 Solution

Choose a smooth compactly supported mollifier of integral one. On every compact subset of $\Omega$, and for sufficiently small $\epsilon$, each translated mollifier is an allowed test, so the assumption gives $w\ast {\rho }_{\epsilon }=0$. The standard mollifier approximation theorem gives $w\ast {\rho }_{\epsilon }\to w$ in local $L^1$. Thus $w=0$ almost everywhere.

For a multi-index $\alpha$, the weak derivative $v=D^{\alpha }u$ is the locally integrable function satisfyingTwo 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 $\varphi$ the defining identity applied to $D^{\alpha }\varphi$ givesThus 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 $D^{\alpha }(D^{\beta }u)=D^{\alpha +\beta }u$ 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 there is a first weak derivative but no locally integrable second weak derivative.