Vtyjkyuk

Say we have a problem such as minimize $f(x)$ such that $h(x)=0$ and $g(x) leq0$. Let the minimum achieved under these constraints be $f(x^*) = p^*$. My question is:

If $f(x)$ is convex, are $p^*$ and $x^*$ unique? Why, why not? If in the general case they are not, are there any special/notable cases or any conditions on $f, g, h$ under which the solution is unique?

The minimizer $x^*$ is not in general unique. Try $f(x) = c$ where $c$ is a constant function. In fact, the minimum $p^*$ is not necessarily unique in general. The constraints $h(x)$ and $g(x)$ give a description of the set where you search for your minimum. If this $mathbfset$ is convex, minimizing a convex function $f(x)$ over this set leads to a unique minimum $p^*$.

The problem will be convex (that is to say, you'll be minimizing over a convex set) if, for example:

• The $g(x)$ are convex functions.


• The $h(x)$ are affine functions.

As for when the minimizer is unique, there are a whole host of conditions that can make this true. When $f(x)$ is "strongly convex", which is a stronger condition than convexity, and you minimize it over a convex set (such as, for example, constraints with the above criterion) the minimizer $x^*$ is unique. A unique minimizer, of course, guarantees a unique minimum.

Strong convexity of $f(x)$ isn't the only way to get a unique minimizer, but its a common one and has a nice geometric intuition. For a second differentiable function $f$, strong convexity is equivalent to having the eigenvalues of the Hessian bounded from below by a positive number. In general, it essentially says that at every point on the domain, $f(x) $ can be bounded from below by a term that's quadratic in $x$ (aka a parabola).