Suppose I have the following optimization problem

$$ \min\limits_{\mathbf{x},\mathbf{y}} f(\mathbf{x},\mathbf{y}) \tag{1} $$

It is already known that the target function $f$ is continuous and differentiable, which has a unique stationary point, and this stationary point is also a global minimum. However, $f$ is not convex. Here, I guess that $f$ is a strict quasi-convex function, and my question is based on this.

From the KKT condition, I find that the necessary condition for the optimal point is $ \mathbf x=g(\mathbf y) $.

  1. Is my guess about this function correctly (just based on the properties discussed above)? If it is, then, we can use convex optimization algorithms to find a global optimal point of $f$ since $f$ is strictly quasi-convex.

  2. Can we substitute $\mathbf x=g(\mathbf y)$ into the target function and solve the consequent optimization problem. In words, is the solution to the following optimization problem equivalent to the original optimization problem in $(1)$?

$$ \min\limits_{\mathbf{y}} f(g(\mathbf y),\mathbf{y}) $$

Any helpful comments are appreciated! ^_^


2 Answers 2


Suppose you have the optimization problem

$$ \min_{x,y} f(x,y),\quad (x,y)\in X\subset\mathbb{R}^n\times \mathbb{R}^m \tag{1} $$

Suppose also that the function $f$ is continuous and differentiable, which has a unique stationary point, and this stationary point is also a global minimum.

Since $f$ is diferentiable you can find that $$\frac{\partial f(x,y)}{\partial x}=0$$ gives the necessary condition for the optimal point. Sometimes it gives you one function $ x=g( y)$. But it can gives you more the one implicit functions.

If you have only one function $x=g(y)$, you can substitute into the function and solve the problem $(1)$ as $$ \min_{y} \phi(y),\qquad \phi(y)=f(g( y),{y}),$$ (please see this topic on math.stackexchange).

You can find more related discussions searching for "\(\min_yf(x^*(y),y)\) " on SearchOnMath, like this topic on math.stackexchange.


  1. Theorem. Let $f:X\to \mathbb{R}$ be differentiable on the open convex set $X\subset \mathbb{R}^n\times \mathbb{R}^m$. Then $f$ is quasiconvex on $X$ if and only if $$u,v\in X,\,f(u)\leq f(v)\Longrightarrow \nabla f(v)\cdot(u-v)\leq 0. $$

a) If you can verify the previous theorem then $f$ is a quasi-convex function.

b) If $f$ was a quasi-convex function. It is not clear why shoud be $\phi(y)=f(g( y),{y})$ convex.

  • $\begingroup$ Thanks for your time! I understand the sequential optimization procedure now. So in my problem, I can substitute $x=g(y)$ into the original problem, but the subsequent problem may not be equivalent to the original one. This depends on my problem and the structure of $x=g(y)$. $\endgroup$
    – HiNull
    May 6 at 3:14

The function $$ f(\mathbf{x},\mathbf{y})=-\cos(\mathbf{x}^2+\mathbf{y}^2)\,e^{-\mathbf{x}^2-\mathbf{y}^2} $$ has a global minimum at $\mathbf{x}=\mathbf{y}=0$ but is not strictly quasi-convex because it is not even quasi convex. For $c\in(-1,+1)$ the set $\{f\le c\}$ is a union of sets of the form $\{\mathbf{x}^2+\mathbf{y}^2<R\}\setminus\{\mathbf{x}^2+\mathbf{y}^2<r\}\,.$

  • $\begingroup$ Thanks for your comment. Yes, the function you write is not quasi-convex, but it does not have a unique stationary point. The point x=y=0 is the global minimizer, but there are also other stationary points. In my function, the stationary point is unique. $\endgroup$
    – HiNull
    Apr 14 at 13:58

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