Showing existence of solution by positive definiteness/convexity

Question

For a physics problem, I am considering the following problem: I have a certain function, $S: \mathbb{R}^M \rightarrow \mathbb{R}$, of which the critical points, given by $$ \frac{\partial S}{\partial\lambda_{\gamma}}=0 $$ ($\lambda\in\mathbb{R}^M$ so M equations, I denote a component of this vector by $\lambda_{\gamma}$). In fact, the critical points form an equation I am actually interested in. I want to prove that these equations have a unique, existing solution. The reference I am following, shows that $$ \sum_{\alpha,\beta}v_{\alpha}v_{\beta}\frac{\partial^2 S}{\partial\lambda_{\alpha}\partial\lambda_{\beta}}>0¸ $$ i.e. S is positive definite in lambda. It then concludes that the equations thus have a unique, existing solution.

I am somewhat confused by this. I know of a theorem that states that if you have a local minimum of a function and that function is strictly convex ($\Leftrightarrow$ it is positive definite), then the local minimum is in fact a global minimum. But in this case we only know that the equations are extrema of S. Can anyone shed his/her light on this?

Answer 1

You can't yet quite conclude $S$ has a unique global minimum: for instance the one-dimensional function $S(\lambda) = e^\lambda$ has positive-definite Hessian but no global minimum.

You can conclude any of the following:

• Every critical point of $S$ is a strict local minimum.
• $S$ has at most one such critical point. In addition to the positive-definiteness of $S$, this requires that $S$ be defined on a convex set (in this case $\mathbb{R}^n$). Let $\lambda^1$ and $\lambda^2$ be two distinct critical points, and draw the line $\gamma(t) = t\lambda_2 + (1-t)\lambda_1$ between them. Then $$\frac{d^2 S(\gamma(t))}{dt^2} = (\lambda_2-\lambda_1)^T\, HS(\gamma[t])\, (\lambda_2-\lambda_1) \geq 0$$ so $$\frac{d S(\gamma(t))}{dt}(t) = \frac{d S(\gamma(t))}{dt}(0) + \int_0^t \frac{d^2 S(\gamma(t))}{dt^2}(s)\,ds \geq \frac{d S(\gamma(t))}{dt}(0) = 0.$$ Similarly $$\frac{d S(\gamma(t))}{dt} \leq \frac{d S(\gamma(t))}{dt}(1) = 0$$ and $S(\gamma)$ is constant, contradicting the fact that $\lambda^1$ must be a strict minimum.
• Any local minimum of $S$ is a global minimum. A calculation along a path much as the one above shows that if $\lambda^1$ is a local minimum and $\lambda^2$ is any point, $S(\lambda^2) \geq S(\lambda^1)$.
• Let $g(a)$ be the sub-level-set of $S$, i.e. $$g(a)=\{\lambda\in\mathbb{R^n}\vert S(\lambda) \leq a\}.$$ If $g(a)$ is nonempty and bounded for some $a$, then $S$ has exactly one critical point, which is the global minimum. This is because $S$ must attain its minimum on the compact set $g(a)$. Depending on the form of $S$, you may be able to show that $S(\lambda) \to \infty$ as $\|\lambda\| \to \infty$ and then use this fact to prove that a global minimum of $S$ must exist.