# Solving constrained minimisation problem using unconstrained optimization of the generalized Lagrangian

My textbook, Deep Learning by Goodfellow, Bengio, and Courville, says the following in a section on constrained optimization:

The Karush-Kuhn-Tucker (KKT) approach provides a very general solution to constrained optimization. With the KKT approach, we introduce a new function called the generalized Lagrangian or generalized Lagrange function.

To define the Lagrangian, we first need to describe $$\mathbb{S}$$ in terms of equations and inequalities. We want a description of $$\mathbb{S}$$ in terms of $$m$$ functions $$g^{(i)}$$ and $$n$$ functions $$h^{(j)}$$ so that $$\mathbb{S} = \{ \boldsymbol{\mathcal{x}} \mid \forall i, g^{(i)}(\boldsymbol{\mathcal{x}}) = 0 \ \text{and} \ \forall j, h^{(j)} (\boldsymbol{\mathcal{x}}) \le 0 \}$$. The equations involving $$g^{(i)}$$ are called the equality constraints, and the inequalities involving $$h^{(j)}$$ are called inequality constraints.

We introduce new variables $$\lambda_i$$ and $$\alpha_j$$ for each constraint, these are called the KKT multipliers. The generalized Lagrangian is then defined as

$$L(\boldsymbol{\mathcal{x}}, \boldsymbol{\lambda}, \boldsymbol{\alpha}) = f(\boldsymbol{\mathcal{x}}) + \sum_i \lambda_i g^{(i)} (\boldsymbol{\mathcal{x}}) + \sum_j \alpha_j h^{(j)}(\boldsymbol{\mathcal{x}}) \tag{4.14}$$

We can now solve a constrained minimisation problem using unconstrained optimization of the generalized Lagrangian. As long as at least one feasible point exists and $$f(\boldsymbol{\mathcal{x}})$$ is not permitted to have value $$\infty$$, then

$$\min_{\boldsymbol{\mathcal{x}}} \max_{\boldsymbol{\mathcal{\lambda}}} \max_{\boldsymbol{\mathcal{\alpha, \alpha}}\ge 0} L(\boldsymbol{\mathcal{x}}, \boldsymbol{\mathcal{\lambda}}, \boldsymbol{\mathcal{\alpha}}) \tag{4.15}$$

has the same optimal objective function value and set of optimal points $$\boldsymbol{\mathcal{x}}$$ as

$$\min_{\boldsymbol{\mathcal{x}} \in \mathbb{S}} f(\boldsymbol{\mathcal{x}}). \tag{4.16}$$

This follows because any time the constraints are satisfied,

$$\max_{\boldsymbol{\mathcal{\lambda}}} \max_{\boldsymbol{\mathcal{\alpha, \alpha}}\ge 0} L(\boldsymbol{\mathcal{x}}, \boldsymbol{\mathcal{\lambda}}, \boldsymbol{\mathcal{\alpha}}) = f(\boldsymbol{\mathcal{x}}),$$

while any time a constraint is violated,

$$\max_{\boldsymbol{\mathcal{\lambda}}} \max_{\boldsymbol{\mathcal{\alpha, \alpha}}\ge 0} L(\boldsymbol{\mathcal{x}}, \boldsymbol{\mathcal{\lambda}}, \boldsymbol{\mathcal{\alpha}}) = \infty$$

these properties guarantee that no infeasible point can be optimal, and that the optimum within the feasible points is unchanged.

I'm having difficulty understanding how $$\min_{\boldsymbol{\mathcal{x}}} \max_{\boldsymbol{\mathcal{\lambda}}} \max_{\boldsymbol{\mathcal{\alpha, \alpha}}\ge 0} L(\boldsymbol{\mathcal{x}}, \boldsymbol{\mathcal{\lambda}}, \boldsymbol{\mathcal{\alpha}})$$

has the same optimal objective function value and set of optimal points $$\boldsymbol{\mathcal{x}}$$ as

$$\min_{\boldsymbol{\mathcal{x}} \in \mathbb{S}} f(\boldsymbol{\mathcal{x}})$$

Specifically, I am not seeing how the latter claim that any time the constraints are satisfied,

$$\max_{\boldsymbol{\mathcal{\lambda}}} \max_{\boldsymbol{\mathcal{\alpha, \alpha}}\ge 0} L(\boldsymbol{\mathcal{x}}, \boldsymbol{\mathcal{\lambda}}, \boldsymbol{\mathcal{\alpha}}) = f(\boldsymbol{\mathcal{x}}),$$

while any time a constraint is violated,

$$\max_{\boldsymbol{\mathcal{\lambda}}} \max_{\boldsymbol{\mathcal{\alpha, \alpha}}\ge 0} L(\boldsymbol{\mathcal{x}}, \boldsymbol{\mathcal{\lambda}}, \boldsymbol{\mathcal{\alpha}}) = \infty$$

I would greatly appreciate it if people could please take the time to clarify this.

Starting with:

$$L(\boldsymbol{\mathcal{x}}, \boldsymbol{\lambda}, \boldsymbol{\alpha}) = f(\boldsymbol{\mathcal{x}}) + \sum_i \lambda_i g^{(i)} (\boldsymbol{\mathcal{x}}) + \sum_j \alpha_j h^{(j)}(\boldsymbol{\mathcal{x}}) \tag{4.14}$$

If the constraints are satisfied, then $$g^{(i)}(x)=0$$ and $$h^{(j)}(x)\leq0$$. Therefore, the terms with $$\lambda$$ all vanish, and the terms with $$\alpha$$ attain their maximum over $$\alpha_j$$ at $$\alpha_j=0$$ (because the last term cannot be positive), so also those terms vanish, leaving you with $$f(x)$$.

On the other hand, suppose a constraint is not satisfied. If $$g^{(i)}(x)\neq 0$$ for some $$i$$, you can let $$\lambda_i g^{(i)}(x)$$ go to infinity by letting $$\lambda_i$$ go to $$\infty$$ if $$g^{(i)}(x)>0$$, and $$\lambda_i \to -\infty$$ otherwise. Similarly, if $$h^{(j)}(x)>0$$ for some $$i$$, you can let $$\alpha_j h^{(j)}(\boldsymbol{\mathcal{x}})$$ go to $$\infty$$ by letting $$\alpha_j \to \infty$$.

If $$x\in S$$ then $$f(x) \ge L(x,\lambda,\alpha)$$ for all $$\lambda,\alpha$$ with $$\alpha\ge0$$.

If $$x\not\in S$$ then one of the constraints is violated,i.e., $$h_i(x)\ne0$$ or $$g_j(x)\>0$$ for some $$i$$ or $$j$$. By taking the corresponding multiplier $$\lambda_i$$ or $$\alpha_j$$ large enough, one see $$\sup_{\lambda,\alpha\ge0} L(x,\lambda,\alpha)=+\infty$$.

• But how do you reconcile your first claim with the claim that $\max_{\boldsymbol{\mathcal{\lambda}}} \max_{\boldsymbol{\mathcal{\alpha, \alpha}}\ge 0} L(\boldsymbol{\mathcal{x}}, \boldsymbol{\mathcal{\lambda}}, \boldsymbol{\mathcal{\alpha}}) = f(\boldsymbol{\mathcal{x}})$? Oct 8, 2019 at 14:05
• But that still contradicts $\max_{\boldsymbol{\mathcal{\lambda}}} \max_{\boldsymbol{\mathcal{\alpha, \alpha}}\ge 0} L(\boldsymbol{\mathcal{x}}, \boldsymbol{\mathcal{\lambda}}, \boldsymbol{\mathcal{\alpha}}) = f(\boldsymbol{\mathcal{x}})$, doesn't it? That's why I'm confused. Oct 8, 2019 at 14:08
• Also with regards to your second claim, why must it be true? It isn't obvious to me that a violation constraint necessarily results in $\max_{\boldsymbol{\mathcal{\lambda}}} \max_{\boldsymbol{\mathcal{\alpha, \alpha}}\ge 0} L(\boldsymbol{\mathcal{x}}, \boldsymbol{\mathcal{\lambda}}, \boldsymbol{\mathcal{\alpha}}) = \infty$, as is claimed? Oct 8, 2019 at 14:10
• If $x$ is feasible the $f(x) \ge L(x,\dots)$ and equality is attained for $\lambda$ and $\alpha$ being zero
– daw
Oct 8, 2019 at 14:10
• Equality is between $f(x)$ and $\max_{\alpha,\lambda} L$ not between $f(x)$ and $L$ for arbitrary values of $\lambda,\alpha$.
– daw
Oct 10, 2019 at 11:00