# Worked causal inverses, examples and exercises

*Written by GPT-6.1 Sol (OpenAI), Ultra reasoning effort, October 2026.  Public domain (CC0 1.0).*

The preceding chapter separates two ways an inverse series can exist. A positive delay makes it locally finite. A continuous error at the vertex instead requires a convergent local construction. The examples below show the distinction, including a compact differential-and-delay kernel, a noncompact kernel with a finite differentiability threshold, and an inverse that grows exponentially.

Basic references are Richard Melrose's [Differential Analysis](https://ocw.mit.edu/courses/18-155-differential-analysis-fall-2004/), Gerd Grubb's [distribution Fourier chapter](https://web.math.ku.dk/~grubb/dist5.pdf), and Lars Hörmander's *The Analysis of Linear Partial Differential Operators*, volumes I and II. The actual proof prerequisites are the preceding chapter, Convolution as addition of supports, Theorems 1.1, 2.1, 3.1 and 3.2, and its Proposition 3.3. The support-frequency comparison uses Support cones force reciprocal bounds, Theorem 1.1, and the explicit cone convention of Logarithmic Fourier graphs construct a cone-supported inverse, Theorem 1.1.

## 1. A line has no positive clock

**Example 1.** In one dimension let
\[
C=\mathbb R,\qquad R=\delta_1+\delta_{-1}.
\tag{E1.1}
\]
The support of \(R\) misses zero. Nevertheless
\[
R^{*2j}
=\sum_{k=0}^{2j}\binom{2j}{k}\delta_{2k-2j}
\tag{E1.2}
\]
has a mass \(\binom{2j}{j}\delta_0\) for every \(j\). This follows by expanding the finite convolution product: choose \(k\) copies of \(\delta_1\) and \(2j-k\) copies of \(\delta_{-1}\), and add their locations.

Choose a nonnegative compact smooth test supported in \((-1/2,1/2)\), with value one at zero. Its pairing with every even power is \(\binom{2j}{j}\ge1\), and its pairing with every odd power is zero. Thus the proposed sum of powers diverges on this one test. It is neither locally finite nor a distributional sum.

Every finite convolution here exists, because all factors are compact. The failure is the infinite sum. Pointedness is what prevents repeated contributions from returning to one bounded region. By contrast, for \(C=\{0\}\), a distribution in \(\mathcal A_C\) that vanishes near zero is zero, and its delayed inverse series is just \(\delta_0\).

## 2. A differential kernel with two transverse delays

**Example 2.** Use coordinates \((t,x)\), and put
\[
\begin{gathered}
C=\{(t,x):t\ge|x|\},\qquad
a=(3/5,-3/10),\\
p=(1,1),\quad q=(1,-1),\quad
b=1/5,\quad c=3/10,\\
\mu=\tau_a(\delta_0+\partial_t\delta_0),\qquad
v=\mu-b\delta_{a+p}-c\delta_{a+q}.
\end{gathered}
\tag{E2.1}
\]
This is a compact kernel. Its inverse without the delays is
\[
E=\tau_{-a}E_0,\qquad
E_0(t,x)=H(t)e^{-t}\otimes\delta_0(x),
\tag{E2.2}
\]
where \(H\) is the indicator of the nonnegative half-line. Integration by parts gives \((1+\partial_t)(H(t)e^{-t})=\delta_0\), so \(\mu*E=\delta_0\). Both the kernel and the inverse have the translated cone supports required by Theorem 2.2.

Let \(d=b\delta_p+c\delta_q\). The error is \(R=d*E_0\). Its clock \(t\) is at least one. Every compact time interval therefore receives only finitely many powers. The inverse is
\[
F=\tau_{-a}F_0,\qquad
F_0=\sum_{j=0}^\infty d^{*j}*E_0^{*(j+1)}.
\tag{E2.3}
\]
This formula has a complete explicit form. For \(\ell\ge1\), define
\[
g_\ell(s)=H(s)e^{-s}\frac{s^{\ell-1}}{(\ell-1)!}.
\tag{E2.4}
\]
Then
\[
E_0^{*\ell}=g_\ell(t)\otimes\delta_0(x),
\qquad
F_0(t,x)=
\sum_{j=0}^\infty\sum_{k=0}^j
\binom jk b^kc^{j-k}
g_{j+1}(t-j)\otimes\delta_{2k-j}(x).
\tag{E2.5}
\]
To verify the first identity, the convolution of \(g_r\) and \(g_s\) is zero at negative time and, at \(t>0\), equals
\[
\frac{e^{-t}}{(r-1)!(s-1)!}
\int_0^t u^{r-1}(t-u)^{s-1}\,du
=e^{-t}\frac{t^{r+s-1}}{(r+s-1)!}.
\tag{E2.6}
\]
The integral formula follows by substituting \(u=tz\) and integrating by parts \(s-1\) times; its last integral is \(\int_0^1 z^{r+s-2}\,dz=1/(r+s-1)\). The boundary terms vanish at each earlier step, and the resulting factorials give the displayed value. The second identity in (E2.5) now follows by the finite binomial expansion of \(d^{*j}\).

For a direct check of the equation,
\[
(1+\partial_t)g_1=\delta_0,\qquad
(1+\partial_t)g_{\ell+1}=g_\ell\quad(\ell\ge1).
\tag{E2.7}
\]
The first formula includes the jump at zero. The other \(g_{\ell+1}\) vanish there, so their distributional derivative has no delta term. Applying \(1+\partial_t-d*\) to (E2.3) cancels successive terms, with only \(\delta_0\) left. This is a valid locally finite cancellation on each compact test support, not a formal multiplication of unrestricted series.

All coefficients in (E2.5) are positive. Its exact support is
\[
\operatorname{supp}F_0
=\{(t,x):x\in\mathbb Z,\ t\ge|x|\}.
\tag{E2.8}
\]
Indeed, the term of level \(j\) lies at \(x=2k-j\), with \(t\ge j\). These locations satisfy \(j\ge|x|\). Conversely, for any integer \(x\), take \(j=|x|\) and the corresponding endpoint \(k\). Its density is positive for \(t>j\), and its closure contains the initial point. The union of these rays is closed, and positivity prevents cancellation.

The closed convex hull of this support is the whole cone \(C\): the initial points at all nonnegative integers fill both boundary rays by convex combinations, and convexity fills the wedge between them. Translation gives the hull \(C-a\) for \(F\).

![Delayed contribution families and a compact clock cutoff in a pointed wedge](figures/delayed-powers-and-clock-cutoff.png)

*Figure 1. Coordinates are \(\sigma=t+3/5\) and \(\chi=x-3/10\), so the inverse's vertex is the origin in this plot. The first four contribution families in (E2.5) start at \((\sigma,\chi)=(j,2k-j)\) and continue upward. Later rays can overlap earlier ones. The dashed line \(\sigma=12/5\) receives only levels \(j=0,1,2\). The ray arrows and the wedge indicate unbounded support; only a finite part is drawn. Formula (E2.5), support identity (E2.8), and Theorem 2.1 give the exact argument.*

## 3. One derivative at the vertex suffices

**Example 3.** On the nonnegative half-line take the noncompact kernel and its point-supported inverse
\[
\mu(t)=H(t)e^{-t},\qquad E=\delta'_0+\delta_0.
\tag{E3.1}
\]
The calculation in (E2.7) gives \(\mu*E=\delta_0\). The local order of \(E\) is at most one. Change the kernel by
\[
w(t)=t_+^2e^{-2t},\qquad v=\mu+w.
\tag{E3.2}
\]
This \(w\) is \(C^1\) on the whole line, although its second derivative jumps at zero. Theorem 3.2 applies with \(m=1\).

The error can be written without an unknown distributional pairing:
\[
R=-w*E=-(w'+w)
=H(t)e^{-2t}(t^2-2t).
\tag{E3.3}
\]
There is no delta term in \(w'\), because \(w\) is continuous and zero at zero. The resulting \(R\) is continuous.

Choose \(0\le\chi\le1\), smooth, equal to one for \(|t|\le1/8\), and supported in \(|t|<1/4\). Such a cutoff is made from the flat function \(e^{-1/s}\) for \(s>0\), extended by zero for \(s\le0\). Put \(r_0=\chi R\), \(R_1=R-r_0\). Since \(|R(t)|\le2t\) for \(0\le t\le1/4\),
\[
\|r_0\|_1\le\int_0^{1/4}2t\,dt=1/16.
\tag{E3.4}
\]
Thus \(T=\delta_0+\sum_{j\ge1}r_0^{*j}\) converges in \(L^1\) after its delta term. The support of \(R_1\) lies in \([1/8,\infty)\), so the same is true of \(Q=T*R_1\). Consequently
\[
S_1=\sum_{j=0}^\infty Q^{*j},\qquad
F=(\delta'_0+\delta_0)*T*S_1
\tag{E3.5}
\]
is a complete construction of the inverse of \(v\). On time intervals ending at \(U\), only powers with \(j/8\le U\) can contribute to \(S_1\). The factorization (3.5) proves \(v*F=\delta_0\), and every factor has nonnegative support.

The construction works despite noncompactness of both \(\mu\) and \(v\), and it uses only the one derivative dictated by the local order of \(E\). Different admissible cutoffs produce the same \(F\), by uniqueness in the cone algebra.

## 4. A continuous feedback with an exponentially growing inverse

**Example 4.** Let
\[
r(t)=4tH(t)e^{-t},\qquad v=\delta_0-r.
\tag{E4.1}
\]
Here \(r\) is continuous at zero. It does not vanish on any right neighborhood of zero, and every nonzero convolution power has support \([0,\infty)\). The powers are not locally finite. Nevertheless their local convergence can be computed exactly:
\[
r^{*j}(t)=
H(t)e^{-t}\frac{4^j t^{2j-1}}{(2j-1)!}
\quad(j\ge1),
\tag{E4.2}
\]
by the same elementary integral as (E2.6). On each bounded time interval the factorials make the sum uniformly convergent, and
\[
F=\delta_0+H(t)e^{-t}\,2\sinh(2t)
=\delta_0+H(t)(e^t-e^{-3t}).
\tag{E4.3}
\]

One can check the equation independently of the series. With \(f(t)=H(t)(e^t-e^{-3t})\), ordinary integration at \(t>0\) gives
\[
\begin{aligned}
(r*f)(t)
&=4e^t\int_0^t s e^{-2s}\,ds
-4e^{-3t}\int_0^t s e^{2s}\,ds\\
&=e^t-4te^{-t}-e^{-3t}=f(t)-r(t).
\end{aligned}
\tag{E4.4}
\]
At negative time all terms vanish. They are continuous at zero, so this is also an identity of distributions. It proves
\[
(\delta_0-r)*(\delta_0+f)=\delta_0.
\tag{E4.5}
\]

The inverse is not tempered. Let \(\psi\ge0\) be a nonzero compact smooth function supported in \((0,1)\). Pairing \(f\) with \(\psi(t-N)\) gives
\[
e^N\int_0^1 e^s\psi(s)\,ds
-e^{-3N}\int_0^1 e^{-3s}\psi(s)\,ds.
\tag{E4.6}
\]
This grows exponentially in \(N\). Every fixed finite collection of Schwartz seminorms of those translates grows at most polynomially, by the product rule and the bound \(|t|\le N+1\) on their support. A tempered distribution has a bound by some such finite collection, contradicting (E4.6).

The inverse is therefore a well-defined causal distribution with no ordinary tempered Fourier transform. The whole construction uses compact tests and proper-support convolution.

![Locally convergent continuous feedback powers compared with their exact inverse density](figures/continuous-feedback-and-local-convergence.png)

*Figure 2. The solid curve is the regular part \(e^t-e^{-3t}\) of (E4.3); the delta mass at zero is specified in the formula and is not drawn as a function value. The other curves are the first one, two and three terms of (E4.2), on \(0\le t\le3/2\). Every power begins at time zero, so convergence here is factorial convergence on compact intervals, rather than the finite level cutoff of Figure 1. The shaded difference is the positive tail after three terms. Equations (E4.2)–(E4.5) supply the proof.*

## 5. Exercises and full solutions

Each exercise is worth 10 points, for 100 points in total.

### Exercise 1. A sharp clock in a wedge

For \(C=\{(t,x):t\ge|x|\}\), show that \(\tau=(1,0)\) and \(\kappa=1/\sqrt2\) work in (1.2), and that this value of \(\kappa\) cannot be increased. Bound all coordinates in a contributing \(r\)-fold sum with total clock at most \(U\).

**Solution.** For \((t,x)\in C\), \(t\ge0\) and \(t^2+x^2\le2t^2\). Therefore \(t\ge|(t,x)|/\sqrt2\). At the nonzero boundary point \((1,1)\), equality holds, proving sharpness for this unit clock. For \(z_j=(t_j,x_j)\in C\) with \(\sum_jt_j\le U\), one has \(\sum_j|z_j|\le\sqrt2U\), so each \(|z_j|\le\sqrt2U\). If \(U<0\), there are no such coordinates. The bound does not depend on the number of factors. Award 4 points for the inequality, 2 for sharpness and 4 for the sum estimate.

### Exercise 2. A curved set still has a translated cone bound

Let \(K=\{(x,y):y\le-x^2\}\) and \(\theta=(0,1)\). Compute its support function where finite, describe every high slice, and verify (4.2) with \(x_0=(0,1)\), \(A=\sqrt2\).

**Solution.** If \(\eta_y>0\), maximizing first in \(y\) gives
\[
H_K(\eta_x,\eta_y)
=\sup_x(\eta_xx-\eta_yx^2)=\frac{\eta_x^2}{4\eta_y}.
\]
The maximum occurs at \(x=\eta_x/(2\eta_y)\). At \(\eta=0\) the support value is zero. If \(\eta_y<0\), letting \(y\to-\infty\) gives \(+\infty\); if \(\eta_y=0\), \(\eta_x\ne0\), letting \(x\) tend with its sign gives \(+\infty\). For \(c\le0\), the high slice is \(c\le y\le-x^2\), so \(|x|\le\sqrt{-c}\); it is closed and bounded. For \(c>0\) it is empty. Finally \(x^2\le-y\) and \(1-y\ge1\), whence
\[
x^2+(1-y)^2\le (1-y)+(1-y)^2\le2(1-y)^2.
\]
Taking square roots gives (4.2). Award 4 points for the support function, 3 for the slices and 3 for the cone inequality.

### Exercise 3. The exact finite contribution budget

Suppose \(\operatorname{supp}R\subset C\cap\{\tau\cdot x\ge b\}\), \(b>0\). Determine how many powers can pair nontrivially with a test supported in \(L\), where \(M=\max_L\tau\cdot x\), and justify the telescoping inverse on that test.

**Solution.** For \(j\ge1\), support inclusion gives clock at least \(jb\). Thus \(j>M/b\) contributes zero. If \(M\ge0\), only \(j=0,\ldots,\lfloor M/b\rfloor\) need be retained; some of these can also vanish. If \(M<0\), all terms vanish, including \(\delta_0\). The equality case \(jb=M\) cannot be discarded without further information about the test and support, so the strict inequality is necessary for this support argument. For finite partial sums, (2.3) is an exact identity. Choose \(N\) with \((N+1)b>M\); its remainder vanishes on the test. Fixed proper-support continuity transfers the partial-sum identity to \(S\). Award 5 points for the budget and endpoint, and 5 for the limit justification.

### Exercise 4. One line of the delayed inverse

In Example 2, find the coefficient density of the atom at \(\chi=0\), for \(2\le\sigma<3\), and evaluate it at \(\sigma=12/5\). Here \((\sigma,\chi)=(t+3/5,x-3/10)\).

**Solution.** Only \(j=0,1,2\) can contribute before time three. A zero transverse coordinate requires \(2k-j=0\), which occurs at \(j=0,k=0\) and \(j=2,k=1\). The density is
\[
g_1(\sigma)+2bc\,g_3(\sigma-2)
=e^{-\sigma}+\frac3{25}g_3(\sigma-2).
\]
At \(\sigma=12/5\), \(g_3(2/5)=2e^{-2/5}/25\), so the value is
\[
e^{-12/5}+\frac6{625}e^{-2/5}.
\]
This is the scalar density multiplying the point mass in \(\chi\), not a two-dimensional function value of \(F\). Award 4 points for the contributing indices, 4 for the value and 2 for its interpretation.

### Exercise 5. The boundary term in a causal derivative

Prove (E2.7) by testing against a compact smooth function. Explain why the case \(g_1\) differs from the others.

**Solution.** For a test \(\phi\),
\[
\langle(1+\partial_t)g_1,\phi\rangle
=\int_0^\infty e^{-t}(\phi(t)-\phi'(t))\,dt=\phi(0),
\]
by integrating the derivative of \(e^{-t}\phi(t)\). For \(\ell\ge1\), \(g_{\ell+1}(0+)=0\). Integration by parts has no endpoint term, and ordinary differentiation on \(t>0\) yields
\[
g'_{\ell+1}=e^{-t}\frac{t^{\ell-1}}{(\ell-1)!}
-e^{-t}\frac{t^\ell}{\ell!}=g_\ell-g_{\ell+1}.
\]
All functions vanish for negative time. Thus the displayed distributional formulas follow. Award 5 points for the delta term and 5 for the higher powers.

### Exercise 6. A finite-regularity perturbation with one extra derivative

Keep \(\mu,E\) of Example 3, but take \(w(t)=t_+^3e^{-2t}\). Compute \(R=-w*E\), determine the regularity of \(w\) at zero, and give a small \(L^1\) bound for the cutoff error on \(0\le t\le1/4\).

**Solution.** The function \(w\) is \(C^2\), while its third derivative jumps from zero to six at zero. Since \(w\) is continuous and zero at zero, differentiation introduces no delta term, and
\[
R=-(w'+w)=H(t)e^{-2t}(t^3-3t^2).
\]
It is continuous near zero. For \(0\le t\le1/4\), \(|R(t)|\le3t^2\). Any cutoff \(0\le\chi\le1\) supported in that positive interval after multiplication by \(R\) gives
\[
\|\chi R\|_1\le\int_0^{1/4}3t^2\,dt=1/64<1.
\]
Choose the cutoff equal to one near zero. The remaining error has a positive delay. The order-one threshold in Theorem 3.2 is already met, and Theorem 3.1 then gives the inverse. Award 4 points for the derivative calculation, 3 for regularity and 3 for the quantitative bound.

### Exercise 7. A uniform bound on the continuous-series tail

Let \(P_J=\sum_{j=1}^J r^{*j}\) in Example 4. Show, for \(0\le t\le T\) and \(J\ge0\), that
\[
0\le f(t)-P_J(t)
\le 2e^{2T}\frac{(2T)^{2J+1}}{(2J+1)!}.
\tag{E5.1}
\]

**Solution.** Formula (E4.2) gives
\[
f-P_J
=2e^{-t}\sum_{j=J+1}^\infty
\frac{(2t)^{2j-1}}{(2j-1)!}.
\]
Its terms are nonnegative. Enlarge the odd-power sum to all powers \(k\ge m=2J+1\), and use \((m+\ell)!\ge m!\ell!\). The resulting upper bound is
\[
2\frac{(2T)^m}{m!}\sum_{\ell=0}^\infty
\frac{(2T)^\ell}{\ell!}
=2e^{2T}\frac{(2T)^m}{m!}.
\]
We used \(e^{-t}\le1\). For \(T=0\) both sides are zero. This proves uniform convergence on every fixed compact positive interval. Award 4 points for the exact tail and 6 for the factorial estimate.

### Exercise 8. Why exponential growth defeats temperedness

Complete the translated-test contradiction after (E4.6), using the standard finite-seminorm bound for a tempered distribution.

**Solution.** Continuity on Schwartz space gives some integers \(M,K\) and a constant \(B\) such that
\[
|\langle T,\phi\rangle|
\le B\max_{j\le K}\sup_t(1+|t|)^M|\phi^{(j)}(t)|.
\]
For \(\phi_N(t)=\psi(t-N)\), all derivatives have fixed sup norms and support in \((N,N+1)\). The right side is at most a fixed constant times \((N+2)^M\). The first integral in (E4.6) is strictly positive, and the second decays exponentially. Hence the left side grows like a positive constant times \(e^N\), contradicting that bound. For large \(N\), the delta part of \(F\) pairs to zero with \(\phi_N\), so \(F\) is also not tempered. Award 5 points for the seminorm estimate and 5 for the contradiction.

### Exercise 9. A dimension matters in a reciprocal diagnostic

Explain why the all-direction derivative counterexample in Example 5.3 of the preceding chapter uses one dimension. For \(\mu=\partial_t\delta_0\) on \(\mathbb R^2\), test the zero-free premise on \(\Gamma=\mathbb R^2\).

**Solution.** The transform is \(i\zeta_t\). At \(\zeta=(0,iR)\), \(R\ge2\), it is zero, while the imaginary norm is \(R\) and the total complex norm is \(R\). For any fixed logarithmic barrier constant \(D\), arbitrarily large \(R\) satisfy \(R>D\log(R+2)\). Thus the all-direction zero-free premise fails in two dimensions. In one dimension the only zero of \(i\zeta\) is zero itself, which lies below every positive logarithmic barrier; the reciprocal premise can hold there. Award 5 points for the exact zero family and 5 for the quantifier comparison.

### Exercise 10. An unrestricted vertex change can destroy the inverse

Take \(\mu=\delta_0\), \(E=\delta_0\), \(a=0\), and any pointed cone \(C\). Set \(v=0\). Identify which perturbation hypotheses fail and whether \(v\) can have an inverse.

**Solution.** All support inclusions are satisfied, since the empty support of \(v\) lies in \(C\). But \(v-\mu=-\delta_0\) neither vanishes near zero nor is a function near zero, so it satisfies neither Theorem 2.2 nor Theorem 3.2. The error \(R=(\mu-v)*E\) is \(\delta_0\); its clock delay is zero, and every power is \(\delta_0\). Finally \(v*F=0\) for every allowed \(F\), so it cannot equal \(\delta_0\). The support condition alone cannot preserve an inverse under arbitrary changes at the vertex. Award 4 points for the failed hypotheses, 3 for the error and 3 for the impossibility.

## References

- Richard B. Melrose, *Differential Analysis*, MIT OpenCourseWare, Fall 2004. [Open course materials](https://ocw.mit.edu/courses/18-155-differential-analysis-fall-2004/).
- Gerd Grubb, *Fourier transformation of distributions*. [Author-hosted chapter](https://web.math.ku.dk/~grubb/dist5.pdf).
- Convolution as addition of supports, especially its proper-support convolution, finite-regularity and cone-algebra statements.
- Support cones force reciprocal bounds, Theorem 1.1.
- Logarithmic Fourier graphs construct a cone-supported inverse, Theorem 1.1, with its stated cone convention.
- Lars Hörmander, *The Analysis of Linear Partial Differential Operators*, volumes I and II, Springer. Background reference.