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Let $g=Du\in L^p(0,1)$ be the weak derivative and defineThe fundamental theorem of calculus for Lebesgue integration makes $F$ an absolutely continuous function, differentiable almost everywhere, with $F^{\prime }=g$ almost everywhere. The distributional derivative of $u-F$ is zero. A locally integrable function with zero distributional derivative on a connected interval is equal almost everywhere to a constant $C$. Consequentlyis an absolutely continuous representative of $u$, and ${\tilde{u}}^{\prime }=g\in L^p(0,1)$ almost everywhere.

Solved by gpt-5.6-sol high.

Each continuous $f_{\alpha }$ of polynomial growth defines a regular tempered distribution byChoosing an integer $L>M_{\alpha }+n$ giveswhich is bounded by finitely many Schwartz space seminorms. Its distributional derivative satisfiesand is therefore tempered. A finite sum of continuous linear forms is continuous, so

Solved by gpt-5.6-sol high.

The regularity from part (i) makes $u\omega$ and its first weak derivatives square-integrable. The product rule therefore holds in the distributional derivative sense:Since $u={\nabla }^{\perp }\Psi$, equality of mixed weak derivatives gives $\nabla \cdot u=0$. Both remaining expressions belong to $L_{\mathrm{per}}^2$, so their distributional equality is an equality in that space:

Solved by gpt-5.6-sol high.

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.