# Convolution and joint frequency carriers

Original exposition, examples and complete solutions: GPT-6.1 Sol (OpenAI), Ultra. CC0 1.0.

Multiplying entire Fourier transforms adds their logarithmic profiles. The resulting carrier is a Minkowski sum, provided both input profiles are selected by the **same frequency sequence**. If one input collapses on that sequence, the product collapses as well.

This lesson proves the full joint-family identity and explains its synchronization requirement. The four examples include atomic cancellation, two singular factors with a smooth convolution, and a family whose output has both an empty and a nonempty carrier.

Basic references are Tao's *246B, Notes 2*, Hörmander's *The Analysis of Linear Partial Differential Operators I*, and its second volume. The actual prerequisites are the preceding proofs of joint logarithmic-frequency compactness, indicator additivity, and compact Fourier inversion. These linked proofs provide what the argument needs.

## 4. Multiplication keeps one common frequency

Let \(u\) be a compact distribution on \(\mathbb R^n\). We use

\[
 F_u(\zeta)=\langle u(x),e^{-ix\cdot\zeta}\rangle,\qquad
 L_u(z,c)=\frac{\log|F_u(c+z\log|c|)|}{\log|c|},
 \quad |c|>2,\quad c\in\mathbb R^n. \tag{4.1}
\]

Use \(D=-i\partial\). The logarithm at a transform zero is \(-\infty\). A proper profile is a plurisubharmonic local \(L^1\) limit; a collapsed profile tends to \(-\infty\) uniformly on compact parameter sets.

Let \(\mathcal J(u_1,\ldots,u_k)\) be the set of indicator tuples realized by simultaneous profile convergence along one escaping real frequency sequence. A proper indicator \(h_j\) is the support function of a nonempty compact convex carrier \(C_j\). A collapsed indicator has empty carrier.

For compact convolution,

\[
 F_{a*b}=F_aF_b,\qquad L_{a*b}=L_a+L_b. \tag{4.2}
\]

There is no extra \(2\pi\) factor in this forward identity. The complete compact-distribution pairing proof is Lemma 2.1 below.

**Theorem 4.1.** For compact distributions \(u'_j,u''_j\), \(1\leq j\leq k\),

\[
 \begin{split}
 &\mathcal J(u'_1*u''_1,\ldots,u'_k*u''_k)\\
 &\quad=
 \left\{(h'_1+h''_1,\ldots,h'_k+h''_k):
 (h'_1,\ldots,h'_k,h''_1,\ldots,h''_k)
 \in\mathcal J(u'_1,\ldots,u'_k,u''_1,\ldots,u''_k)\right\}.
 \end{split} \tag{4.3}
\]

The indicator sum is \(-\infty\) if either summand is \(-\infty\). Every proper output carrier is \(C'_j+C''_j\); with an empty input carrier, the output carrier is empty. Theorem 1.1 gives the full proof.

The displayed right side selects all factors jointly. Selecting one indicator tuple from the primed factors and an unrelated tuple from the double-primed factors would allow incompatible frequencies. Worked example 3 shows that this larger Cartesian choice can predict a nonempty carrier when the convolution is smooth.

## 5. The two possible limiting behaviors

If both factors have proper profiles \(v'\) and \(v''\) on the same sequence, their normalized logarithms add in local \(L^1\):

\[
 \|L_{u'*u''}-v'-v''\|_{L^1(B)}
 \leq\|L_{u'}-v'\|_{L^1(B)}+\|L_{u''}-v''\|_{L^1(B)}
 \longrightarrow0. \tag{5.1}
\]

The sum is proper because the two profiles are finite on a common set of full measure. Their canonical sum is PSH and satisfies the same kind of global linear imaginary-growth bound. The exact additivity theorem in Directional averages and additivity of growth indicators gives \(h_{v'+v''}=h_{v'}+h_{v''}\).

If one input collapses, every compact parameter set has

\[
 \sup_B L_{u'*u''}\leq\sup_B L_{u'}+M_B\longrightarrow-\infty, \tag{5.2}
\]

where \(M_B\) is an upper bound for the other input normalized logarithms. The other factor need not converge for this estimate. Compact Fourier growth provides the bound.

The empty carrier here records logarithmic collapse. It does not say that the distribution is zero, or that its ordinary support is empty. A nonzero compact smooth function has empty singular carrier profiles and a nonempty ordinary support.

## 6. Why equality needs both constructions

To start with an output tuple, keep its witnessing frequency sequence and extract profiles of all finitely many factors on a common subsequence. The original output profiles persist there. Proper input pairs add; a collapsed input forces its product to collapse. This puts every output tuple on the right side of (4.3).

To start with a joint input tuple, retain its witnessing sequence. Each product coordinate converges on that sequence, by (5.1) or (5.2), and its indicator is the required sum. This gives the opposite inclusion. The two arguments also cover mixed tuples, with some proper coordinates and some collapsed coordinates.

Neither construction replaces a witnessing sequence by unrelated frequencies. Neither identifies an indicator with a finite-height supremum. These are the two synchronization points of the proof.

## Worked example 1. Derivative orders add, while the carrier translates

On \(\mathbb R\), let \(a,b\in\mathbb R\) and integers \(m,r\geq0\). Set

\[
 u=D^m\delta_a,\qquad v=D^r\delta_b.
 \tag{7.1}
\]

Their transforms are \(\zeta^m e^{-ia\zeta}\) and \(\zeta^r e^{-ib\zeta}\). On every escaping real frequency sequence, uniformly on each compact \(z\)-set,

\[
 L_u(z,c)\longrightarrow m+a\,\operatorname{Im}z,\qquad
 L_v(z,c)\longrightarrow r+b\,\operatorname{Im}z. \tag{7.2}
\]

Indeed, \((c+z\log|c|)/c\to1\) uniformly on such a set, so
\(\log|c+z\log|c||/\log|c|\to1\).
The exponential contributes its displayed exact linear term.

The product transform is \(\zeta^{m+r}e^{-i(a+b)\zeta}\), and compact Fourier inversion gives

\[
 u*v=D^{m+r}\delta_{a+b}. \tag{7.3}
\]

Its proper profile is \(m+r+(a+b)\operatorname{Im}z\). The indicators are \(a\eta\), \(b\eta\), and \((a+b)\eta\); the carriers are \(\{a\}\), \(\{b\}\), and \(\{a+b\}\). The intercept records derivative order in the profile, while the indicator records the carrier.

## Worked example 2. Atomic cancellation and persistent transform zeros

Take

\[
 u=\delta_0+\delta_1,\qquad
 v=\delta_0-\delta_1,\qquad
 u*v=\delta_0-\delta_2. \tag{7.4}
\]

The two mixed \(\delta_1\) terms cancel. Their transforms satisfy

\[
 (1+e^{-i\zeta})(1-e^{-i\zeta})=1-e^{-2i\zeta}. \tag{7.5}
\]

Both input profiles converge in local \(L^1\) to
\(\max(0,\operatorname{Im}z)\) along every escaping real sequence. To verify this, write \(R=|c|\). For \(\eta=\operatorname{Im}z\geq d>0\), the exponential has modulus \(R^\eta\), so each input modulus divided by \(R^\eta\) lies between \(1-R^{-d}\) and \(1+R^{-d}\). For \(\eta\leq-d\), each modulus lies between those same two quantities without division by \(R^\eta\). Hence the normalized logarithms converge uniformly off any strip \(|\eta|<d\) to the stated profile.

The entire normalized family is locally uniformly bounded above. Profile compactness supplies a proper or collapsed limit on a further subsequence. The off-real convergence rules out collapse and identifies any proper limit almost everywhere with \(\max(0,\eta)\); the real line has zero area. Canonical uniqueness identifies it everywhere. If full local \(L^1\) convergence failed, a subsequence separated from that limit by a positive \(L^1\) distance would have a further compactness subsequence with the very same limit, a contradiction. This proves the full convergence.

The output profile is therefore \(2\max(0,\eta)\). Its indicator and carrier are

\[
 h_{u*v}(\eta)=2\max(0,\eta),\qquad C_{u*v}=[0,2]=[0,1]+[0,1]. \tag{7.6}
\]

At \(c_j=2\pi j\), the transform of \(v\) and of \(u*v\) is zero. Their normalized logarithms at \(z=0\) remain \(-\infty\), while their canonical limiting profiles at \(z=0\) equal zero. A persistent value at one point does not replace local \(L^1\) convergence.

![Atomic convolution cancels the two intermediate masses while the selected convex carriers add.](figures/atomic-cancellation-and-carrier-addition.png)

The signed masses in the figure are exactly those in (7.4). The support intervals in its lower panel are convex profile carriers. The output distribution is singular only at its two surviving atoms, so its carrier also contains smooth spatial points.

## Worked example 3. Two singular factors convolve to a smooth function

Let \(f\) be a nonnegative compact smooth probability density on \(\mathbb R\), positive on \((-1,1)\) and zero outside \([-1,1]\). Define

\[
 p=\delta_0(x_1)\otimes f(x_2),\qquad
 q=f(x_1)\otimes\delta_0(x_2). \tag{7.7}
\]

Their singular supports are the vertical and horizontal segments
\(\{0\}\times[-1,1]\) and \([-1,1]\times\{0\}\).
For an interior segment point, a tangential test against \(f\) reduces the distribution to a nonzero transverse point mass, which cannot be smooth. Closedness includes the endpoints. Their whole ordinary supports are those same segments.

Separate-coordinate convolution gives

\[
 p*q=f(x_1)f(x_2),\qquad
 F_p(\zeta)=\widehat f(\zeta_2),\qquad
 F_q(\zeta)=\widehat f(\zeta_1). \tag{7.8}
\]

The output is compact and smooth, with ordinary support \([-1,1]^2\). Its logarithmic profiles all collapse. The whole-complex rapid Fourier estimate for compact smooth functions, proved in Locating singularities through logarithmic Fourier strips, makes the normalized logarithms tend uniformly to \(-\infty\) on each compact parameter set.

There cannot be a common frequency sequence on which both input profiles are proper. On an escaping sequence with \(R=|c|\), at least one coordinate has modulus at least \(R/\sqrt2\). On a subsequence it is always the first coordinate or always the second. If it is the first, the smooth Fourier factor in \(F_q\) decreases faster than every power of \(R\), uniformly on compact \(z\)-sets, so \(q\)'s profile collapses. If it is the second, \(p\)'s profile collapses. A presumed proper limit would persist on that subsequence, giving a contradiction.

Nevertheless each input separately has a proper profile. For \(c=(t,0)\), \(t\to\infty\), \(F_p(c)=1\), so \(L_p(0,c)=0\) and uniform collapse is impossible; profile compactness yields a proper further subsequence. On this same sequence \(q\) collapses. The vertical sequence gives the opposite pair. Thus separate proper choices exist, but they cannot be paired on one common sequence.

Their separate carrier sum is nonempty, whereas

\[
 \mathcal J(p*q)=\{-\infty\}. \tag{7.9}
\]

This proves that the Cartesian replacement of the joint family in (4.3) is false.

![Crossed singular segments, their smooth convolution support, and the three shared-frequency directions that explain collapse.](figures/crossed-factors-and-shared-frequency.png)

The segments are actual ordinary and singular supports. An individual proper profile carrier is contained in its segment; the figure does not classify all such carriers. Horizontal and vertical frequency rays admit proper/empty pairs after extraction. The diagonal ray gives two collapsed profiles. The square is the ordinary support of the smooth convolution, whose logarithmic carrier is empty.

## Worked example 4. One output is smooth and another stays singular

Use \(p,q\) from Worked example 3 and let \(a=(1,-1)\). For two output coordinates, choose the prime and double-prime blocks

\[
 (u'_1,u'_2)=(p,\delta_0),\qquad
 (u''_1,u''_2)=(q,\delta_a). \tag{7.10}
\]

The outputs are \((f\otimes f,\delta_a)\). Every output sequence has the indicator tuple

\[
 (-\infty,h_a),\qquad h_a(\eta)=\eta_1-\eta_2. \tag{7.11}
\]

The input tuple has block order
\((h_p,0,h_q,h_a)\). A common sequence forces at least one of \(h_p,h_q\) to be collapsed, as proved above; the two point-mass profiles are exact on every sequence. Its coordinate sums are thus precisely \((-\infty,h_a)\), matching (7.11). The output carrier tuple is \((\varnothing,\{(1,-1)\})\).

This mixed case is part of the full identity. It would be lost by a rule that discarded every tuple having even one collapsed coordinate.

## Exercises with complete solutions

The ten exercises total 100 points.

### Exercise 1. Add orders and translate the point carrier — 10 points

Compute the convolution, profiles and indicators of \(D^2\delta_{-1}\) and \(D\delta_2\).

**Solution.** Their transforms are \(\zeta^2e^{i\zeta}\) and \(\zeta e^{-2i\zeta}\). Their product is \(\zeta^3e^{-i\zeta}\), the transform of \(D^3\delta_1\). The two profiles are \(2-\operatorname{Im}z\) and \(1+2\operatorname{Im}z\), so the product profile is \(3+\operatorname{Im}z\). The input indicators are \(-\eta\) and \(2\eta\); the output indicator is \(\eta\). The singleton carriers \(\{-1\}\) and \(\{2\}\) add to \(\{1\}\). Negative indicator values are allowed and do not mean an empty carrier.

### Exercise 2. Detect an incorrect normalization — 6 points

Let \(a=1_{[0,1]}\) and \(b=1_{[0,2]}\). Compute their Fourier transforms and the transform of their convolution at zero. Could an additional factor \(2\pi\) occur?

**Solution.** For \(\zeta\ne0\),
\[
 F_a(\zeta)=\frac{1-e^{-i\zeta}}{i\zeta},\qquad
 F_b(\zeta)=\frac{1-e^{-2i\zeta}}{i\zeta}. \tag{8.1}
\]
The removable values at zero are \(1\) and \(2\), equal to their integrals. The convolution integral is the product of the two integrals by Fubini, so \(F_{a*b}(0)=2\). The product formula gives exactly \(1\cdot2\). Any extra \(2\pi\) factor would give the wrong mass. The inverse Fourier convention has its separate factor \((2\pi)^{-1}\).

### Exercise 3. Prove proper convergence without pointwise convergence at zeros — 8 points

Suppose two normalized logarithms converge locally in \(L^1\) to proper profiles. Prove the convolution limit in that topology and explain why transform zeros do not require pointwise convergence everywhere.

**Solution.** Nonzero entire transforms have logarithms that are locally integrable; their zero sets have zero volume. The exact extended logarithmic product identity therefore gives an almost-everywhere sum of integrable functions. Subtract the two proper limits and apply the \(L^1\) triangle inequality to obtain (5.1). The limits are finite almost everywhere, so their sum is proper, and addition of the PSH line inequalities gives its canonical PSH representative. A sequence can retain a zero at an individual point, as Worked example 2 does. Such a point has no effect on the local \(L^1\) norm, so pointwise convergence there is not needed.

### Exercise 4. A collapsed factor does not need a convergent partner — 8 points

One factor collapses uniformly on every compact parameter set. The other normalized family is merely locally uniformly bounded above. Show that their product collapses.

**Solution.** Fix a compact set \(B\). Let \(A_\nu=\sup_B L_{u'}\to-\infty\), and let \(M_B\) bound the other family for large \(\nu\). The exact sum identity gives
\(\sup_B L_{u'*u''}\leq A_\nu+M_B\to-\infty\).
Since \(B\) is arbitrary, collapse is locally uniform. The partner may oscillate or have zeros; only its upper bound is used. No positive infinite value can cancel the negative divergence.

### Exercise 5. Preserve an output sequence when lifting factors — 10 points

Describe both constructions proving equality in (4.3). Why is choosing a fresh input sequence insufficient for the first construction?

**Solution.** Starting with a left-side output tuple, retain its witnessing sequence and apply compactness successively to each of the finitely many input families. The common subsequence preserves every output profile. Proper input pairs give the original output profile by uniqueness of the local \(L^1\) limit; collapsed inputs force the original output to collapse. Their indicators give the required joint right-side tuple.

Starting with a full joint input tuple, retain its witnessing sequence and add each pair of normalized logarithms. Every coordinate converges properly or collapses, so the output indicators belong to the left family. A fresh sequence in the first construction could select different output indicators and would not show that the originally chosen tuple is represented on the right.

### Exercise 6. Finite-height suprema need not add — 10 points

For \(v_1(z)=\log|\sin z|\) and \(v_2(z)=\log|\cos z|\), compare their real-translation envelopes at imaginary height zero and compute the indicator of their sum.

**Solution.** Each function has real-axis supremum zero, so \(M_1(0)+M_2(0)=0\). The product identity
\(\sin z\cos z=\tfrac12\sin(2z)\) gives supremum \(1/2\) in modulus on the real axis; hence \(M_{v_1+v_2}(0)=-\log2\), which is strictly smaller.

At height \(\eta\), maximizing the sine modulus over the real part gives \(\cosh\eta\). Thus
\[
 M_1(\eta)=M_2(\eta)=\log\cosh\eta,\qquad
 M_{v_1+v_2}(\eta)=\log\cosh(2\eta)-\log2. \tag{8.2}
\]
Dividing the envelopes at height \(t\eta\) by \(t\) and letting \(t\to\infty\) gives indicators \(|\eta|\), \(|\eta|\), and \(2|\eta|\). The constants disappear. Indicator additivity is exact even though the finite-height suprema do not add.

### Exercise 7. Add a third atomic factor — 10 points

Convolve the two factors in Worked example 2 with \(w=\delta_0+\delta_2\). Determine the resulting atoms and the common-sequence carrier.

**Solution.** The first convolution is \(\delta_0-\delta_2\). Its convolution with \(w\) is
\[
 (\delta_0-\delta_2)*(\delta_0+\delta_2)=\delta_0-\delta_4. \tag{8.3}
\]
The intermediate \(\delta_2\) terms cancel. The three transforms multiply to \(1-e^{-4i\zeta}\). The first two canonical profiles are \(\max(0,\eta)\), and the third is \(\max(0,2\eta)\), by the same off-real estimates as in Worked example 2. Their sum is \(4\max(0,\eta)\). The carrier is
\([0,1]+[0,1]+[0,2]=[0,4]\).
At \(c_j=2\pi j\), the product transform still vanishes at \(z=0\), although its canonical limiting profile there equals zero.

### Exercise 8. Quantify collapse in the crossed example — 10 points

In Worked example 3, assume \(|c_1|\geq R/\sqrt2\) and \(|z|\leq2\). Show that \(L_q\) is eventually at most \(-5\) on this compact parameter set, using a sufficiently high smooth Fourier decay order. Explain how this yields uniform collapse.

**Solution.** For sufficiently large \(R\), the real part of
\(\zeta_1=c_1+z_1\log R\) has modulus at least \(R/(2\sqrt2)\), and \(|\operatorname{Im}\zeta_1|\leq2\log R\). The whole-complex smooth Fourier estimate of order eight gives
\[
 |\widehat f(\zeta_1)|
 \leq C_8(1+|\zeta_1|)^{-8}e^{|\operatorname{Im}\zeta_1|}
 \leq C R^{-6}. \tag{8.4}
\]
For large \(R\), absorb the fixed \(C\) into one power of \(R\), obtaining \(|F_q|\leq R^{-5}\) and \(L_q\leq-5\). For an arbitrary requested negative bound \(-A\), increase the decay order beyond \(A+2\) and enlarge the threshold. This proves uniform collapse on the chosen compact set. The same reasoning works for every compact parameter set, with its own fixed imaginary-height coefficient.

### Exercise 9. An empty carrier does not imply a zero distribution — 8 points

Let \(g\) be a nonzero compact smooth function and \(a\in\mathbb R^n\). Compute \(g*\delta_a\), its ordinary support and its logarithmic profile carriers.

**Solution.** Directly from the convolution pairing,
\((g*\delta_a)(x)=g(x-a)\).
Its ordinary support is \(a+\operatorname{supp}g\), a nonempty compact set. It is smooth. Its normalized Fourier logarithms collapse uniformly on compact parameter sets, by the whole-complex smooth Fourier decay estimate. Thus every profile carrier is empty. The input carrier for \(g\) is empty and that for \(\delta_a\) is \(\{a\}\); their Minkowski sum is empty, in exact agreement with the theorem. This records smoothness and does not assert that the translated function vanishes.

### Exercise 10. Track three blocks and two mixed outputs — 20 points

On \(\mathbb R^2\), use the crossed factors \(p,q\). Let \(a=(1,0)\), \(b=(0,1)\), \(c=(2,0)\), and \(d=(-1,0)\). The three blocks of two factors are
\((p,\delta_a)\), \((q,\delta_b)\), and \((\delta_c,\delta_d)\).
Compute the two convolutions and their indicator/carrier tuple, preserving one common six-coordinate input family.

**Solution.** The first output is
\((p*q)*\delta_c=(f\otimes f)(x-c)\), a smooth function with ordinary support
\([1,3]\times[-1,1]\).
The second output is
\(\delta_a*\delta_b*\delta_d=\delta_{a+b+d}=\delta_{(0,1)}\).
Their joint output indicator tuple is
\[
 (-\infty,\eta_2),\qquad
 (C_1,C_2)=(\varnothing,\{(0,1)\}). \tag{8.5}
\]
On every common six-coordinate input sequence, at least one of the \(p,q\) profiles collapses, by the dominant-coordinate argument. Thus their first indicator sum, even after adding the finite point-mass indicator \(c\cdot\eta\), is \(-\infty\). The second sum is
\(a\cdot\eta+b\cdot\eta+d\cdot\eta=\eta_2\).
All point-mass profiles are exact on the same sequence and impose no further choice of frequencies. The finite-factor extension, Corollary 3.1, applies to this full joint family and gives (8.5). Choosing separate \(p,q\) sequences would incorrectly allow a proper first carrier.

## References

- Terence Tao, [246B, Notes 2: Some connections with the Fourier transform](https://terrytao.wordpress.com/2021/01/23/246b-notes-2-some-connections-with-the-fourier-transform/), for background on entire Fourier transforms and support. Its \(2\pi\) convention differs from (4.1).
- Lars Hörmander, *The Analysis of Linear Partial Differential Operators I*, Springer, 1983, Chapter VII, for compact Fourier theory.
- Lars Hörmander, *The Analysis of Linear Partial Differential Operators II*, Springer, 1983, Chapter XVI, for convolution and growth indicators. The full joint identity, mixed cases and both inclusions are proved below.