# Weak Duality

• Lagrange dual problem

• Weak duality and minimax inequality

• Examples

## Lagrange dual problem

### Primal problem

In this section, we consider a possibly non-convex optimization problem

where the functions We denote by the domain of the problem (which is the intersection of the domains of all the functions involved), and by its feasible set.

We will refer to the above as the primal problem, and to the decision variable in that problem, as the primal variable. One purpose of Lagrange duality is to find a lower bound on a minimization problem (or an upper bounds for a maximization problem). Later, we will use duality tools to derive optimality conditions for convex problems.

### Lagrangian

To the problem we associate the Lagrangian , with values

where the new variables , are called Lagrange multipliers, or dual variables.

We observe that, for every feasible , and every , is bounded below by :

The Lagrangian can be used to express the primal problem (ref{eq:convex-pb-L11}) as an unconstrained one. Precisely:

where we have used the fact that, for any vector , %, we have

### Lagrange dual function

We then define the Lagrange dual function (dual function for short) as the function with values

Note that, since is the point-wise minimum of affine functions ( is affine for every ), it is concave. Note also that it may take the value .

From the bound above, by minimizing over in the right-hand side, we obtain

which, after minimizing over the left-hand side, leads to the lower bound

### Lagrange dual problem

The best lower bound that we can obtain using the above bound is , where

We refer to the above problem as the dual problem, and to the vector as the dual variable. The dual problem involves the maximization of a concave function under convex (sign) constraints, so it is a convex problem. The dual problem always contains the implicit constraint .

### Case with equality constraints

If equality constraints are present in the problem, we can represent them as two inequalities. It turns out that this leads to the same dual, as if we would directly use a single dual variable for each equality constraint, which is not restricted in sign. To see this, consider the problem

We write the problem as

Using a multiplier for the constraint , we write the associated Lagrangian as

where does not have any sign constraints.

Thus, inequality constraints in the original problem are associated with sign constraints on the corresponding multipliers, while the multipliers for the equality constraints are not explicitly constrained.

## Weak duality and minimax inequality

### Weak duality theorem

We have obtained:

Theorem: weak duality

For a general (possibly non-convex) problem

weak duality holds: .

### Geometric interpretation

Assume that there is only one inequality constraint in the primal problem (), and let (

{cal G} := left{ (f_1(x),f_0(x))  :  x in mathbf{R}^n right}. ) We have

and

If the minimum is finite, then the inequality defines a supporting hyperplane, with slope , of at . (See Figs. 5.3 and 5.4 in BV,p.233.)

### Minimax inequality

Weak duality can also be obtained as a consequence of the following minimax inequality, which is valid for any function of two vector variables , and any subsets , :

To prove this, start from

and take the minimum over on the right-hand side, then the maximum over on the left-hand side.

Weak duality is indeed a direct consequence of the above. To see this, start from the unconstrained formulation (ref{eq:pb-primal-unconstr-L11}), and apply the above inequality with the Lagrangian of the original problem, and the vector of Lagrange multipliers.

### Interpretation as a game

We can interpret the minimax inequality result in the context of a one-shot, zero-sum game. Assume that you have two players A and B, where A controls the decision variable , while B controls . We assume that both players have full knowledge of the other player’s decision, once it is made. The player A seeks to minimize a payoff (to player B) , while B seeks to maximize that payoff. The right-hand side in (ref{eq:min-max-theorem-L11}) is the optimal pay-off if the first player is required to play first. Obviously, the first player can do better by playing second, since then he or she knows the opponent’s choice and can adapt to it.

## Examples

### Linear optimization problem, inequality form

Consider the LP in standard inequality form

where , , and the inequality in the constraint is interpreted component-wise.

The Lagrange function is (

{cal L}(x,lambda) = c^Tx + lambda^T(Ax-b) ) and the corresponding dual function is

The dual problem is an LP in standard (sign-constrained) form, just as the primal problem was an LP in standard (inequality) form.

Weak duality implies that

for every such that , . This property can be proven directly, by replacing by in the left-hand side of the above inequality, and exploiting and .

### Linear optimization problem, standard form

We can also consider an LP in standard form:

The equality constraints are associated with a dual variable that is not constrained in the dual problem.

The Lagrange function is

and the corresponding dual function is

This is an LP in inequality form.

### Minimum Euclidean distance problem

Consider the problem of minimizing the Euclidean distance to a given affine space:

where , . We assume that is full row rank, or equivalently, . The Lagrangian is

and the Lagrange dual function is

In this example, the dual function can be computed analytically, using the optimality condition . We obtain , and

The dual problem expresses as

The dual problem can also be solved analytically, since it is unconstrained (the domain of is the entire space ). We obtain , and

We have thus obtained the bound .

### A non-convex boolean problem

For a given matrix , we consider the problem

In this maximization problem, Lagrange duality will provide an upper bound on the problem. This is called a ‘‘relaxation’’, as we go above the true maximum, as if we’d relax (ignore) constraints.

The Lagrangian writes

where .

To find the dual function, we need to maximize the Lagrangian with respect to the primal variable . We express this problem as

The last inequality holds if and only if

Hence the dual function is the optimal value of an SDP in one variable:

We can solve this problem explicitly:

The dual problem involves minimizing (that is, getting the best upper bound) the dual function over the variable :

The above is an SDP, in variable . Note that is automatically enforced by the PSD constraint.

The Lagrange relaxation of the primal problem can be interpreted geometrically, as follows.

 Geometric interpretation of dual problem in the boolean quadratic problem. For , , consider the ellipsoids The primal problem amounts to find the smallest for which the ellipsoid contains the ball . Note that for every , contains the ball . To find an upper bound on the problem, we can find the smallest for which there exist such that . The latter condition is precisely , .