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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.

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  • $\begingroup$ 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

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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.

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