Pullback of a distribution by a diffeomorphism

Let $n\ge 1$ be an integer, let $F:U\to V$ be a smooth diffeomorphism between open subsets of ${\mathbb{R}}^n$, and let $u\in {\mathcal{D}}^{\mathrm{\prime }}(V)$ in the convention of Distribution. Put $J_F(x)=\mathrm{\mid }\det DF(x)\mathrm{\mid }$. The chain rule applied to $F^{-1}\circ F$ makes $DF(x)$ invertible, so $J_F>0$. It is smooth: the determinant is smooth and nonzero, and its sign is locally constant. Define $⟨F^{\ast }u,\phi ⟩=⟨u,\left(\frac{\phi }{J_F}\right)\circ F^{-1}⟩(\phi \in \mathcal{D}(U))\mathrm{.}$ The transformed test has support in $F(\mathrm{supp}\phi )$, a compact subset of $V$. Multiplication by the fixed smooth reciprocal Jacobian and diffeomorphic composition are continuous linear test operations by Test function operations are continuous, so their transpose defines a distribution. This definition is choice-free. For the identity map the Jacobian is one and the pullback is the identity; the zero distribution pulls back to zero. Empty diffeomorphic domains give zero test and distribution spaces. The absolute value of the determinant handles orientation reversal; arbitrary smooth maps are not covered by this definition.

For compatibility with functions, use the regular-functional convention of Regular distribution from a locally integrable function and assume the Axiom of Countable Choice (The Axiom of Countable Choice (${\mathrm{A}\mathrm{C}}_{\omega }$)). If $f\in L_{\mathrm{l}\mathrm{o}\mathrm{c}}^1(V)$ and $K\subseteq U$ is compact, put $h_K={\mathbf{1}}_{F(K)}\mathrm{\mid }f\mathrm{\mid }$. This is an $L^1(V)$ function. The exact formula of A C^1 diffeomorphism satisfies the change-of-variables formula for L^1 functions gives ${\int }_K\mathrm{\mid }f(F(x))\mathrm{\mid }{J}_F(x)dx={\int }_{F(K)}\mathrm{\mid }f(y)\mathrm{\mid }dy<\mathrm{\infty }\mathrm{.}$ The formula also makes the transformed integrand measurable; since $J_F$ is positive and its reciprocal is bounded on $K$, this proves $f\circ F\in L_{\mathrm{l}\mathrm{o}\mathrm{c}}^1(U)$. If $f=g$ almost everywhere, apply the same formula to ${\mathbf{1}}_{F(K)}\mathrm{\mid }f-g\mathrm{\mid }$, whose integral is zero, and use A nonnegative measurable function has integral $0$ exactly when it vanishes almost everywhere to obtain $f\circ F=g\circ F$ almost everywhere on each compact $K$. Thus composition respects local almost-everywhere classes. For a fixed test $\phi$, the function $f(y)(\phi \mathrm{/}{J}_F)(F^{-1}(y))$ is integrable, since its smooth factor is bounded with compact support. Applying the same change-of-variables formula gives $⟨F^{\ast }{u}_f,\phi ⟩={\int }_{V}f(y)\frac{\phi (F^{-1}(y))}{J_F(F^{-1}(y))}dy={\int }_{U}f(F(x))\phi (x)dx\mathrm{.}$ Thus $F^{\ast }{u}_f=u_{f\circ F}$, with both regular functionals distributions by Locally integrable functions embed in distributions. Countable Choice is used through the published Lebesgue change-of-variables and embedding results, not through the definition of $F^{\ast }u$.