A ball rebuilt everywhere but its centre

A function can be rebuilt from its frequencies, and in one dimension the rebuilding has a pleasant property: whether it succeeds at a point depends only on what the function does near that point. Riemann noticed this in 1854 and it is called the localisation principle. A square wave has jumps, and the series built from round waves overshoots beside each jump by 9% for ever; but in the middle of a flat stretch, where the function is perfectly smooth, the partial sums settle on the right value. Trouble at a jump stays at the jump.

In three dimensions this fails. Take the simplest solid there is, the indicator function of a ball: 1 inside, 0 outside. At the centre it is constant for a whole radius in every direction, as smooth as anything can be. Rebuild it from its frequencies, keeping all frequencies below a cut-off $R$ in size, and read off the result at the centre. As $R$ grows the answer does not settle. It swings between about 0.36 and 1.64, for ever, and the cause is the surface of the ball, a full radius away from where the failure happens.

Keeping the frequencies inside a ball #

The Fourier transform takes a function $f$ on $d$-dimensional space and gives its amplitude $\hat f(\xi)$ at every frequency vector $\xi$; the inversion formula rebuilds $f$ as the integral of $\hat f(\xi) e^{2\pi i x\cdot\xi}$ over all frequencies. That integral may not converge outright for a function with a jump, which is why a cut-off is needed: keep the frequencies with $|\xi| < R$ and let $R$ grow,

$$S_R f(x) = \int_{|\xi| < R} \hat f(\xi)\, e^{2\pi i x\cdot \xi}\, d\xi .$$

In one dimension that is the integral version of a partial sum, and when the period grows without bound showed how the sum over separate frequencies turns into this integral. In more dimensions there is a choice of what shape of region to keep, and the ball $|\xi| < R$ is the natural one, since it treats every direction alike. It is what a physicist does when a lens or a detector passes every spatial frequency below a threshold.

For the indicator of the unit ball, everything at the centre can be computed exactly, because the function depends only on distance from the centre and so does its transform. With $\nu = d/2$ and $J_\nu$ a Bessel function, the value at the centre is

$$S_R f(0) = \frac{1}{2^{\nu-1}\Gamma(\nu)} \int_0^{2\pi R} u^{\nu - 1} J_\nu(u)\, du .$$

In one dimension this is $(2/\pi)\,\mathrm{Si}(2\pi R)$, the sine integral, which tends to 1 with ripples dying away like $1/R$. In two it is $1 - J_0(2\pi R)$, which tends to 1 with ripples dying like $R^{-1/2}$. In three it is

$$S_R f(0) = \frac{2}{\pi}\big(\mathrm{Si}(2\pi R) - \sin 2\pi R\big),$$

and the second term does not die at all. Since $\mathrm{Si}$ tends to $\pi/2$, the centre value is close to $1 - (2/\pi)\sin 2\pi R$ for large $R$, and it keeps swinging between $1 - 2/\pi \approx 0.363$ and $1 + 2/\pi \approx 1.637$. In the hero the ball’s curve reaches 1.637 at $R = 7.75$ and returns to 0.987 at $R = 8$, a quarter-period later. The phenomenon was described by Mark Pinsky and his colleagues in the early 1990s, and it carries his name.

A dip that narrows and never fills #

Is the whole ball rebuilt badly, or just its centre? The value at any point at distance $s$ from the centre can be computed in the same way, since averaging a plane wave over all directions turns it into a single function of the frequency’s size.

Away from the centre the rebuilt ball settles towards 1 in the usual way, and at the surface it has the familiar overshoot of a jump. The centre is a separate event. At each of the three cut-offs the value there is 0.363, and the dip around it is narrower as $R$ doubles, its width shrinking like $1/R$. So the failure is confined to a smaller and smaller region, and the partial inversions still converge in the averaged sense that the mean-square theory guarantees for every function whose square has a finite integral. Pointwise, one point is lost entirely.

It is worth being clear how strange this is. The function is constant on a ball of radius 1 around the centre. In one dimension, a function constant on an interval around a point has partial integrals converging to that constant at the point, whatever it does elsewhere. Here the function does nothing at all near the centre, and the rebuilding fails there and only there.

Where the failure comes from #

The partial inversion is an average of the function against a fixed kernel: $S_R f(0)$ is the integral of $f(y) D_R(y)$, where $D_R$ is the inverse transform of the ball of frequencies — the multidimensional version of the Dirichlet kernel whose ripples make a series run away. For the ball, that average is simply the kernel’s total mass inside radius 1. So the question is how the kernel’s mass accumulates as the radius grows.

The kernel is radial and ripples, positive and negative in alternating thin spherical shells about $1/R$ thick. In the plane the cumulative mass swings round 1 with a shrinking amplitude, so wherever the edge of the disc falls, the mass inside it is close to 1. In space the swings never shrink: at every radius the mass inside is somewhere between $1 - 2/\pi$ and $1 + 2/\pi$, and which depends on where in its last ripple the radius falls. The ball’s surface is at radius 1, and the phase of the kernel’s ripple there is $2\pi R$, which keeps turning as $R$ grows.

Why does space keep the swings? Each ripple of the kernel is a shell, and its mass is its density times its area. The kernel’s density at radius $r$ falls off like $r^{-(d+1)/2}$, while a shell’s area grows like $r^{d-1}$, so a ripple’s mass behaves like $r^{(d-3)/2}$. In the plane that is $r^{-1/2}$ and the ripples weaken. In space it is $r^0$: every ripple, however far out, carries the same mass. The ball’s surface is a whole sphere of points at exactly the same distance from the centre, so all of it sits in one ripple, and nothing averages its contribution away.

The term that is multiplied by the cut-off #

The same mechanism can be followed through the formula. Because the ball and the kernel are both radial, the centre value reduces to a single integral over the radius, of the ball’s radial profile against the kernel’s mass per unit radius. In three dimensions that mass per unit radius is

$$\frac{2}{\pi r}\big(\sin 2\pi R r - 2\pi R r \cos 2\pi R r\big),$$

and its second part is $-4R\cos 2\pi R r$: an oscillation whose size grows with the cut-off. Riemann’s argument in one dimension rests on the fact that a fixed smooth function multiplied by a faster and faster oscillation integrates to almost nothing, which is the Riemann–Lebesgue lemma. Here the oscillation is faster, but it is also taller in exact proportion. Integrated from 0 up to the ball’s edge it gives $-(2/\pi)\sin 2\pi R$ exactly, a contribution that comes entirely from the endpoint $r = 1$, where the profile jumps from 1 to 0. A smooth profile would let an integration by parts move the factor $R$ onto a derivative and kill it. A jump leaves a boundary term, and the boundary term is the swing.

In one dimension the corresponding kernel has no such growing part, because there is no area factor to supply the extra power of $r$, and the Riemann–Lebesgue lemma wins. In two the growing part is present but only as $R^{1/2}$ against a decay of the same order, and it loses by a margin of $R^{-1/2}$. Three dimensions is the first place where the area of a sphere grows fast enough to balance the decay of the kernel exactly.

Why every other point is rebuilt #

From a point at distance $s$ from the centre, the ball’s surface is not at one distance but at every distance between $1 - s$ and $1 + s$. Each piece of the surface still sends a contribution oscillating with its own distance, but now the phases are spread over an interval of length $2s$, and for $R s$ much larger than 1 they cancel one another, much as the many partial reflections from a rough surface cancel to leave no echo. The cancellation needs $R s$ to be large, which is why the dip in the profile has width proportional to $1/R$: inside that width the phases are still nearly aligned.

So the failure is a matter of exact symmetry. Move the evaluation point by any fixed amount, however small, and for large enough $R$ the partial inversion settles there. Replace the ball by an ellipsoid that is not a ball, and the centre is no longer equidistant from the surface: the distances run between the shortest and longest semi-axes, the phases spread, and the swing at the centre dies away too. The ball is the worst case precisely because it is the most symmetric shape there is.

Three dimensions is the threshold #

The exponent $(d-3)/2$ predicts more than the three-dimensional case. In four dimensions the ripples’ mass grows with the radius, and the centre value should swing with an amplitude growing like $R^{1/2}$; in five, like $R$.

The slopes measured from the exact formula between $R = 24$ and $90$ are $-0.504$, $0.000$, $0.496$ and $0.989$. Three is precisely where the error stops decaying. Above three the partial inversions at the centre of a ball are unbounded: in five dimensions the error passes 100 before $R$ reaches 100. Dimension three is not special in the formulas, which run smoothly through every $d$; it is special because $(d-3)/2$ changes sign there. That is the same kind of threshold as the one that makes a random walk come home in two dimensions and not in three, though the arithmetic is different.

The same jump, as a sound #

The ball’s surface delivering its whole jump to the centre at once has a physical twin, and it is exact. Solve the wave equation in space starting from a displacement equal to the ball’s indicator and no velocity. Kirchhoff’s formula gives the solution at the centre as the time derivative of $t$ times the average of the initial data over a sphere of radius $t$. That average is 1 while $t < 1$ and 0 after, so the solution at the centre is 1 until $t = 1$, then an infinitely sharp negative spike at exactly $t = 1$, then nothing. The whole surface of the ball sends its jump inwards at unit speed, and every part of it arrives at the centre at the same instant. A plucked string carries its corners unchanged in one dimension; a ball focuses its edge into a point in three.

In the plane there is no such spike. Waves in two dimensions leave a trailing wake, so the disc’s edge reaches the centre spread out over time, and its effect there stays finite. The frequencies and the waves are telling the same story: in three dimensions a sphere of discontinuity is focused, and the focus is the centre.

Averaging over the cut-off #

The swing is essentially a sine of $2\pi R$, and a sine averages to nothing over long stretches, so the running average of the centre value over all cut-offs settles on the right answer. At $R = 12$ it is 0.9916 and still closing. This is the trick of Fejér’s averages in one dimension, which tame the overshoot of a square wave, applied to the cut-off rather than to the terms. Salomon Bochner’s version smooths the edge of the ball of frequencies instead of averaging over its size: weight each frequency by $(1 - |\xi|^2/R^2)^\delta$ rather than cutting it off sharply. For $\delta$ large enough the problem at the centre disappears, and how large is large enough is the subject of the open problem below.

The series on a torus, summed two ways #

Everything so far concerned the integral over all of space. The same failure appears for a Fourier series, where the frequencies are whole-number vectors and the function lives on a cube whose opposite faces are glued.

Summed over balls of frequency vectors, the series at the centre follows the prediction closely, swinging between about 0.36 and 1.64 with a little extra noise from how the lattice points fall on spheres. Summed over cubes instead — all vectors with every coordinate at most $N$ in size — the same coefficients behave quite differently: the values are 0.910 at $N = 10$, 0.982 at 40 and 0.989 at 60, and computed further they reach 0.993 at $N = 100$, closing on 1 roughly in proportion to $1/N$.

So the convergence of a multiple Fourier series at a point is not a property of the function alone. It depends on the order in which the terms are added, which in one dimension never arises because there is only one sensible order. The cube of frequencies has flat faces, and a flat face cannot focus. The ball of frequencies has a curved face, and its kernel has spherical ripples that do. In a grid that packs the copies the shape of a region of frequencies decided how efficiently a signal could be sampled; here it decides whether the signal can be rebuilt at a point at all.

An endless swing read from a formula, not a picture #

The figures compute exact formulas, or numerical integrals of them, at finitely many cut-offs, and an endless swing is inferred from the closed form $(2/\pi)(\mathrm{Si}(2\pi R) - \sin 2\pi R)$, not from the pictures. A reader shown only the hero could not tell whether the swings would eventually die out at a cut-off of a million. The closed form says they do not.

The convergence of the cubic sums on the torus is a computation to $N = 100$, not a theorem proved here, and the noise in the spherical sums from lattice points has not been analysed. The figures also say nothing about functions that are not radial. Pinsky’s phenomenon needs the discontinuity to lie on a sphere centred at the point in question, and a ball seen from any point other than its centre has its surface at many different distances, which is why the profile settles everywhere except at $s = 0$.

Centre values computed, Pinsky’s theorem quoted #

Computed here: the centre value in dimensions 1 to 5 from the Bessel-integral formula, by Simpson’s rule, checked at three cut-offs against $(2/\pi)\mathrm{Si}(2\pi R)$, $1 - J_0(2\pi R)$ and $(2/\pi)(\mathrm{Si}(2\pi R) - \sin 2\pi R)$ to within $2 \times 10^{-4}$; the profile along a radius from the radial transform and the spherical average of a plane wave; the kernel’s cumulative mass; the swing amplitudes and their slopes; the running average; and the torus sums over all lattice points within radius 60, and over cubes to half-side 100. Kirchhoff’s formula was applied to the ball by hand, as written above.

Quoted: Riemann’s localisation principle; the attribution to Pinsky and to Pinsky, Stanton and Trapa for the torus; Bochner’s means; and the results named in the next section.

Still open: the Bochner–Riesz problem #

The sharp cut-off at a ball of frequencies fails in a much stronger way than at one point. Charles Fefferman proved in 1971 that in two or more dimensions the operation “keep the frequencies inside a ball” does not even keep functions of finite $p$-th power integral bounded, for any $p$ other than 2. Bochner’s smoothed cut-offs repair this, and the Bochner–Riesz conjecture states exactly how much smoothing is needed: for a given $p$, the means with exponent $\delta$ should be bounded whenever $\delta$ exceeds $\max\{0,\; d\,|1/p - 1/2| - 1/2\}$. It was proved in the plane by Lennart Carleson and Per Sjölin in 1972. In three and more dimensions it is open, and it is tied to the Kakeya problem about sets containing a segment in every direction, which is also unresolved there.

Pinsky’s phenomenon is a small visible piece of this. The sharp cut-off fails because the boundary of a ball is curved, and the same curvature, which makes spherical ripples focus, is what the Bochner–Riesz exponent is measuring.

One point, a radius away from the trouble #

Rebuilding a solid ball from its frequencies below a cut-off fails at its centre in three dimensions, where the result swings between $1 - 2/\pi$ and $1 + 2/\pi$ for ever, though the ball is constant there. The whole spherical surface is the same distance from the centre, and the averaging kernel’s spherical ripples keep the same mass at every radius, so the surface’s jump arrives at the centre undiminished and in step. In the plane the ripples weaken and the centre is rebuilt; in four or more dimensions the error grows without bound. Averaging over the cut-off, or summing over cubes of frequencies, restores the centre.