Notation:
$J[y]$ is a functional of $y=y(x) $
$||y||_0 = \max_I |y|$ for some closed interval $I$
$||y||_1 = \max_I |y| + \max_I |y'|$
$\mathbb{F} $ is the set of continuous functions
$\mathbb {D}_1$ is the set of differentiable functions with continuous fist derivative.
$J [y]$ has a weak extremum for $y=\hat{y} $ if there exists an $\epsilon > 0$ s.t. $J[y]-J [\hat {y}] $ has the same sign for all $y $ in the domain of $J $ which satisfy $||y-\hat{y}||_1 < \epsilon $
$J [y]$ has a strong extremum for $y=\hat{y} $ if there exists an $\epsilon > 0$ s.t. $J[y]-J [\hat {y}] $ has the same sign for all $y $ in the domain of $J $ which satisfy $||y-\hat{y}||_0 < \epsilon $
This makes very little sense to me. It's clear that if $||y-\hat{y}||_1 < \epsilon$, then $||y-\hat{y}||_0<\epsilon $
so it seems that the weak condition is actually implying the strong condition. The only rationale I can come up with is that the $\mathbb{D}_1 \subset \mathbb{F} $, so the strong condition has more potential functions to 'ween out'.
That is to say, for an arbitrary $\hat {y} $, any function that has the potential to satisfy $||y-\hat {y}||_0 < \epsilon $ also has the potential to satisfy $||y-\hat{y}||_1 < \epsilon$, despite the second condition being a stronger test for closeness (I suppose the function is our concern for extrema, not closeness)
As you already observed, the implication $\|y-\hat y\|_1<\epsilon$ implies $\|y-\hat y\|_0$. Hence, the set of test functions for 'weak extremum' is a subset of the test functions for 'strong extremum'. Hence, the latter is a stronger condition.