I have this definition of a local minimum:
We say that $u$ is a local minimum of $f$ is there exist a neighborhood $V$ of $u$ such that for all $v\in V$ $f(v)\geq f(u).$
So we say that $u$ is not a local minimum if for all neighborhood $V$ of $u$ there exist $v\in V$ such that $f(v)<f(u).$ right ?
But look at this proof, the authors proved that $0$ is not a minimum, but they say there exist $\delta$ so it is the same as to say that there exist a neighborhood, where is the problem ?
My question is What they mean by "non-minimum" a non global minimum or a non local minimum ?
Thank you.
They are providing points $u$ arbitrarily close to $0$ (but not arbitrarily close to $\theta$, it seems) with $J(u)<J(\theta)$. So they are showing that $\theta$ cannot be a global minimum point. Their notation is quite confusing, as $u$ is being to represent both a function and a real number in $[-\delta,\delta]$.