# Question concerning SVMs in machine learning course CS229 by Andrew Ng

On page 12 in https://see.stanford.edu/materials/aimlcs229/cs229-notes3.pdf, the author uses the claim that the gradient of the lagrangian with respect to the non-constraint variables is zero. Why is this true? When we're trying to minimize the lagrangian for fixed $$\alpha$$ - a constraint variable, how does it follow that the minimum is at a point of local minima/stationary point, and the function isn't perhaps unbound? Does it follow from convexity of the lagrangian (convexity implies that a local minima is global)? Or does it have something to do with KKT conditions?

Could someone give me a counterexample of type:

$$\textrm{max}_{\lambda} \textrm{min}_x f(x,\lambda) \neq \textrm{max}_{\lambda} \tilde{f}(\lambda)$$

where $$\tilde{f}{(\lambda)}$$ is chosen to be $$f(x_0,\lambda)$$ for some $$x_0$$ local minima of $$f$$ with respect to $$\lambda$$ being fixed (assuming that a local minima exists for each $$\lambda$$). I think a function like $$f(x,\lambda) = x^3$$ could be an example.

Alternatively, in the equation above, is there some sort of neat condition on $$\tilde{f}$$ could imply an equality? Suppose that the actual extrema is reached at $$f(x^*,\lambda^*)=M$$ and that we're interested in somehow defining $$\tilde{f}$$ based on $$f$$, using $$\tilde{f}(\lambda) = f(x_0,\lambda)$$ for some choice of $$x_0$$.

Working with $$x_0$$ being chosen as some (existing) stationary point with respect to a fixed $$\lambda$$, and assuming that $$x^*$$ is the unique stationary point with respect to $$\lambda^*$$ where minimum is reached with respect to fixed $$\lambda^*$$, all that's necessary is for the value of $$f$$ in all the other stationary points to be lower than $$M$$. But I'm not sure if there's some sort of neat condition that would imply that.