1 - \lambda^{*})y)$. WHich implies that $\lambda - \epsilon \in A$, contradict to the selection of $\lambda$.