Weak and strong extremum

Remarks

Since $\norm{y}_0 \le \norm{y}_1$, every strong local minimum is a weak local minimum.

Example: let $J(y) = \int_0^1 y'^2 \, dx$ on $\mathcal{A} = \set{y \in C^1([0, 1]) : y(0) = 0, \ y(1) = 1}$, and $y_0(x) = x$. For $y \in \mathcal{A}$ write $y' = 1 + \eta$; then $\int_0^1 \eta \, dx = 0$, so $J(y) = 1 + \int_0^1 \eta^2 \, dx \ge J(y_0)$. So $y_0$ is a global, hence strong, minimum.

Non-example: let $J(y) = \int_0^1 y'^3 \, dx$ on the same class, and $y_0(x) = x$. With $y' = 1 + \eta$ one gets $J(y) = 1 + \int_0^1 \eta^2 (3 + \eta) \, dx$, which is at least $1$ when $\abs{\eta} < 3$; so $y_0$ is a weak local minimum. It is not a strong local minimum. Fix $h > 0$, let $N$ be large and $\delta = h/N$, and let $y$ have slope $-N$ on $[0, \delta]$ and slope $(1 + h)/(1 - \delta)$ on $[\delta, 1]$, with corners rounded off. Then $\norm{y - y_0}_0 \le h + \delta$, while $J(y)$ is close to $-N^2 h + (1 + h)^3/(1 - \delta)^2$, which tends to $-\infty$ as $N \to \infty$.