= Solution
Choose a smooth compactly supported <mollifier> of integral one. On every compact subset of $\Omega$, and for sufficiently small $\varepsilon$, each translated <mollifier> is an allowed test, so the assumption gives $w*\rho_\varepsilon=0$. The standard <mollifier> approximation theorem gives $w*\rho_\varepsilon\to w$ in local $L^1$. Thus \b[$w=0$ <almost everywhere>].
For a multi-index $\alpha$, the <weak derivative> $v=D^\alpha u$ is the locally integrable function satisfying
$$
\boxed{\int_\Omega uD^\alpha\phi=(-1)^{|\alpha|}\int_\Omega v\phi\quad(\phi\in C_0^\infty(\Omega)).}
$$
Two such functions have zero pairing difference against every test, so the first result proves uniqueness. A compactly supported smooth test on an open subset $E$ extends by zero to one on $\Omega$, proving restriction of the derivative to $E$.
Put $w=D^\beta u$. For every test $\phi$ the defining identity applied to $D^\alpha\phi$ gives
$$
(-1)^{|\alpha|}\int wD^\alpha\phi=(-1)^{|\alpha|+|\beta|}\int uD^{\alpha+\beta}\phi.
$$
Thus these two <distributions> are identical; one is represented by a locally integrable function if and only if the other is. Their <weak derivatives> then agree, proving \b[$D^\alpha(D^\beta u)=D^{\alpha+\beta}u$] with exactly the asserted existence equivalence.
On $(0,2)$ the displayed piecewise affine function is continuous at one. <Integration by parts> separately on the two intervals shows its first <weak derivative> is one on $(0,1)$ and minus one on $(1,2)$; the first boundary terms cancel by continuity. The derivative of that step function is $-2\delta_1$. This <distribution> is not represented by a locally integrable function: tests of height one supported in shrinking neighborhoods of one have pairing $-2$ with it, whereas their integrals against any fixed local $L^1$ function tend to zero. Hence \b[there is a first <weak derivative> but no locally integrable second <weak derivative>].
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