For a strictly convex function $f:\mathbb{R}^n\to \mathbb{R}$ critical point is unique. Definition. A smooth function $f:\mathbb{R}^n\to \mathbb{R} $ is strictly convex 
if its Hessian matrix $[\frac{\partial^2f}{\partial x^i \partial x^j}] $ as a quadratic form is strictly positive definite.
I want to show that if f has a critical point then the critical point is unique. 
I would appreciate any help. Thank you. 
 A: It is straightforward to verify that a differentiable convex function 
on $\mathbb{R}^n$ satisfies
$f(x+h) -f(x) \ge {\partial f(x) \over \partial x} h$ for all $h$.
In particular, $x$ is a global $\min$ of $f$ iff ${\partial f(x) \over \partial x} = 0$.
Suppose $x,x'$ are two critical points, then ${\partial f(x) \over \partial x} = {\partial f(x') \over \partial x} = 0$ and
so $x,x'$ are global $\min$.
Since $f$ is convex, every point along $[x,x']$ is also a global $\min$ and hence ${\partial f(y) \over \partial x} = 0$ for all
$y \in [x,x']$.
By considering the 2nd order Taylor expansion of $\phi(t) = f(x+ t(x'-x))$ we see that
$\phi(1) -\phi(0) = 0 = {1 \over 2}\langle x'-x, {\partial^2 f(x+\xi(x'-x)) \over \partial x^2} (x'-x) \rangle $, which 
contradicts the positive definiteness of the Hessian.
Note: If the function is strictly convex, then the Hessian argument
is unnecessary. If $f(x)=f(y)$ and $x \neq y$ then
$f({x+y \over 2}) < f(x)$. Hence there cannot be two distinct minimisers and hence there cannot be two distinct critical points.
