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.

  • $\begingroup$ Not really my field, but I'd say assume there are two and then restrict $f$ to the line connecting the two critical points. By positive definiteness, you should get that the restriction has an everywhere non-zero second derivative, which would be a contradiction. $\endgroup$ Commented Dec 31, 2017 at 5:23
  • $\begingroup$ What do you mean by a critical point? Zero derivative? $\endgroup$
    – copper.hat
    Commented Dec 31, 2017 at 5:24
  • $\begingroup$ The definition is wrong, according to yours, $x^4$ is not strictly convex? $\endgroup$
    – max_zorn
    Commented Dec 31, 2017 at 5:28
  • $\begingroup$ @copper.hat Yes $\endgroup$
    – GouldBach
    Commented Dec 31, 2017 at 5:34
  • $\begingroup$ @Callus Thanks then use M.V.T? $\endgroup$
    – GouldBach
    Commented Dec 31, 2017 at 5:35

1 Answer 1


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.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .