I have been reading about convex optimization. We have:

minimize $f(x)$ s.t. $h(x) = 0$, $g(x) \le 0$, $x \in X$

It's Lagrangian dual is:

maximize $\phi(\lambda,\mu)$ s.t. $\mu \ge 0$, where $\phi(\lambda,\mu) = \inf[f(x) + \lambda' h(x) + \mu 'g(x)]$

I don't understand why $\mu$ must be greater than zero. Can anyone please explain?


2 Answers 2


For this post: $\mathcal{L} = \mathcal{L}(\lambda, \mu) = f(x) + \lambda h(x) + \mu g(x)$

You want $\mathcal{L}(\lambda,\mu) \le f(x)$ for all valid $x$.
A valid $x$ is one that satisfies $h(x) = 0$, $g(x) \le 0$.

If $x$ satisfies $h(x) = 0$, then the term $\lambda h(x)$ in $\mathcal{L}$ is zero regardless of $\lambda$.

If $x$ satisfies $g(x) \le 0$, the the term $\mu g(x)$ in $\mathcal{L}$ is negative or zero as long as $\mu \ge 0$.

This is required for $\mathcal{L} = f(x) + \lambda h(x) + \mu g(x) \le f(x)$ for valid $x$. This makes the Lagrangian a lower bound on $f(x)$ for valid $x$.


Ignore the equality constraint for a moment and only think of the inequality constrained problem (assuming all functions are at least once continuously differentiable). Now think about what it means for a point $x^\ast$ to be a minimizer of your problem:

  • If it lies in the interior of the domain described by $g(x)\le 0$ (i.e.: $g(x^\ast)<0$), then $x^\ast$ can only be a minimizer if $\nabla f(x^\ast)=0$ because otherwise going in direction $-\nabla f$ would lower the function value. So, in this case, the optimality condition $\nabla f(x^\ast) + \mu^\ast \nabla g(x^\ast)=0$ is satisfied only if $\mu^\ast=0$.

  • The consider the case where the minimizer $x^\ast$ lies at the boundary of the feasible domain, i.e., $g(x^\ast)=0$. In this case, for $x^\ast$ to be a minimizer, the gradient of $f$ at $x^\ast$ must point perpendicularly to the boundary and into the feasible set since, otherwise, going into the domain or along some direction along the boundary a little bit would lower the function value. On the other hand, the direction perpendicular to the boundary in the outward direction is given by $\nabla g(x^\ast)$. Consequently, we can express the optimality condition as saying that $\nabla f(x^\ast)$ must be equal to a negative multiple of $\nabla g(x^\ast)$. In other words, $\nabla f(x^\ast) + \mu \nabla g(x^\ast) = 0$ with some positive multiplier $\mu$.

Makes sense? It's actually a pretty intuitive argument if you think of it in terms of geometry.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.