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For $\mathrm{Re}p>0$, let $r_{-}$ and $r_1,r_2$ be the characteristic roots of $r^3+r+p=0$, split by the signs of their real parts. Put $d_j=3r_j^2+1$. The real-line Green function of $p+{\partial }_x+{\partial }_x^3$ isThe reciprocal-polynomial root identities imply continuity of $R_p,R_p^{\prime }$ and jump $R_p^{\prime \prime }(0+)-R_p^{\prime \prime }(0-)=1$. The distributional derivative therefore gives $(p+{\partial }_s+{\partial }_s^3)R_p={\delta }_0$. The transport term changes the roots: the pure Airy resolvent kernel with $r^3+p=0$ cannot be substituted unchanged.

For $a<b$ strictly inside a one-dimensional domain, the indicator $u=1_{[a,b]}$ has distributional derivative $Du={\delta }_a-{\delta }_b$, hence $\mathrm{TV}(u)=2$. Directly from the test-function definition, $\int u{\phi }^{\prime }=\phi (b)-\phi (a)\le 2$, and two disjoint normalized smooth bump functions attain the bound. Values at the endpoints themselves do not change the $L^1$ function.

A distribution is invariant under every translation parallel to the $i$th coordinate axis exactly when its $i$th distributional derivative vanishes. Differentiate the translated pairing for one direction; integrate that derivative in the translation parameter for the converse.

A locally integrable function $g$ is the $i$th weak derivative of $u$ whenfor every test function $\phi \in C_c^{\infty }(U)$. Thus a weak derivative is a distributional derivative that is represented by a locally integrable function.