Multiplication of a distribution by a smooth function

For a distribution $u$ and a smooth function $a$, the product $au$ is defined by $⟨au,\phi ⟩=⟨u,a\phi ⟩$. It obeys the Leibniz rule; in one dimension, $(au{)}^{\prime }=a^{\prime }u+a{u}^{\prime }$.

If a smooth function $a$ on $\mathbb{R}$ has finitely many zeros $r_j$, all simple, then $aw=0$ exactly when $w={\sum }_j{c}_j{\delta }_{r_j}$. Away from the zeros, divide a test function by the nonvanishing $a$ to show $w$ vanishes. Near $r_j$, write $a(x)=(x-r_j)b_j(x)$ with smooth nonzero $b_j$. The kernel of multiplication by a coordinate, after translation and multiplication by $b_j$, says that the local distribution is a multiple of ${\delta }_{r_j}$. A partition of unity gives the global sum. Simplicity of the zeros excludes derivatives of deltas.

In one dimension, $xw=0$ if and only if $w=c{\delta }_0$. Choose a cutoff function $\eta =1$ near zero and write every test function as $\phi =\phi (0)\eta +x\psi$. The pairing with $x\psi$ vanishes, leaving $⟨w,\phi ⟩=\phi (0)⟨w,\eta ⟩$.

For smooth $a$ and a point $r$,These follow by applying both sides to a test function and using the Leibniz rule. In particular $(x-r){\delta }_r^{\prime }=-{\delta }_r$.

The Cauchy principal valuedefines a distribution. Near zero the odd constant contribution cancels, and the remaining numerator is $O(x)$. It satisfies $x\mathrm{pv}(1/x)=1$. Hence all solutions of $xv=1$ are $v=\mathrm{pv}(1/x)+c{\delta }_0$ by the kernel of multiplication by a coordinate.

For a continuous even function $f$ on $\mathbb{R}$, the Cauchy principal valuedefines a distribution of order of a distribution at most one. For test functions supported in $[-R,R]$, the absolute value is at most $2R{\max }_{[0,R]}|f|\Vert {\varphi }^{\prime }{\Vert }_{\infty }$, since the numerator is bounded by $2x\Vert {\varphi }^{\prime }{\Vert }_{\infty }$. When $\varphi (0)=0$, the expression equals the absolutely convergent ordinary integral $\int f\varphi /x$.

It is a regular distribution exactly when ${\int }_0^r|f(x)|/xdx<\infty$ for some $r>0$. Necessity follows by testing away from zero: any representing locally integrable function must equal $f/x$ almost everywhere off zero. Sufficiency follows from absolute convergence. Continuity with $f(0)=0$ alone does not suffice: $f(x)=1/\log (1/|x|)$ near zero gives the divergent integral ${\int }_1^{\infty }ds/s$.

Target: L1-normalized dilation of a distribution. Parent: Dilation of a distribution. Intended file: distribution-theory.bigb.