Consider the problem of maximising a smooth function subject to the inequality constraint that $g(x) \leq b$. The complementary slackness condition says that
$$ \lambda[g(x) - b] = 0$$
It is often pointed out that, if the constraint is slack at the optimum (i.e. $g(x^*) < b$), then this condition tells us that the multiplier $\lambda = 0$. I agree with this. However, it has also been said that, if the constraint 'binds' (which implies that $g(x^*) - b = 0$), we must have $\lambda > 0$. Is this true? As a logical matter, it is not immediately implied by the complimentary slackness condition: we could have both $g(x^*) - b = 0$ and also $\lambda = 0$.
Edit: it has been demonstrated here why we can have both $\lambda = 0$ and $g(x^*) - b = 0$ (thanks to @markleeds for the pointer). I am wondering, however, whether we can have $\lambda = 0$ while the constraint also binds (i.e. makes a difference to the solution -- note that this is subtly different from the constraint holding with equality). I suspect that the answer is 'no' given that $\lambda$ reflects the effect of slightly relaxing the constraint on the objective function. However, I would appreciate confirmation of this.