(December 21, 2014) Harmonic analysis on spheres

Paul Garrett [email protected] http://www.math.umn.edu/egarrett/ [This document is http://www.math.umn.edu/˜garrett/m/mfms/notes 2013-14/09 spheres.pdf] 1. Calculus on spheres 2. Spherical Laplacian from Euclidean 3. Eigenvectors for the spherical Laplacian 4. Invariant integrals on spheres 5. L2 spectral decompositions on spheres 6. Sup-norms of on Sn−1 7. Pointwise convergence of Fourier-Laplace series 8. Irreducibility of representation spaces for O(n) 9. Hecke’s identity • Appendix: Bernstein’s proof of Weierstraß approximation

Harmonic analysis on the circle S1 ≈ R/Z uses Fourier series expansions of functions and generalized functions X  Z  f ∼ f(t) e−2πint dt e2πinx / n∈Z R Z The exponential functions are simple, special functions, in several regards. Exponentials are eigenfunctions d iξx of the translation-invariant differential operator dx , and the maps x → e for ξ ∈ C are continuous group homomorphisms to C×. The senses in which Fourier expansions converge or diverge play a key role. Harmonic analysis on the line uses Fourier inversion expansions of functions and generalized functions Z  Z  f ∼ f(t) e−2πiξt dt e2πiξx dξ R R This is more complicated than on the circle, because the line is not compact. On R the exponential functions, d × 2 still eigenfunctions for dx and still giving group homomorphisms to C , are no longer in L (R). Similarly, Fourier inversion expresses functions as integrals of exponential functions, not as sums. Senses of convergence of Fourier integrals are commensurately more complicated than Fourier series.

Both the circle and the real line are abelian groups. The abelian-ness simplifies the harmonic analysis. Unsurprisingly, harmonic analysis related to non-abelian groups is more complicated. One immediate complication is that for non-abelian G many subgroups H are not normal, equivalently, quotients G/H are not groups. By contrast, the quotient R/Z of the real line by the presents the circle as a group, not merely as a quotient space of a group.

n−1 n [1] A tractable non-abelian, but still compact, situation is spheres S ⊂ R , which are quotients Sn−1 ≈ SO(n − 1) \ SO(n)

of rotation groups SO(n). Spheres themselves are rarely groups, but are acted-upon transitively by rotation groups. Specifying a rotation-invariant measure on Sn−1 gives a space

Z 1/2 2 n−1 o n−1  2 L (S ) = completion of C (S ) with respect to |f|L2 = |f| Sn−1 With corresponding hermitian inner-product Z hf, F i = f · F Sn−1

[1] Rotation groups as orthogonal groups and special orthogonal groups are reviewed below.

1 Paul Garrett: Harmonic analysis on spheres (December 21, 2014)

the translation action f → g · f of g ∈ SO(n) on functions f by (g · f)(x) = f(xg) is unitary, meaning that it preserves the inner product: using the translation-invariance of the measure, Z Z Z hg · f, g · F i = f(xg) F (xg) dx = f(x) F (x) d(g−1x) = f(x) F (x) dx = hf, F i Sn−1 Sn−1 Sn−1 For example, Weyl’s equidistribution criterion for point sets on Sn−1 requires identification of a convivial orthonormal basis for L2(Sn−1). In fact, we will prove an orthogonal decomposition

∞ 2 n−1 M L (S ) = (completion of) Hd (orthogonal) d=0

into finite-dimensional O(n)-stable subspaces Hd. In particular,

n Hd = {harmonic, degree d homogeneous on R } where the harmonic condition is ∆f = 0, with the usual Euclidean Laplacian

2 2 n ∂ ∂ R ∆ = ∆ = 2 + ... + 2 ∂x1 ∂xn 2 n−1 2 n−1 Letting projd be the orthogonal projection of L (S ) to Hd, expansions of f ∈ L (S ) as ∞ ∞ X X f = fd = projd(f) (with fd ∈ Hd) d=0 d=0 are sometimes called Fourier-Laplace series. On S1, these expansions are readily compared to the usual Fourier expansions, as done below. While L2 convergence follows from simple generalities about Hilbert spaces, on Sn−1 with n−1 > 1, pointwise convergence of Fourier-Laplace series is somewhat more complicated than for Fourier series, because the the n−1 n n−2 worst-case sup-norms of orthonormal bases of Hd for S ⊂ R grow like d as d increases. We prove 2 n−1 P a Sobolev imbedding theorem: let f be in L (S ) with Fourier-Laplace expansion d fd with fd ∈ Hd. Suppose X s 2 n−1 (1 + d) · |fd|L2 < +∞ (for some s > dim S ) d Then f is continuous, and its Fourier-Laplace series converges uniformly pointwise to f. The left-hand side of the displayed inequality is the sth Sobolev (squared).

Some ideas from the discussion of sup-norms can be used to show that the spaces Hd are irreducible for the action of SO(n), that is, given a non-zero f ∈ Hd, the collection of finite linear combinations of translates g · f of f by g ∈ SO(n) is all of Hd. That is, Hd is an irreducible representation of SO(n).

The irreducibility of Hd enables an easy proof of Hecke’s identity Z −πhx,xi −2πihξ,xi −d −πhξ,ξi f(x) e · e dx = i · f(ξ) e (for f ∈ Hd) Rn d by reducing to the case f(x1, . . . , xn) = (x1 ± ix2) , which allows direction computation. Hecke’s identity is essential to ascertain the behavior of harmonic theta series

X −πi(m2+...+m2 ) z θ(z) = θn,f (z) = f(m) e 1 n (with z ∈ H) n m=(m1,...,mn)∈Z m under inversion z → −1/z, to see that these theta series are modular forms of weight 2 + d. Indeed, for m odd, the weight is half-integral, a significantly more complicated case than the integral-weight case. For example, these theta series arise when carrying out a Weyl-criterion equidistribution analysis for lattice points projected to the unit sphere, as we see later.

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1. Calculus on spheres

To use the rotational symmetry of spheres, we want eigenfunctions for rotation-invariant differential operators on spheres, and expect that these eigenfunctions will be the analogues of exponential functions on the circle or line. Thus, we must identify/construct rotation-invariant differential operators on spheres. Also, we need rotation-invariant measures or integrals on spheres. Rather than writing formulas in coordinates, we first describe these objects by their desired properties, then give constructions in terms of the ambient Euclidean space.

For 1 ≤ n ∈ Z, the usual unit (n − 1)-sphere is n−1 n 2 2 S = S = {(x1, . . . , xn) ∈ R : x1 + ... + xn = 1}

n−1 Write ∆S = ∆S for the desired rotation-invariant second-order [2] differential operator on functions on R S S, and let f → S f denote the desired rotation-invariant (positive) integral. We call ∆ the Laplacian on the sphere. All functions here are indefinitely differentiable. [3] Two desired properties are R S R S Sn−1 (∆ f) · ϕ = Sn−1 f · (∆ ϕ)(self-adjointness)

R S Sn−1 (∆ f) · f ≤ 0 (definiteness) with equality only for f constant. We also assume that ∆S has real coefficients, in the abstracted sense that ∆Sf = ∆S f There is the natural complex hermitian inner product Z hf, F i = f · F Sn−1

A typical linear algebra conclusion, via a typical argument:

[1.0.1] Corollary: Granting existence of invariant ∆S and invariant measure on Sn−1, with the self- adjointness and definiteness properties, eigenvectors f, F for ∆S with distinct eigenvalues are orthogonal with respect to the inner product h, i. Any eigenvalues are non-positive real numbers.

Proof: Let ∆Sf = λ · f and ∆SF = µ · F . Assume λ 6= 0 (or else interchange the roles of λ and µ). Then 1 Z 1 Z λ Z hf, fi = (∆Sf) · f = f ∆Sf = f f λ S λ S λ S Since λ 6= 0, f is not identically 0, the integral of f · f is not 0, and λ = λ, so λ is real. The negative definiteness of ∆S and positive-ness of the invariant measure on S give Z λ · hf, fi = (∆Sf) · f < 0 S

[2] Unlike the circle R/Z and line R with d/dx, there are no rotation-invariant first-order scalar-valued differential operators on higher-dimensional spheres. Fortunately, invariant second-order operators are ubiquitous, and can be derived from the Euclidean Laplacian.

[3] The notion of differentiability for functions on a sphere can be given in several ways, all equivalent. At one extreme, the most pedestrian is to declare a function f on Sn−1 differentiable if the function F (x) = f(x/|x|) on n n n−1 R − 0 is differentiable on R − 0. At the other extreme one gives S its usual structure of smooth manifold, which incorporates a notion of differentiable function. Happily, the choice of definition doesn’t matter much, since we won’t be attempting to directly compute derivatives, but only use properties of differentiation.

3 Paul Garrett: Harmonic analysis on spheres (December 21, 2014) so λ < 0. Next, 1 Z 1 Z µ Z hf, F i = (∆Sf) · F = f · ∆SF = f · F λ S λ S λ S The eigenvalues λ, µ are real, so for µ/λ 6= 1 necessarily the integral is 0. ///

The standard special orthogonal group [4] is

> SO(n) = {g ∈ GLn(R): g g = 1n and det g = 1} and acts on S by right [5] matrix ,

k × x −→ xk (for x ∈ Sn−1 and k ∈ SO(n)) considering elements of Rn as row vectors. Refer to the action of elements of SO(n) on S = Sn−1 as rotations or translations. [6] It is useful that for g ∈ SO(n), inverting both sides of g>g = 1 gives g−1(g>)−1 = 1, and then 1 = gg>.

The usual inner product h, i on Rn

hx, yi = xy> (for row vectors x, y) is preserved by SO(n), so SO(n) preserves lengths and angles:

hxk, yki = (xk)(yk)> = x (kk>) y> = x y> = hx, yi

For any k such that k>k = 1

2 > > (det k) = det k · det k = det(k k) = det 1n = 1

Thus, det k = ±1. Recall:

[1.0.2] Claim: The action of SO(n) on Sn−1 is transitive. [7]

Proof: We show that, given x ∈ S there is k ∈ SO(n) such that e1k = x, where e1, . . . , en is the standard basis for Rn. That is, we construct k ∈ SO(n) such that the top row of k is x. Indeed, complete x to n [8] an R-basis x, x2, x3, . . . , xn for R . Then apply the Gram-Schmidt process to find an orthonormal (with n > respect to the standard inner product) basis x, v2, . . . , vn for R . The condition kk = 1, when expanded, is

[4] > The full orthogonal group is defined by g g = 1n without the condition. The determinant condition preserves orientation, however orientation is defined. The modifier special refers to the determinant condition.

[5] The choice of right matrix multiplication on row vectors is not essential, but right action on vectors has minor notational advantages when discussing functions on spaces of vectors.

[6] Proof that SO(n) is exactly all the rotations would require a precise definition of rotation, which we avoid.

[7] Recall that transitivity means that for all x, y ∈ S there is g ∈ O(n) such that gx = y. If SO(n) is all rotations of the sphere, then physical intuition makes the transitivity plausible. But this requires two assumptions: that all rotations are given by SO(n), and that intuition about rotations is accurate in higher dimensions. We can do better.

[8] Recall that, given a basis v1, . . . , vn for a (real or complex) with an inner product (real-symmetric or complex hermitian), the Gram-Schmidt process produces an orthogonal or orthonormal basis, as follows. Replace v1 by v1/|v1 to give it length 1. Then replace v2 first by v2 − hv2, v1iv1 to make it orthogonal to v1 and then by v2/|v2| to give it length 1. Then replace v3 first by v3 − hv3, v1iv1 to make it orthogonal to v1, then by v3 − hv3, v2iv2 to make it orthogonal to v2, and then by v3/|v3| to give it length 1. And so on.

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the assertion that the rows of k form an orthonormal basis, so taking x, v2, . . . , vn as the rows of k, we have k such that e1k = x. As noted just above, the determinant of this k is ±1. To ensure that it is 1, replace vn by −vn if necessary. This still gives e1k = x, giving the transitivity. ///

The isotropy group SO(n)en of the last standard basis vector en = (0,..., 0, 1) is  A 0  (isotropy group) = SO(n) = { : A ∈ SO(n − 1)} ≈ SO(n − 1) en 0 1

Thus, by transitivity, as SO(n)-spaces

Sn−1 ≈ SO(n − 1)\SO(n)

n−1 n [9] The action of g ∈ SO(n) on functions f on the sphere S = S (or on the ambient R ) is (g · f)(x) = f(xg)

The rotation invariance conditions are  Z Z  g · f = f (for g ∈ SO(n)) Sn−1 Sn−1  ∆S(g · f) = g · (∆Sf) (for g ∈ SO(n))

[1.0.3] Remark: To prove existence of an invariant integral and invariant Laplacian, there are several approaches, of varying expense and quality. For immediate purposes, we choose a pedestrian ad hoc approach, despite its ugliness. A better argument will be given later.

2. The spherical Laplacian

One direct proof of existence of the invariant Laplacian on S = Sn−1 uses the imbedding of the sphere S = Sn−1 in Rn. The usual Euclidean Laplacian  2  2 n ∂ ∂ ∆ = ∆R = + ... + (on Rn) ∂x1 ∂xn

[10] n n−1 is SO(n)-invariant on differentiable functions on R . For a function f on S , create a function F on Rn − 0 by F (x) = f(x/|x|), and define ∆Sf = (restriction to S of ) ∆F

[9] If we let the group act by left multiplication, and then try to define (g · f)(x) = f(gx), we have a problem, as follows. For g, h ∈ SO(n), associativity fails, since

((gh) · f)(x) = f((gh) x) = f(g(hx)) = (g · f)(hx) = (h · (g · f))(x)

instead of g · (h · f). A left action on the set requires defining (g · f)(x) = f(g−1x). Avoiding the inverse is a pleasant minor economy.

[10] To verify that the usual Euclidean Laplacian is SO(n)-invariant is a direct computation. In fact, the Laplacian th n n is invariant under the full orthogonal group O(n). Let k ∈ O(n) with ij entry kij. For F on R and x ∈ R , X ∂ X X ∂ X X ∆(k · F )(x) = ( )2F (..., x k ,...) = k F (..., x k ,...) ∂x i ij ∂x `s s i ij ` ` i ` ` s j

th where Fs is the of F with respect to its s argument. Taking the next derivative gives

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[11] n The map f → F that creates from f on S the degree-zero positive-homogeneous function F on R − 0 commutes with the action of SO(n). [12] From the definition,

∆S f = ∆S f

The SO(n)-invariance of the spherical Laplacian follows from the SO(n)-invariance of the usual Laplacian: for k ∈ SO(n) S ∆ (k · f) = (∆(k · F ))|S = (k · (∆F ))|S = k · (∆F )|S since restriction to the sphere commutes with SO(n), as does f → F . Thus, ∆S is SO(n)-invariant.

3. Eigenvectors for the spherical Laplacian

n As usual, a function on Rn is harmonic when it is annihilated by the Euclidean Laplacian ∆ = ∆R .

[3.0.1] Proposition: For f positive-homogeneous of degree s and harmonic, the restriction f|S of f to the sphere Sn−1 is an eigenfunction for ∆S,

S ∆ (f|S) = −s(s + n − 2) · (f|S) with eigenvalue −s(s + n − 2).

Proof: First, the effect of ∆S on a positive-homogeneous function f of degree s is directly computed:

th Let r = |x| and let fi be the partial derivative with respect to the i argument:

2 S −s  X ∂ 2 − s  ∆ (f| ) = ∆f(x/|x|) = ∆ |x| · f = (r ) 2 · f S ∂x2 i i

X ∂  s 2 −( s +1) 2 −s/2  X ∂  2 −( s +1) 2 −s/2  = − (2x )(r ) 2 f + (r ) f = −sx (r ) 2 f + (r ) f ∂x 2 i i ∂x i i i i i i

X  2 −( s +1) s 2 −( s +2) 2 −( s +1) = −s (r ) 2 f + sx ( + 1)(2x )(r ) 2 f − sx (r ) 2 f i 2 i i i i

s 2 −( s +1) 2 −s/2  − (2x )(r ) 2 f + (r ) f 2 i i ii 2 −( s +1) 2 2 −( s +2) 2 −( s +1) 2 −s/2 = −ns (r ) 2 f + sr (s + 2) (r ) 2 f − s(r ) 2 sf + (r ) ∆f

P P P P th ` s,t k`sk`tFst(..., i xikij,...). Interchange the order of the sums and observe that ` k`sk`t is the (s, t) > th entry of k · k, which is the (s, t) entry of 1n. Thus, the whole is

X X Fss(..., xikij,...) = (∆F )(xk) = (k · (∆F ))(x) s i

[11] m m s A function F on R is positive-homogeneous of degree s ∈ C if, for all t > 0 and for all x ∈ R , F (tx) = t · F (x). We make F (x) = f(x/|x|) positive-homogeneous of degree 0, rather than |x|s f(x/|x|) positive-homogeneous of degree n s, so that constant functions on the sphere become constant functions on R −0, and are annihilated by any differential operator.

[12] The map f → F commutes with the action (k · F )(x) = F (xk) of SO(n) on functions because F (xk) = f(xk/|xk|) = f((x/|x|) · k) = (kf)(x/|x|).

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by using Euler’s identity [13] that for positive-homogeneous f of degree s,

X xi fi(x) = s · f i

P 2 2 as well as the obvious i xi = r . Simplifying,

∆(|x|−s · f) = −ns r−(s+2) f + s(s + 2) r−(s+2) f − 2s r−(s+2) sf + r−s ∆f

= −s(n − (s + 2) + 2s) r−(s+2) f + r−s ∆f = −s(n + s − 2) r−(s+2) f + r−s ∆f Then for ∆f = 0, restriction to the sphere makes r = 1, giving the eigenfunction property. ///

The most tractable positive-homogeneous functions are homogeneous polynomials. Let

Hd = {homogeneous (total) degree d harmonic polynomials in C[x1, . . . , xn]}

(d) Let C[x1, . . . , xn] be the homogeneous polynomials of degree d. Introduce a temporary complex-hermitian form [14] (, ): C[x1, . . . , xn] × C[x1, . . . , xn] −→ C by   (P,Q) = Q(∂)P (x) |x=0

where Q(∂) means to replace xi by ∂/∂xi in a , and R|x=0 means to evaluate R at x = 0. The main point of this construction is the immediate identity

(∆f, g) = (f, r2g)

2 2 2 2 where r = x1 + ... + xn. That is, multiplication by r is adjoint to application of ∆ with respect to (, ). This will be useful after proof that (, ) is non-degenerate:

[3.0.2] Claim: The form (, ) is positive-definite hermitian.

Proof: Observe that (P,Q) = 0 for homogeneous polynomials P,Q unless P,Q are of the same degree. (d) When restricted to the homogeneous polynomials C[x1, . . . , xn] of degree d, the form (, ) has an orthogonal basis of distinct , since  m1 mn e1 en 0 (if any mi 6= ei) (∂x1 . . . ∂xn )(x1 . . . xn )|x=0 = m1! . . . mn! (if every mi = ei)

[13] Euler’s identity is readily proven by considering the function g(t) = G(tx) = tsG(x) for t > 0, differentiating with respect to t, and evaluating at t = 1:

X ∂G(tx) sts−1G(x) = g0(t) = x i ∂x i i

gives s · G(x) = P x ∂G(x) . i i ∂xi

[14] This hermitian form is not as ad hoc as it may look. From a slightly more sophisticated viewpoint: it pairs polynomials against Fourier transforms of polynomials, which are derivatives of Dirac delta at 0, which are compactly- supported distributions, so can be evaluated on polynomials, which are smooth. One need not think in these terms to use the properties here, which also admit elementary proofs.

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(d) Looking at the orthogonal basis of monomials, (, ) is hermitian and positive definite on C[x1, . . . , xn] . ///

[3.0.3] Claim: The map (d) (d−2) ∆ : C[x1, . . . , xn] −→ C[x1, . . . , xn] (d) 2 is surjective. Harmonic polynomials f in C[x1, . . . , xn] are orthogonal to polynomials r h (with h ∈ (d−2) C[x1, . . . , xn] ) with respect to (, ).

(d−2) (d) Proof: For h ∈ C[x1, . . . , xn] , if (∆f, h) = 0 for all f ∈ C[x1, . . . , xn] , then

0 = (∆f, h) = (f, r2h) (for all f)

so r2h = 0, so h = 0, by the positive-definiteness of (, ). This also proves the second assertion. ///

Iterating the claim just proven,

(d) 2 d−2 4 d−4 C[x1, . . . , xn] = Hd ⊕ r H ⊕ r H + ...

[3.0.4] Claim: Polynomials restricted to the n-sphere are equal to linear combinations of harmonic polynomials restricted to the sphere.

Proof: Use (d) 2 d−2 4 d−4 C[x1, . . . , xn] = Hd ⊕ r H ⊕ r H + ... to write a as

2 4 f = f0 + r f2 + r f4 + ... (with each fi harmonic)

Restricting to the sphere,

2 4 f|S = (f0 + r f2 + r f4 + ...)|S = (f0 + f2 + f4 + ...)|S

since r2 = 1 on the sphere. ///

[3.0.5] Remark: Again, from computations above,

S ∆ f = −d(d + n − 2) · f (for f ∈ Hd)

Since the degree d is non-negative, n − 2 n − 2 λ = −d(d + n − 2) = −(d + )2 + ( )2 d 2 2

the eigenvalues λd = −d(d+n−2) are non-positive, and 0 only for degree d = 0. As d → +∞, the eigenvalues 0 go to −∞. Indeed, λd0 < λd ≤ 0 for d > d, so the spaces Hd are distinguished by their eigenvalues for the spherical Laplacian.

[3.0.6] Remark: For the circle S1, the 0-eigenspace is 1-dimensional and for d > 0 the (−d2)-eigenspace is 2-dimensional, with basis (x ± iy)d. By contrast, for n > 1 the dimensions of eigenspaces are unbounded as the degree d goes to +∞. Specifically,

[3.0.7] Claim: The dimension of Hd is

n + d − 1 n + d − 3 dim H = dim [x , . . . , x ](d) − dim [x , . . . , x ](d−2) = − C d C 1 n C 1 n n − 1 n − 1

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Proof: From above, (d) (d−2) ∆ : C[x1, . . . , xn] −→ C[x1, . . . , xn]

is surjective, so the dimension of Hd is the indicated difference of dimensions. The dimension of the space of degree d polynomials in n variables can be determined by counting the number e1 en P of monomials x1 . . . xn with i ei = d, as follows. Imagine each exponent as consisting of the corresponding number of marks, lined up, with n−1 additional marks to separate the marks corresponding to the n distinct variables xi, for a total of n+d−1. The choice of location of the separating marks is the binomial coefficient. ///

n−2 [3.0.8] Corollary: The dimension of Hd grows like d as d → +∞. ///

[3.0.9] Corollary: The dimensions of the polynomial λ-eigenspaces for the spherical Laplacian ∆ grow like n −1 |λ| 2 . ///

4. Invariant integrals and measures on spheres

We used the assumed existence of an SO(n)-invariant integral on Sn−1 to be sure that eigenvalues for the spherical Laplacian ∆S are non-positive, in determining all eigenvectors. To prove existence of an invariant integral we can write a formula [15] as follows, using the fact that the usual measure on Rn is SO(n)-invariant, since the of the determinant of an element of SO(n) is 1. For γ a fixed smooth non-negative function on [0, ∞) with Z γ(|x|2) dx = 1 Rn for a f on S, define

Z Z  x  f = γ(|x|2) f dx n−1 n |x| S R For convenience, we may at some moments suppose that γ has compact and vanishes identically on a neighborhood of 0. [16] For k ∈ SO(n) we have the SO(n)-invariance of the integral:

Z Z xk Z x Z x Z k · f = γ(|x|2) f( ) dx = γ(|xk−1|2) f( ) dx = γ(|x|2) f( ) dx = f S Rn |xk| Rn |x| Rn |x| S

by changing variables to replace x by xk−1, and using |xk−1| = |x|. Proof of the desired integration-by- parts-twice result from this clunky viewpoint is a little more trouble:

[4.0.1] Proposition: For differentiable functions f, ϕ on S = Sn−1, Z Z (∆Sf) · ϕ = f · ∆Sϕ Sn−1 Sn−1

[15] If we were already happy with Haar measure on SO(n) and invariant measures on quotient spaces such as Sn−1, then we would not need another construction. We rarely need more than the existence (and essential uniqueness) of such an integral.

[16] Compact support and the vanishing condition may be convenient, since then there are no boundary terms in n integrating by parts on R .

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Further, ∆S is negative-definite in the sense that Z (∆Sf) · f ≤ 0 S with equality only for f constant.

Proof: Let F (x) = f(x/r) and Φ(x) = ϕ(x/r), where r = |x|. By definition, Z Z (∆Sf) · ϕ = γ(r2) r2 · (∆F )(x) Φ(x) dx S Rn where the r2 is inserted so that r2∆F is positive-homogeneous of degree 0 as required by the integration formula. [17] Integrating by parts, this becomes

Z X ∂F ∂ − r2 · γ(r2) Φ(x) dx n ∂x ∂x R i i i

Let δ(r2) = r2γ(r2). Then

∂  2 2  ∂  2  0 2 2 ∂Φ r · γ(r ) Φ(x) = δ(r ) Φ(x) = 2xiδ (r )Φ(x) + δ(r ) ∂xi ∂xi ∂xi The whole is Z   Z X ∂F 0 2 2 ∂Φ X ∂F 2 ∂Φ − 2xiδ (r )Φ(x) + δ(r ) dx = − δ(r ) dx n ∂x ∂x n ∂x ∂x R i i i R i i i since Euler’s identity asserts that X ∂F x = 0 i ∂x i i because F is positive-homogeneous of degree 0. That last expression for the integral is symmetric in F and Φ, giving the desired integration-by-parts result. Finally, with Φ = F , the last expression for the integral is visibly non-positive, and is 0 only if ∂F/∂xi = 0 for all i, only if F is constant, only if f is constant. ///

[4.0.2] Remark: This argument is unsatisfying, since it does not extend to more general situations. A more universal existence argument for G-invariant measures on quotient spaces H\G will be given later.

0 [4.0.3] Corollary: For distinct degrees d, d , the spaces Hd and Hd0 are mutually orthogonal.

S S Proof: Earlier, we saw that the eigenvalues of ∆ on distinct Hd’s are distinct. Distinct eigenspaces of ∆ are orthogonal. ///

5. L2 spectral decompositions on spheres

The idea of spectral decomposition on the sphere is that functions on the sphere should be sums of eigenfunctions for the spherical Laplacian. For L2 functions the convergence should be in L2. For smooth functions, the sum should converge well in a uniform pointwise sense. As usual, L2 convergence does not imply pointwise convergence.

[17] Since F (x) = f(x/r) is positive-homogeneous of degree 0, ∆F is positive-homogeneous of degree −2, so we need to adjust it by r2.

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[5.0.1] Remark: The proof here is relatively elementary, but sadly fails to suggest a causal mechanism that might apply in other situations. A better, somewhat more costly, proof will be given later.

[5.0.2] Theorem: The collection of finite linear combinations of homogeneous harmonic polynomials restricted to Sn−1 is dense in L2(Sn−1).

Proof: By Weierstraß’ approximation theorem, as in the appendix, polynomials are dense in Co(Rn), meaning in the topology given by sups on compacts. Even though it has no interior, the sphere is compact, so restrictions of polynomials to Sn−1 approximate continuous functions on the sphere in sup norm. Our description of L2(Sn−1) is as the L2 completion of Co(Sn−1), so Co(Sn−1) is dense in L2(Sn−1) in the L2 topology. Since the sphere has finite total measure, density in sup norm implies density in L2 norm, so restrictions of polynomials to Sn−1 are dense in L2(Sn−1). We showed that every homogeneous degree d polynomial f is expressible as

2 4 f = fd + r · fd−2 + r · fd−4 + ... (where fj ∈ Hj)

Restricting to the sphere, r = 1, so

f = fd + fd−2 + fd−4 + ... (restricted to the sphere)

Thus, harmonic polynomials are dense in L2. ///

[5.0.3] Corollary: 2 n−1 M L (S ) = completion of Hd d≥0

Proof: Given f in the orthogonal complement to that completion, let ϕ be a harmonic polynomial 2 P approximating f to within ε > 0 in L norm. Write ϕ = d ϕd where ϕd is homogeneous of degree S d. The various spaces Hd are mutually orthogonal, since they have distinct eigenvalues for ∆ . Thus, orthogonality of f to every Hd assures that ϕd = 0 for all d. Thus, |f|L2 < ε. This holds for every ε > 0, so f = 0 in L2. ///

[5.0.4] Corollary: Thus, every L2 function f on Sn has at least an L2 Fourier-Laplace expansion

∞ X 2 f = fd (in L sense) d=0

2 n−1 where fd is the orthogonal projection in L (S ) of f to the space Hd of homogeneous degree d harmonic polynomials restricted to the sphere.

[5.0.5] Remark: All this discussion applies to the circle S1 ⊂ R2, presented as {(x, y): x2 + y2 = 1}. For the circle, there are also parametrizations θ → e2πiθ or θ → eiθ, corresponding to the presentations S1 ≈ R/Z or R/2πZ. The comparison is that the exponentials θ → e2πinθ with n ∈ Z are exactly the restrictions to {x2 + y2 = 1} of the harmonic polynomials (x + iy)n for n ≥ 0, and (x − iy)|n| for n ≤ 0. Thus, the Weierstraß approximation theorem and the above discussion yield yet-another proof that exponentials form an orthonormal basis for L2(R/Z).

11 Paul Garrett: Harmonic analysis on spheres (December 21, 2014)

6. Sup-norms of spherical harmonics on Sn−1

To eventually assess the pointwise convergence of Fourier series on the sphere, we must be aware that, unlike inθ 1 n [18] the bounded functions e on the circle S , for f ∈ Hd on S with n > 1 there is no trivial comparison of the two norms Z 1/2 2 |f|Co = sup |f(x)| |f|L2 = |f(x)| dx x∈Sn−1 S Nevertheless, a little work gives a useful comparison: [19]

[6.0.1] Proposition: Let f ∈ Hd. Then s dim Hd |f| o ≤ · |f| 2 C vol (Sn−1) L

And the estimate is sharp, in the sense that there is a function in Hd for which equality holds.

[6.0.2] Remark: The occurrence of the square root of the total measure of Sn−1 compensates for the fact that the square root of the total measure enters in the L2-norm, and does not enter in the sup norm.

[6.0.3] Remark: Using the dimension computation from above, as d → +∞

s    p n + d − 1 n + d − 3 n −1 dim H ∼ − ∼ d 2 d n − 1 n − 1

Proof: For x ∈ S, from the Riesz-Fischer theorem, the functional f → f(x) is necessarily given by

f(x) = hf, Fxi (for some Fx ∈ Hd)

The Cauchy-Schwarz-Bunyakowsky inequality gives

|f(x)| = |hf, Fxi| ≤ |f|L2 · |Fx|L2

2 2 which bounds the value f(x) in terms of its L -norm with constant being the L -norm of Fx.

n−1 We show that Fx(x) is independent of x ∈ S . Let g ∈ SO(n), with action (g · f)(x) = f(xg). Again, by design, the action of g ∈ SO(n) on functions is unitary in the sense that Z Z hg · f, g · F i = f(xg) F (xg) dx = f(x) F (x) dx = hf, F i Sn−1 Sn−1

[18] Since the spaces Hd are finite-dimensional, each has a unique topology compatible with the vector space structure. (The latter fact is not trivial to prove, but is not too hard.) Thus, the fact that the sup-norm and L2-norm are comparable on these finite-dimensional spaces is clear a priori. But this qualitative fact is irrelevant to the issue at hand. What is not clear is growth of the comparison constants in the parameter d, and this greater precision is needed for Sobolev-type estimates.

[19] In fact, the proof of the following proposition uses few of the specifics of this situation, and, indeed, the same argument proves that for a compact group K acting transitively on a set X, with a finite-dimensional K- stable space V of functions on X, for f ∈ V we have the same comparison of sup-norm and L2-norm, namely p |f|Co ≤ (dim V )/vol (X) · |f|L2 . This inequality is interesting even for finite groups.

12 Paul Garrett: Harmonic analysis on spheres (December 21, 2014)

−1 by replacing x by xg in the integral. The functions Fx have natural relations among themselves, namely,

−1 hf, Fxgi = f(xg) = (g · f)(x) = hg · f, Fxi = hf, g · Fxi

Thus, −1 Fxg = g · Fx 2 Since the action of g is unitary, the L -norm of the function Fx is the same for every x ∈ S. In particular,

−1 −1 Fx(x) = hFx,Fxi = hg · Fx, g Fxi = Fxg(xg)

Since SO(n) is transitive on the sphere, Fx(x) is independent of x. Further, 2 |Fx(y)| = |hFx,Fyi| ≤ |Fx|L2 · |Fy|L2 = |Fx|L2 because all the L2 norms are the same. This determines the sup-norm

2 |Fx|Co = Fx(x) = |Fx|L2

2 The L norm is evaluated as follows. Express Fx in terms of an orthonormal basis {fi} for Hd X Fx = hFx, fii · fi i

Evaluating both sides at x gives

X X X 2 Fx(x) = hFx, fii · fi(x) = fi(x) · fi(x) = |fi(x)| i i i

2 Again, the value Fx(x) = |Fx|L2 is independent of x, since for g ∈ SO(n) and x ∈ S

Fxg(xg) = hFxg,Fxgi = hg · Fx, g · Fxi = hFx,Fxi = Fx(x)

P 2 n−1 Integrating Fx(x) = i |fi(x)| over S, using the fact that Fx(x) is independent of x ∈ S ,

n−1 vol (S ) · Fx(x) = dimC Hd Then 2 Fx(x) = |Fx|Co = |Fx|L2 gives s p dim Hd |F | 2 = F (x) = x L x vol (Sn−1)

Combining this with |f(x)| ≤ |Fx|L2 · |f|L2 from above, s dim Hd |f| o ≤ |F | 2 · |f| 2 · |f| 2 C x L L vol (Sn−1) L

as claimed. Further, we saw that s dim Hd |F | o = |F | 2 · |F | 2 = · |F | 2 x C x L x L vol (Sn−1) x L so the estimate is sharp. ///

13 Paul Garrett: Harmonic analysis on spheres (December 21, 2014)

7. Pointwise convergence of Fourier-Laplace series

2 The comparison of sup-norms and L norms on Hd yields useful criteria for pointwise convergence. The following L2 Sobolev criterion proves the most useful in practice.

2 n−1 P [7.0.1] Corollary: (Sobolev imbedding) A function f ∈ L (S ) with Fourier-Laplace expansion f = d fd with fd ∈ Hd satisfying

X s 2 n−1 (1 + d) · |fd|L2 < ∞ (for any s > dim S = n − 1) d

is continuous, and its Fourier-Laplace expansion converges uniformly pointwise to f.

Proof: For any s ∈ R,

X X p X n−2 sup fd(x)  dim Hd · |fd|L2  (1 + d) 2 · |fd|L2 n−1 x∈S d d d

1/2 1/2 X s 1  X s 2   X 1  = (1 + d) 2 · |f | 2 · = (1 + d) · |f | 2 · d L s−(n−2) d L (1 + d)s−(n−2) d (1 + d) 2 d d by Cauchy-Schwarz-Bunyakowsky. For s > n − 1, the latter sum is finite. Thus, for any s > n − 1, if

X s 2 (1 + d) · |fd|L2 < +∞ d P then the Fourier-Laplace series d fd converges uniformly pointwise. That is, the finite partial sums converge uniformly. Since the partial sums are continuous, and a uniform limit of continuous functions is continuous, the limit is continuous. Since f is the L2 limit of the finite partial sums, and the Co limit is also in L2, it must be that f is that Co limit, so is continuous, and the partial sums converge uniformly to f itself. ///

[7.0.2] Remark: One might think of the superficially simpler analogous assertion and proof wherein the L2 norms of the Fourier components fd are not squared: a similar chain of inequalities

X X p X n−2 sup fd(x)  dim Hd · |fd|L2  (1 + d) 2 · |fd|L2 n−1 x∈S d d d

proves uniform convergence when the last sum is finite. However, Hilbert spaces are far more congenial than Banach spaces, in many technical regards.

8. Irreducibility of representation spaces for O(n)

n Rn Certainly O(n) stabilizes each Hd, because the action of O(d) on functions on R commutes with ∆ = ∆ , and the of the action preserves the degree of polynomials.

[8.0.1] Claim: The vector spaces Hd are irreducible for O(n), in the sense that their only subspaces stable under the action of O(n) are {0} and the whole Hd.

Proof: The proof will show that every finite-dimensional O(n)-stable subspace of any Hd contains a non- zero O(n − 1)-fixed vector, where O(n − 1) is the isotropy subgroup of the last standard basis element en.

14 Paul Garrett: Harmonic analysis on spheres (December 21, 2014)

Then we show that each Hd contains a unique O(n − 1)-fixed vector. The O(n)-invariance of h, i assures that the orthogonal complement to any O(n)-stable subspace of Hd is itself O(n)-stable, so if Hd were not irreducible, there would be at least a two-dimensional space of O(n − 1)-fixed vectors in Hd, contradiction. We will need to average over rotations by O(n − 1), so we need an invariant measure/integral on O(n − 1) itself. That is, with the action (h · ϕ)(y) = ϕ(yh) for h, y ∈ O(n − 1) and continuous ϕ on O(n − 1), require Z Z h · ϕ = ϕ O(n−1) O(n−1)

Discussion of the invariant integral is postponed to the end.

First, we show that a non-zero O(n)-stable finite-dimensional subspace V of Hd contains a non-zero O(n−1)- fixed vector. Take not-identically-zero f ∈ V . For f(en) 6= 0, the averaged version Z F = g · f O(n−1) is designed to be O(n − 1)-invariant: for h ∈ O(n − 1), Z Z Z h · F = h · g · f dg = hg · f dg = g · f dg O(n−1) O(n−1) O(n−1) by replacing g by h−1g. Note that we need confidence that the translation action can be moved inside the integral, when the integral is viewed as an integral of an Hd-valued function on O(n − 1). And F is not accidentally identically 0, because its value at en is a non-zero constant multiple of that of f: Z Z Z Z F (en) = (g · f)(en) dg = f(eng) dg = f(en) dg = f(en) 1 dg O(n−1) O(n−1) O(n−1) O(n−1)

If by mischance f(en) = 0, then use the O(n)-stability to rotate f by h ∈ O(n) so that f(enh) 6= 0. This is possible, since O(n) is transitive on Sn−1 and f is not identically 0. Thus, modulo facts about integration on O(n − 1), any O(n)-stable subspace of Hd contains a non-zero O(n − 1)-fixed vector.

Next, let V be a non-zero O(n)-stable subspace of Hd. We claim that the orthogonal complement W of V inside Hd is still O(n)-stable. Indeed, for w ∈ W , v ∈ V , and g ∈ O(n), by unitariness of the action of O(n),

hg · w, vi = hg · w, gg−1 · vi = hw, g−1 · vi = 0 since g−1 · v ∈ V . Thus, if W 6= {0}, then it has a non-zero O(n − 1)-fixed vector.

Now prove that Hd has at most one O(n − 1)-fixed vector, up to scalar multiples. An O(n − 1) invariant function f can be written

q 2 2 f(x1, . . . , xn−1, xn) = F (ρ, xn) (where ρ = x1 + ... + xn−1)

Using subscripts to denote partial derivatives with respect to arguments, the harmonic condition on such functions is 2 2 1 X ∂ ∂ F X ∂  2 · 2xi  0 = 2 F (ρ, xn) + 2 = F1 + F22 ∂x ∂x ∂xi ρ 1≤i≤n−1 i n 1≤i≤n−1

X 1 x2 x2  n − 2 = F − i F + i F + F = F + F + F ρ 1 ρ3 1 ρ2 11 22 ρ 1 22 11 1≤i≤n−1

15 Paul Garrett: Harmonic analysis on spheres (December 21, 2014)

Note that the O(n−1)-invariance entails further that F (ρ, y) is even as a function of ρ. Using the homogeneity, F (0, 1) = f(en) = 1, and the even-ness, write

d d−2 2 d−4 4 F (ρ, y) = y + c2y ρ + c4y ρ + ... and use the harmonic condition to solve recursively for the coefficients: n − 2   0 = F + F + F yd + c yd−2ρ2 + c yd−4ρ4 + ... ρ 1 22 11 2 4  n − 2  = d(d − 1)yd−2 + c (d − 2)(d − 3)yd−4ρ2 + 2yd−2ρ1 + 2yd−2 · 1 + ... 2 ρ

d−2  d−4 2 d−2 d−2  = d(d − 1)y + c2 (d − 2)(d − 3)y ρ + (n − 2)2y · 1 + 2y · 1 + ...

For example, equating the yd−2 coefficients gives   0 = d(d − 1) + c2 (d − 2)(d − 3) + 2(n − 2) + 2

The coefficient of c2 is strictly positive, so is non-zero, and c2 is completely determined. Similarly, all coefficients c2j are determined recursively:   0 = c2j(2j)(2j − 1) + c2j+2 (2j − 2)(2j − 3) + (d − 2j + 2)(n − 2) + (d − 2j + 2)(d − 2j + 1)

The coefficient of c2j+2 is a sum of three products of non-negative integers, and at least one of the three products is strictly positive. Thus, up to normalizing constant, there is at most one O(n − 1)-invariant element in Hd. Existence was already proven, by averaging over O(n − 1). ///

[8.0.2] Remark: Note that we did not use any feature of the invariant integral on O(n) except its existence, and the possibility of moving the translation action from outside an integral to inside, when the integrand is vector-valued, with values in the finite vector space Hd. The sanest argument for existence would be to invoke general existence results for Haar (invariant) measures on topological groups. Nevertheless, a more elaborate version of our contriving a rotation-invariant measure on Sn−1 by using the standard (translation and rotation-invariant) measure on the ambient Euclidean space, can be made to give an invariant measure 2 on any orthogonal group O(n) from the standard measure on the n -dimensional Euclidean space Mn of n-by-n real matrices, as follows.

First, we would note that G = GLn(R) is of full measure inside Mn. Second, from the spectral theorem for self-adjoint operators on real vector spaces, one shows that every g ∈ G has a unique expression g = s · k 1 with k ∈ O(n) and s positive-definite symmetric. Letting s 2 of a symmetric positive-definite matrix s refer to its unique positive-definite square root, observe that

− 1 − 1 − 1  > 2  > 2  >  2 2 − 1 gg · g = (sk)(sk) · sk = skk s · sk = (s ) 2 · sk = k

n2 With dx denoting the usual measure on Mn ≈ R ,

Z >  − 1  f −→ e−tr(xx ) · f xx> 2 · x dx GLn(R) gives a right O(n)-invariant integral on functions f on O(n).

[8.0.3] Remark: For a continuous vector-valued integrand f on a compact (finite-measure) space X, and for a continuous T : V → W from the topological vector space V where f takes its values to another topological vector space W , it is natural to hope that the obvious relation  Z  Z T f = T ◦ f X X

16 Paul Garrett: Harmonic analysis on spheres (December 21, 2014)

Indeed, under very mild assumptions on V,W , for continuous, compactly-supported V -valued f, this relation holds, as shown by Gelfand and Pettis independently in the 1930s. In the present situation, we only needed the result for V = W finite-dimensional, and various more primitive arguments give the result, depending on one’s characterization of integrals in the first place. With considerable hindsight, for purposes similar to R those here, and even many far more sophisticated, it may be best to characterize a V -valued integral X f as being the element of V such that

 Z  Z λ f = λ ◦ f (for all continuous linear λ : V → C) X X

This is a Gelfand-Pettis integral. [20] It reduces properties of vector-valued integrals to properties of scalar- valued ones. For finite-dimensional vector-valued integrands, elementary considerations suffice, and oughtn’t be entangled with more serious general considerations, but it is reassuring that even in considerable generality (integrands with values in locally convex, quasi-complete topological vector spaces), Gelfand-Pettis integrals of continuous, compactly supported functions do exist and are unique.

[8.0.4] Remark: For non-abelian groups of linear operators on a vector space such as L2(Sn−1), and for non-commutative rings of such operators, it is unreasonable to hope for simultaneous eigenvectors. The correct generalization, or substitute, is irreducible representations, just as Hd is irreducible for O(n). That is, instead of hoping to express general vectors as linear combinations of eigenvectors, we should only hope to express general vectors as linear combinations of vectors from irreducible subspaces, such as Hd. Typically, for non-abelian groups, these irreducibles are not one-dimensional, as we see for Hd. Despite the loss of the relevance of the simpler notion of simultaneous eigenvector, irreducible subspaces are a very useful substitute, as illustrated below in proof of Hecke’s identity about Fourier transforms.

2 n−1 [8.0.5] Remark: We proved that each of the irreducibles Hd of O(n) occurring in L (S ) has a unique (up to scalars) O(n − 1)-fixed vector. If we could prove that every irreducible of O(n) contains at most one O(n − 1)-fixed vector, then the two compact groups O(n − 1),O(n) would be said to be a Gelfand pair.

9. Hecke’s identity

[9.0.1] Theorem: (Hecke) For homogeneous degree d harmonic polynomial f on Rn,

2 2 Fourier transform f(x) e−π|x| = i−d · f(x) e−π|x|

That is, Z 2 2 f(x) e−π|x| · e−2πihξ,xi dx = i−d · f(ξ) e−π|ξ| Rn

d Proof: We reduce to the case f(x1, x2, x3, . . . , xn) = (x1 + ix2) by using the irreducibility of Hd as O(n)- d representation space, as follows. Certainly ϕ(x) = (x1 + ix2) is in Hd. Suppose for the moment that it satisfies Hecke’s identity. By irreducibility, the collection of finite linear combinations of translates of ϕ by O(n) is the whole space Hd. Recall that Fourier transform commutes with the action of g ∈ O(n): with hξ, xi = xξ> for row vectors x, ξ,

Z Z −1 Z F (xg) · e−2πihξ,xi dx = F (x) · e−2πihξ,xg i dx = F (x) · e−2πihξg,xi dx = Fb(ξg) Rn Rn Rn

[20] Sometimes Gelfand-Pettis integrals are called weak integrals, but in this use weak refers to the hypothesis, not the conclusion. A different approach to integration of vector-valued functions, imitating the Riemann or Lebesgue constructions, is the Bochner integral, which we do not need here.

17 Paul Garrett: Harmonic analysis on spheres (December 21, 2014)

by changing variables, since

hξ, xg−1i = xg−1ξ> = xg>ξ> = x(ξg)> = hξg, xi

d Thus, it suffices to prove Hecke’s identity for (x1 + ix2) . Indeed, since the Fourier transform factors over the coordinates, and

−π|x|2 −π(x2+...+x2 ) −π(x2+x2) −π(x2+...+x2 ) e = e 1 n = e 1 2 · e 3 n it suffices to treat the two-dimensional case. Let z = x1 + ix2 and z = x1 − ix2, and also w = ξ1 + iξ2 and 1 w = ξ1 − iξ2. Since x1ξ1 + x2ξ2 = 2 (zw + zw), Z  d −πzz d −πzz −πi(zw+zw) Fourier transform z e (w) = z e · e dx1 dx2 C Since  ∂ d (−πiz)d = e−πi(zw+zw) ∂w we have

   ∂ d Z (−πiz)d · Fourier transform zd e−πzz (w) = e−πzz · e−πi(zw+zw) dx dx ∂w 1 2 C

2 Recall that the Gaussian e−π|x| is its own Fourier transform, so this is

   ∂ d (−πiz)d · Fourier transform zd e−πzz (w) = e−πww = (−πw)d · e−πww ∂w

Cancelling the factors of π gives Hecke’s identity. ///

[9.0.2] Remark: Thus, harmonic-polynomial multiples of Gaussians are eigenfunctions for Fourier transform. There are many other distinguishable eigenfunctions, but the present examples will have applications in construction of theta series relevant to equidistribution problems.

10. Appendix: Bernstein’s proof of Weierstraß approximation

Weierstraß proved that polynomials are dense in Co(C) for compact C of Rn, with respect to sup- norm. Decades later, Stone greatly abstracted this. Prior to Stone, S. Bernstein gave a memorable argument for Weierstraß’ concrete case, with the additional virtue of suggesting similarly intuitive arguments for various function spaces on topological spaces with transitive group actions.

[10.0.1] Theorem: Given f ∈ Co(U) for U open in Rn, for compact C ⊂ U and ε > 0, there is a polynomial P such that sup |f(x) − P (x)| < ε x∈C

n Proof: The idea is to create a sequence P` of polynomials on R whose restrictions ϕ` to a fixed compact, such as the cube C = {x = (x1, . . . , xn): |xi| ≤ 1, for all i} form an approximate identity, in the sense that their masses bunch up at 0.

18 Paul Garrett: Harmonic analysis on spheres (December 21, 2014)

More precisely, we want the restrictions ϕ` to C to be non-negative, to have integrals 1 on the smaller cube 1 C, and to satisfy 2 Z ϕ` lim δC = 1 (for each fixed positive δ ≤ 1 ) `→∞ Z 2 ϕ` 1 2 C

Granting all that, we can show that the mollifications of f by the ϕ` approach f in the sup norm: Z lim sup f(x) − f(x + y) ϕ`(y) dy = 0 (as ` → ∞) `→∞ 1 1 x∈ 2 C 2 C

Indeed, Z Z Z f(x + y) ϕ`(y) dy = f(x + y) ϕ`(y) dy + f(x + y) ϕ`(y) dy 1 1 2 C δC 2 C−δC Z Z Z  = f(x) ϕ`(y) dy + f(x + y) − f(x) ϕ`(y) dy + f(x + y) ϕ`(y) dy 1 δC δC 2 C−δC The first integral goes to 1 as ` → ∞, for fixed δ > 0, by the bunching-up property. The second integral goes to 0 uniformly in x as δ → 0, by the uniform continuity of f on C. The third integral goes to 0 as ` → ∞, since the masses of the ϕ` bunch up inside δC. Thus, assuming we have such polynomials, Z 1 lim f(x + y) ϕ`(y) dy = f(x) (uniformly in x ∈ 2 C) `→∞ 1 2 C

At the same time, Z Z f(x + y) ϕ`(y) dy = f(y) ϕ`(−x + y) dy 1 1 2 C x+ 2 C is a superposition of polynomials of degrees at most that of ϕ`. The space V of such polynomials is finite- dimensional. Thus, this integral of a compactly-supported continuous V -valued function lies in V . That is, this integral is equal to a polynomial, as a function. This would prove the theorem.

To make suitable polynomials P`, it suffices to treat the single-variable case. Let

2 ` P`(x) = (1 − x ) (for x ∈ R)

First, determine where the second derivative vanishes: solve

d   0 = − 2`x(1 − x2)`−1 = 4`(` − 1)x2(1 − x2)`−2 − 2`(1 − x2)`−1 dx   = 2` · (` − 1)x2 − (1 − x2) · (1 − x2)`−2 √ Thus,√ in the√ interior of [−1, 1], the second derivative vanishes at ±1/ `, so the curve bends downward in [−1/√`, 1/ `], and bends upward outside that interval. In particular, the line segments from the points (±1/ `, 0) to (0, 1) are below the graph of P`, so

Z 2 √ P`(x) dx ≥ √ |x|≤1/ ` `

On the other hand, the standard fact that x lim (1 − )` = e−x `→∞ `

19 Paul Garrett: Harmonic analysis on spheres (December 21, 2014)

suggests a certain approach. For example, √ log `  log `` P` √ = 1 − ` `

Since log(1 − x) ≤ −x for x ≥ 0, √ log ` log P` √ ≤ − log ` ` Thus, √ log ` 1 P` √ ≤ ` `

Thus we have a sufficient bunching-up result: obviously √1 < log√ ` , so ` `

Z 2 P`(x) dx ≥ √ log ` |x|< √ ` `

while Z 2 P`(x) dx ≤ log ` √ <|x|<1 ` ` That is, letting P`(x) ϕ`(x) = Z P`(x) dx 1 |x|≤ 2 gives the single-variable approximate identity desired. The product

ϕ`(x1) . . . ϕ`(xn)

is the desired collection for Rn. ///

[10.0.2] Remark: Although it is unnecessary for the above, it is interesting to determine the integral of the single-variable P` over [−1, 1]. Integrating by parts repeatedly, it is

Z 1 ` Z 1 `(` − 1) Z 1 (1 − x)` · (1 + x)` dx = (1 − x)`−1 · (1 + x)`+1 dx = (1 − x)`−2 · (1 + x)`+2 dx −1 ` + 1 −1 (` + 1)(` + 2) −1

`! `! Z 1 `! `! 22`+1 = ... = (1 + x)2` dx = (2`)! −1 (2`)! 2` + 1

[Stein-Weiss 1978] E. Stein, G. Weiss, Introduction to Fourier analysis on Euclidean spaces, Princeton Press, 1978.

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