# Can we substitute this KKT condition into this optimization problem to reformulate the optimization problem?

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 $$(1)$$?

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

• It is dangerous to confuse $f(\mathbf{x},\mathbf{y})$ and $f$. Apr 14, 2022 at 14:52
• Apr 28, 2022 at 18:13

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

• 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. Apr 14, 2022 at 13:58

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.

Note:

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.

• 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 $x=g(y)$. May 6, 2022 at 3:14