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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.

The Schwartz space consists of the smooth functions $\phi$ for whichfor all multi-indices $\alpha ,\beta$. Its topology is generated by these seminorms; ${\phi }_j\to \phi$ means convergence in every seminorm. A tempered distribution is a continuous linear functional on this space. Convergence in ${\mathcal{S}}^{\prime }$ here means weak convergence: $⟨u_j,\phi ⟩\to ⟨u,\phi ⟩$ for every Schwartz function.

Use the unnormalized angular-frequency Fourier transformThe transform is a continuous automorphism of the Schwartz space, so this duality defines a tempered distribution. Put $D_j=-i{\partial }_j$, as required by the stated identities. Integration by parts and differentiation under the integral giveThus, keeping the transpose factor of $D=-i\partial$ explicit,andIterating gives the Fourier derivative identities for distributions

The logarithm defines a tempered distribution because it is locally integrable and grows more slowly than a polynomial. Its derivative is $u^{\prime }(x)=x/(1+x^2)$. The standard transform pair $\widehat{(1+x^2{)}^{-1}}=\pi e^{-|\lambda |}$, obtainable by the residue theorem, givesAs $\widehat{u^{\prime }}=i\lambda \hat{u}$, we have $\lambda \hat{u}=-\pi \mathrm{sgn}(\lambda )e^{-|\lambda |}$. DefineThis integral is absolutely convergent: the numerator is $O(|\lambda |)$ near zero, and the exponential controls the tails. It defines a tempered distribution satisfying the same multiplication equation. The kernel of multiplication by a coordinate therefore gives $\hat{u}=T+c{\delta }_0$.

To determine $c$, use the supplied Gaussian Fourier transform pair with ${\phi }_n(0)=1/(2\sqrt{\pi })$. ThenAlso $⟨T,{\phi }_n⟩\to 0$ by the dominated convergence theorem: the relevant integrand is bounded by a constant times $|\lambda |e^{-|\lambda |}$ for $n\ge 1$. ConsequentlyThe cutoff limit agrees with the absolutely convergent integral; this particular subtraction convention fixes the otherwise ambiguous delta term. The reusable identity is Fourier transform of the logarithm of one plus x squared.

The principal-value reciprocal distribution isThe limit exists because the constant part of the numerator cancels symmetrically at zero. Multiplying by $x$ gives $x\mathrm{pv}(1/x)=1$.

If $v$ is any other solution, $w=v-\mathrm{pv}(1/x)$ obeys $xw=0$. To identify this kernel of multiplication by a coordinate, choose a cutoff function $\eta$ equal to one near zero. Every test function has the form $\phi =\phi (0)\eta +x\psi$ with $\psi \in \mathcal{D}$. Thus $⟨w,\phi ⟩=\phi (0)⟨w,\eta ⟩$, proving $w=c{\delta }_0$. ThereforeFor real distributions the constants are real.

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.