# Local minimum in undefined points in Hessian and first derivative of a function

It might seem naive but consider having a convex function like $$|x|$$ for which $$f'(x)$$ is not defined in 0, (is it? because $$f'(x)= \frac{x}{|x|}$$.) Now what do we say about its local minimum? Because for a local minimum we have $$f'(x) =0$$ and $$f'(x) >=0$$. Which don't hold here.

I think you're mixing up two somewhat-related but distinct theorems:

1. If $$f$$ is a convex function, every local minimum of $$f$$ is also a global minimum.

2. If $$f$$ is $$C^2$$ and $$f'(x)=0; f''(x)>0$$ then $$f$$ has a local minimum at $$x$$.

This second fact is a sufficient condition on the existence of a local minimum, but not a necessary condition. (This sufficient condition generalizes to higher dimension, where the condition instead is that $$\nabla f(x) =0$$ and $$Hf(x)$$ is positive-definite. It is an immediate consequence of Taylor's theorem.)

Even for smooth functions, sometimes a local minimum does not have positive second derivative. Consider for instance $$f(x) = x^4$$, which clearly has a minimum at $$x=0$$ but $$f''(0)=0$$.

The function $$f(x) = |x|$$ is not a contradiction to the above: it is convex so fact (1) applies. It is not in $$C^2$$, but even if it were, it would not be a contradiction to fact (2) since the latter is a sufficient condition, not an "if-and-only-if" condition.

• Thanks it was really helpful. Feb 16, 2021 at 3:22