Distributions Test Functions and Differentiation — Examples

Since $\mathrm{\mid }H\mathrm{\mid }\le 1$, its absolute integral on every compact interval is finite, so F1 makes $u_H$ a distribution. Altering its value only at zero changes no integral by F4. For a test $\phi$, choose $R>0$ beyond its compact support. Then F2 and F3 give $D{u}_H(\phi )=-{\int }_0^R{\phi }^{\mathrm{\prime }}(x)dx=\phi (0)-\phi (R)=\phi (0)={\delta }_0(\phi )$. This calculation applies to complex tests componentwise.

The equality on every test proves the distribution identity. The sign is positive because the negative transpose sign cancels the lower-endpoint sign. A test supported away from zero gives zero; the zero test gives zero; only the finite interval $[0,R]$ enters, so no endpoint at infinity is evaluated. $\mathrm{\square }$

By F1, $u_f$ and $u_g$ are distributions. Fix a test $\phi$ and a closed interval $[b,c]$ containing its support in its interior, with endpoints outside $A$. Local finiteness and compactness imply $A\cap [b,c]$ is finite: take a finite subcover of neighborhoods each meeting finitely many points. List these break points in increasing order. On each intervening open interval $(s,t)$ apply F2 to $f,\phi$ on $[s+\epsilon ,t-\epsilon ]$ for sufficiently small positive $\epsilon$. This gives [given, F1, F2, F5] $-{\int }_{s+\epsilon }^{t-\epsilon }f{\phi }^{\mathrm{\prime }}={\int }_{s+\epsilon }^{t-\epsilon }g\phi +f(s+\epsilon )\phi (s+\epsilon )-f(t-\epsilon )\phi (t-\epsilon )\mathrm{.}$

Let $\epsilon$ decrease to zero, for example through the reciprocal integers once the truncated interval is nonempty. F3 applies to the two integrals, with majorants $\mathrm{\mid }f{\phi }^{\mathrm{\prime }}\mathrm{\mid }$ and $\mathrm{\mid }g\phi \mathrm{\mid }$, integrable on $[b,c]$ by the local integrability assumptions. The boundary values tend to $f(s+)\phi (s)$ and $f(t-)\phi (t)$ by the finite one-sided limits; at $b,c$ the test vanishes. Sum over the finitely many intervals. At each break point $a$, the left interval contributes $-f(a-)\phi (a)$ and the right contributes $f(a+)\phi (a)$. The result is exactly the asserted formula when paired with $\phi$.

On any fixed compact test support $K$, the delta sum is finite and bounded in modulus by $({\sum }_{a\in A\cap K}\mathrm{\mid }f(a+)-f(a-)\mathrm{\mid })p_0(\phi )$. It therefore defines a distribution by F4, so the test equality proves the distribution identity. If there are no break points in $K$ the sum is zero, and a zero jump contributes no delta. If $f$ is $C^1$ up to both sides, $f$ and $g$ are bounded on each of the finitely many compact pieces meeting a compact interval, hence locally integrable, verifying the stated sufficient case. Merely being $C^1$ on the open pieces does not supply local integrability of $g$ at the breaks. $\mathrm{\square }$

For $a>0$, divide $0<\mathrm{\mid }x\mathrm{\mid }\le a$ into shells $2^{-j-1}a<\mathrm{\mid }x\mathrm{\mid }\le 2^{-j}a$, $j\ge 0$. Each lies in a box of side $2^{1-j}a$, and $1\mathrm{/}\mathrm{\mid }x\mathrm{\mid }\le 2^{j+1}\mathrm{/}a$ there. F6 bounds the integral of $1\mathrm{/}\mathrm{\mid }x\mathrm{\mid }$ by ${\sum }_{j\ge 0}16a^2{2}^{-2j}<\mathrm{\infty }$. A singleton is null since it lies in boxes of arbitrarily small volume. Thus $N$ is locally integrable and its value at zero is immaterial. Direct differentiation gives ${\mathrm{\partial }}_{i}N=x_i\mathrm{/}(4\pi \mathrm{\mid }x{\mathrm{\mid }}^3)$ and $\mathrm{\Delta }N=(3\mathrm{\mid }x{\mathrm{\mid }}^{-3}-3\mathrm{\mid }x{\mathrm{\mid }}^2\mathrm{\mid }x{\mathrm{\mid }}^{-5})\mathrm{/}(4\pi )=0$ off zero.

For $\epsilon >0$ put $N_{\epsilon }(x)=-(4\pi {)}^{-1}(\mathrm{\mid }x{\mathrm{\mid }}^2+{\epsilon }^2{)}^{-1\mathrm{/}2}$. This is smooth everywhere, and coordinate differentiation gives [step 1.1, algebra] $h_{\epsilon }(x):=\mathrm{\Delta }{N}_{\epsilon }(x)=\frac{3{\epsilon }^2}{4\pi (\mathrm{\mid }x{\mathrm{\mid }}^2+{\epsilon }^2{)}^{5\mathrm{/}2}}\ge 0.$ For $R>0$ we supply the ball presentation required by F3. In each coordinate direction its base is the closed radius-$R$ disc and its lower and upper functions are $-\sqrt{R^2-\mathrm{\mid }t{\mathrm{\mid }}^2}$ and $\sqrt{R^2-\mathrm{\mid }t{\mathrm{\mid }}^2}$, continuous and strictly ordered on the interior. These descriptions also prove that the ball is Jordan measurable. Use the parametrization $P(\varphi ,\theta )=R(\sin \varphi \cos \theta ,\sin \varphi \sin \theta ,\cos \varphi )$ on the eight rectangles cut at $\varphi =\pi \mathrm{/}2$ and $\theta =\pi \mathrm{/}2,\pi ,3\pi \mathrm{/}2$. It is smooth on neighborhoods of the rectangles and $P_{\varphi }\times P_{\theta }=R\sin \varphi P$, nonzero on each interior. An interior image has three nonzero coordinates; its third coordinate uniquely determines $\varphi \in (0,\pi )$ and its first two uniquely determine the azimuth in its quadrant, so it shares its image with no other point of that closed rectangle. Distinct patches overlap only over rectangle edges, whose preimages have content zero. For each direction sort the four octants with positive coordinate as upper and the four with negative coordinate as lower, with no lateral patches. The corresponding area-vector coordinate has the required strict sign, and projected interiors are the four disjoint open quarter discs. Their omissions are the two diameters and boundary circle, all content zero: diameters admit arbitrarily thin rectangle covers; the circle lies in annuli of content $\pi ((R+h{)}^2-(R-h{)}^2)\to 0$ by F3. Thus all adaptation clauses hold for the same eight-patch list. This proves the ball is elementary using only the definitions, not a B-page supplier. F2 on this ball $B_R$ with functions $1,N_{\epsilon }$ gives ${\int }_{B_R}{h}_{\epsilon }={\int }_{\mathrm{\partial }{B}_R}{\mathrm{\partial }}_n{N}_{\epsilon }$. The supplied parametrization $R(\sin \varphi \cos \theta ,\sin \varphi \sin \theta ,\cos \varphi )$ has area density $R^2\sin \varphi$, by direct cross product. Its total area is $R^2{\int }_0^{2\pi }{\int }_0^{\pi }\sin \varphi d\varphi d\theta =4\pi R^2$, and its outward normal is $x\mathrm{/}R$. Thus [step 1.1, F2, F3, F4] ${\int }_{B_R}{h}_{\epsilon }=\frac{R^3}{(R^2+{\epsilon }^2{)}^{3\mathrm{/}2}}⟶1.$ F4 identifies these compact-region Riemann integrals with Lebesgue integrals. All Green functions are $C^2$ on a neighborhood of the entire closed ball.

Fix a test $\psi$ supported in the interior of $B_R$. F2 for $N_{\epsilon },\psi$ has zero boundary terms since $\psi$ and its derivatives vanish near the sphere. Hence $\int N_{\epsilon }\mathrm{\Delta }\psi ={\int }_{B_R}{h}_{\epsilon }\psi$. On the left, $\mathrm{\mid }{N}_{\epsilon }\mathrm{\mid }\le 1\mathrm{/}(4\pi \mathrm{\mid }x\mathrm{\mid })$ off zero, an integrable bound on $B_R$ by step 1.1, and $N_{\epsilon }\to N$ almost everywhere. F5 gives convergence to $\int N\mathrm{\Delta }\psi$.

For $0<\delta <R$, the difference between ${\int }_{B_R}{h}_{\epsilon }\psi$ and $\psi (0){\int }_{B_R}{h}_{\epsilon }$ is bounded by ${\sup }_{\mathrm{\mid }x\mathrm{\mid }\le \delta }\mathrm{\mid }\psi (x)-\psi (0)\mathrm{\mid }$ times a mass at most one, plus $2\mathrm{\parallel }\psi {\mathrm{\parallel }}_{\mathrm{\infty }}{\int }_{B_R\setminus B_{\delta }}{h}_{\epsilon }$. On the latter region, $h_{\epsilon }\le 3{\epsilon }^2\mathrm{/}(4\pi {\delta }^5)$, so the second term tends to zero by finite box volume. First choose $\delta$ using continuity, then let $\epsilon \to 0$. Together with step 2.1 this proves ${\int }_{B_R}{h}_{\epsilon }\psi \to \psi (0)$. Step 3.1 and F1 now give $(\mathrm{\Delta }{u}_N)(\psi )=\int N\mathrm{\Delta }\psi =\psi (0)$. This proves the identity with its positive sign. The zero test gives zero, and no value of the singular formula at zero is used. $\mathrm{\square }$

Changing $x$ to $-t$ in the negative interval, which follows directly by reflecting its tagged partitions, gives ${\int }_{\mathrm{\mid }x\mathrm{\mid }>\epsilon }\phi (x)\mathrm{/}xdx={\int }_{\epsilon }^R(\phi (t)-\phi (-t))\mathrm{/}tdt$ for $0<\epsilon <R$. The quotient extends continuously to $t=0$ with value $2{\phi }^{\mathrm{\prime }}(0)$ by the definition of derivative. Its modulus is at most $2\sup \mathrm{\mid }{\phi }^{\mathrm{\prime }}\mathrm{\mid }$ by F2. F3 therefore gives an integral on $[0,R]$, and the omitted interval has integral tending to zero, bounded by a constant times $\epsilon \sup \mathrm{\mid }{\phi }^{\mathrm{\prime }}\mathrm{\mid }$. This proves existence of the principal value.

The expression is linear in $\phi$, since each truncation is linear and limits preserve finite sums. For any fixed compact $K\subseteq \mathbb{R}$, choose $R>0$ with $K\subseteq [-R,R]$. The integral representation from step 1.1 gives $\mathrm{\mid }\mathrm{pv}(1\mathrm{/}x)(\phi )\mathrm{\mid }\le 4R{p}_1(\phi )$ for $\phi \in {\mathcal{D}}_K$, by separately bounding real and imaginary integrals; the sharper $2R$ bound also follows from the complex integral triangle inequality but is unnecessary. F1 proves continuity. An even test gives zero because the quotient vanishes. Tests supported away from zero give their ordinary integral against $1\mathrm{/}x$, and the zero test gives zero. The limit requires symmetric removal at zero, with no assertion about independently varying two cutoffs. $\mathrm{\square }$

Suppose $u_f={\delta }_0$ with $f\in L_{\mathrm{l}\mathrm{o}\mathrm{c}}^1({\mathbb{R}}^n)$. On the open domain $U={\mathbb{R}}^n\setminus \{0\}$, all tests evaluate to zero at the origin, so F2 gives $u_{f\mathrm{\mid }U}=0$. F1 applied on this entire open domain, without selecting pointwise neighborhoods, gives $f=0$ almost everywhere on $U$. By F4 the omitted singleton is null, so $f=0$ almost everywhere on ${\mathbb{R}}^n$.

Consequently $u_f=0$, but F3 supplies $\phi$ with $\phi (0)=1$, and F2 gives ${\delta }_0(\phi )=1$. This contradiction proves that the witness is not regular and refutes the proposed universal statement. The zero function does represent the zero distribution; the failure is the nonzero point mass, not a failure of the regular-distribution construction. Dimension zero is excluded, since its singleton has mass one in the library convention. $\mathrm{\square }$

F2 supplies the bump by taking a nonzero nonnegative test with the required support and dividing by its positive finite integral. For $j\ge 1$, the support of $f_j$ lies in $(3\mathrm{/}(2j),5\mathrm{/}(2j))$. Thus $f_j(x)=0$ for every $j\in \mathbb{N}$ when $x\le 0$, and for each fixed $x>0$ it is zero once $j>5\mathrm{/}(2x)$. In particular the pointwise limit is zero even at the origin. Each $f_j$, including $f_0$, is smooth and compactly supported, so F1 applies.

For every test $\phi$ and every $j\ge 1$, the substitution $t=jx-2$ from F3 gives $u_{f_j}(\phi )=\int \rho (t)\phi ((t+2)\mathrm{/}j)dt$. Since $\mathrm{\mid }t+2\mathrm{\mid }\le 5\mathrm{/}2$ on the bump support, [step 1.1, F1, F3] $\mathrm{\mid }{u}_{f_j}(\phi )-\phi (0)\mathrm{\mid }\le {\sup }_{\mathrm{\mid }s\mathrm{\mid }\le 5\mathrm{/}(2j)}\mathrm{\mid }\phi (s)-\phi (0)\mathrm{\mid }⟶0.$ Here positivity and unit mass give the inequality, and continuity at zero gives the limit. Thus the weak limit is ${\delta }_0$. Choose a cutoff test equal to one near zero by F2; its pairings are eventually one, whereas the zero regular distribution pairs to zero. This is the failed conclusion for the explicit witness sequence. Its mass is one for every $j$, concentrated in a shrinking interval; pointwise convergence alone does not control these pairings. $\mathrm{\square }$

F1 immediately gives $\mathrm{\mid }({\mathrm{\partial }}^{\alpha }{\delta }_a)(\phi )\mathrm{\mid }\le {\sup }_{x\in \mathrm{\Omega },\mathrm{\mid }\beta \mathrm{\mid }\le m}\mathrm{\mid }{\mathrm{\partial }}^{\beta }\phi (x)\mathrm{\mid }$, an explicit global estimate of order $m$ and constant one. It vanishes on tests supported away from $a$. To see it is nonzero on every neighborhood of $a$, take by F3 a cutoff equal to one near $a$ and multiply it by $(x-a{)}^{\alpha }\mathrm{/}\alpha !$; its $\alpha$ derivative at $a$ is one. Thus the support is exactly $\{a\}$, consistent with F2's compact-support theorem.

Suppose $m\ge 1$ and fix compact $K$ containing a ball about $a$. Choose $h\in \mathcal{D}({\mathbb{R}}^n)$ supported in the unit ball and equal to $x^{\alpha }\mathrm{/}\alpha !$ near zero, by F3. For sufficiently small $\epsilon >0$, let ${\phi }_{\epsilon }(x)={\epsilon }^{m}h((x-a)\mathrm{/}\epsilon )$, whose support is in $K$. Then ${\mathrm{\partial }}^{\alpha }{\phi }_{\epsilon }(a)=1$, whereas for every integer $0\le q<m$, [step 1.1, F1, F3] $p_q({\phi }_{\epsilon })\le {\max }_{\mathrm{\mid }\beta \mathrm{\mid }\le q}{\epsilon }^{m-\mathrm{\mid }\beta \mathrm{\mid }}\mathrm{\parallel }{\mathrm{\partial }}^{\beta }h{\mathrm{\parallel }}_{\mathrm{\infty }}⟶0.$ An order-$q$ bound would force a pairing of modulus one to tend to zero, a contradiction. Thus no smaller order works on $K$, or globally. For $m=0$, the upper bound and nonvanishing in step 1.1 prove exact order zero, since allowable orders are nonnegative integers. Compacts avoiding $a$ give the zero restriction, and compacts without a neighborhood of $a$ are excluded from the sharp local assertion. $\mathrm{\square }$