James Rowan 18.155 Problem Set 9 December 9, 2016

Collaborators: course notes from the course website, the hints on the pset, Jonathan Tidor, H¨ormander, volume 1, chapter 3, the notes on spectral theory and pseudodi↵erentia operators from Heiko Gimperlein, http://www.macs.hw.ac.uk/~hg94/pdst11/ Problem 1 (Homogeneous extension from the sphere to Rn 0 ). \{ } n 1 n Solution. Let f be a continuous on the sphere S = x R x =1 . n { 2 | | } Suppose there were two continuous functions f1,f2 on R 0 , both homogeneous of degree s n 1 \{ } that are both equal to f when restricted to S , such that for somex0, f1(x0) = f2(x0). We s s 6s have, by homogeneity, f1(x0)= x0 f1(x0/ x0 )= x0 f(x0/ x0 ) and f2(x0)= x0 f2(x0/ x0 )= s | | | | | | | | | | | | x0 f(x0/ x0 ). So f1(x0)=f2(x0) after all, and we have a contradiction. This means that f1 must | | | | n be equal to f2 on all of R 0 , so the homogeneous of degree s extension is unique if it exists. \{ } We now construct a continuous extension (on Rn 0 ) that is homogeneous of degree s. Consider \{ } the function F (x)= x sf(x/ x ) defined on Rn 0 . I claim that this is a that is | | | | \{ } homogeneous of degree s that is an extension of f to Rn 0 .When x = 1, x s = 1 and x/ x = x, n 1 s \{ } | s| s | | | | so that F (x)=f(x) on S . Moreover, F (ax)= ax f(ax/ ax )=a x f(x/ x ) for all a>0, so F is homoeneous of degree s. | | | | | | | | It remains to show that F is continuous. The function x/ x : Rn 0 Rn 0 is a continuous xi | | \{ } ! \{ } function, as each of the coordinate functions xi n 2 is an algebraic function with nonzero 7! p j=1 xj n n n 1 denominator on R 0 . Moreover, the image under x/P x of R 0 is S , as each point in the \{ } n 1 | | \{ } sphere corresponds to an open ray in R going through that point and the image of any x under n n 1 this map must have x/ x = x / x = 1. So x/ x : R 0 S is a continuous function. Since | | n 1| | | | | | \{ } ! n f is a continuous function S C, the composition f(x/ x ) is a continuous function R 0 C. ! | | \{ } ! Since x s is a continuous function Rn 0 C,theproduct x sf(x/ x ) is a continuous function | | \{ } ! | | | | Rn 0 . This completes the proof that this F : Rn 0 C we have constructed is a continuous \{ } \{ } ! extension of f on Rn 0 and homogeneous of degree s. \{ } Problem 2 (Tempered extension of continuous functions on Rn 0 ). \{ } s n Solution. Let f = x u(x/ x ) be a homogeneous distribution of degree s. For CC1(R 0 ), we can write using| polar| coordinates| | 2 \{ }

1 s n 1 f()= r u(!)(r!)r d!dr n 1 Z0 ZS 1 s = r+(r)dr, Z0 where rs = 0 on r 0, exp(s log r) on r>0, and + 

n 1 (r)= u(!)(r!)r d!. n 1 ZS n n This lets us view f as an element of C1(R 0 ). Now for (R ), since u(!) is bounded being a continuous function on a compact set, we\{ can} see that is2 smoothS by di↵erentiating under the integral sign and we have

sup r↵d(r) < x n,↵, k | | 1 2R  because the derivatives of (r!) are sums of bounded numbers of of r` terms times derivatives of

n 1 while multiplying by u(!) at worst multiplies the supremum by maxS u and integrating over ! at worst multiplies the supremum by the surface area of the (n 1)-sphere, so if is Schwartz, so is . So we only need to show that for any homogeneity degree s we can find a distribution supported

1 James Rowan 18.155 Problem Set 9 December 9, 2016

s s in r 0 equal to r for r>0. The distribution r+ works for s not a negative (this was shown in lecture 5). So the only remaining problem is what to do for negative integer homogeneity degree. We define s s ⇢+ to be r+/(s + 1) for the gamma function, Re(s) > 1. When s is not a negative integer, s this is just a scalar multiple of the r+ that we already have, so making this replacement is fine. d s s 1 s Since (s + 1) = s(s), we have that dr ⇢+ = ⇢+ for Re(s) > 1. But since ⇢+ was analytic (as n s a map C 0(R ) for Re s> 1, this formula gives a way to analytically continue ⇢+ to all of C ! S 0 by repeatedly di↵erentiating from some s0 where ⇢+ can be defined. We know that ⇢+ =0when 0 r 0, 1 when r 0, so ⇢+ = H(r), the heaviside function. So for s equal to some negative integer  k k 1 s d d k, we have that the distribution ⇢+ = drk H(r)= drk 1 (r) is a distribution supported in r 0 s n 1 with the right homogeneity degree. This means that the distribution ⇢ ( n 1 u(!)(rx)r d! + S gives a tempered distribution homogeneous of degree s extending x su(x/ x ), so every continuous | | | R| homogeneous function on Rn 0 is the restriction to Rn 0 of a tempered distribution as desired. \{ } \{ }

Problem 3 (Smooth functions on the sphere).

Solution. Consider the 2n coordinate patches xi > 0 and xi < 0 for i =1, 2,...,n,withthe n 1{ } { } coordinates at a point (x1,x2,...,xn) on S in the patch xi > 0 or xi < 0 to be the other n 1 coordinates. Since the points lie on the sphere, we have that the sum of the squares of these other n 1 coordinates must be less than 1, so these coordinate functions give a homeomorphism to a unit n 1 n 1 ball B in R . Moreover, if we take for our smooth functions the restriction to S of smooth functions on Rn, the pullback of any compactly-supported smooth function on B via the coordinate n 1 map is a smooth function that is equal to zero outside some compact of corresponding n 1 § exactly in coordinates other than the ith in S lying within the of the function on B. n 1 And every such smooth function f that is zero outside some compact set K S lying in x > 0, ⇢ i say (all the patches are exactly the same) will have a corresponding element of CC1(B) by taking f(x1,...,xi,...,xn), where xi denotes removing the coordinate xi. n 1 We now show the property. Suppose we have an open cover U↵ of S such that any n 1 continuousb u on S is equalb to some u↵, u↵ smooth, when restricted to any U↵. We can take a n n x conic open cover of R by taking V↵ = x R 0 x U↵ and one additional open set, the 1 { 2 \{ }| | | 2 } open ball of radius 2 around the origin. There exists a partition of unity subordinate to this open n n 1 V cover of R . Taking to be the restriction of the elements of to S , we get a partition of unity on M, since the restrictionU of any element supported in the openV ball of radius .5 around the origin is 0 on Sn and otherwise the elements of are a subset of the elements of and any ⇢ supported U n 1 V in V must have support of ⇢ n 1 V S = U , each point still must have ⇢ (x) = 1, ↵ |S ⇢ ↵ \ ↵ I and each point can still have only finitely many elements of U↵ nonzero (since each such element corresponds to one of and there are only finitely many of those). This means isP a partition of V U unity subordinate to U↵. From this partition of unity ⇢↵ subordinate to U↵, we know that if u = u↵, u↵ smooth on each open set U↵, we can write u = ↵ ⇢↵u↵ in a well-defined way since the sum is n 1 finite at every point in S and we know that if x is in U↵ and U, u↵(x)=u(x)=u(x), so at P points of overlap the fact that ↵ ⇢↵ = 1 guarantees that u takes the correct value. Since the u↵s and ⇢↵s are smooth, the sum of their products is also smooth and thus u must be smooth. So the P n 1 sheaf property is satisfied, and thus we have a C1 structure on S .

Problem 4 (The Euclidean and spherical Laplacians).

n n 1 Solution. Let x = rw be the representaiton of a point in R 0 for r (0, ), ! S . Now let \{ } 2 1 2 a function F on Rn 0 be represented as f(r)g(!), where ! = x/ x . We know that \ | | (fg)=(f)g + 2( f) ( g)+f(g) r · r

2 James Rowan 18.155 Problem Set 9 December 9, 2016 since = . We also know that since f is defined radially and g is defined spherically, the vectorsr · r for f and g are orthogonal; in particular, this means that ( f) ( g) = 0. Now we evaluate f and g. r · r For f, we use the . @ @ @r = + ⇤, @xi @r @xi where the terms in ⇤ are derivatives in terms of the coordinates on the sphere, which will be 0 in the n 2 case we care about since we are evaluating f and f depends on r alone. We know r = j=1 xj , so @r = 2xi = xi .So @ = xi @ .Di↵erentiating again, we get q @xi n 2 r @xi r @r P 2p j=1 xj P @2 @ x @ x @ = i + i + @x2 @x r @r ⇤ r @r 4 i i ✓ ◆ 1 x2 @ x @r @2 = i + i r r3 @r r @x @r2 ✓ ◆ i 1 x2 @ x2 @2 = i + i , r r3 @r r2 @r2 ✓ ◆ where the terms ⇤ and , being not in the radial direction, become zero when applied to f so they again become irrelevant.4 Now we take

n @2 = @x2 j=1 i X n 1 x2 @ x2 @2 = i + i r r3 @r r2 @r2 j=1 X ✓✓ ◆ ◆ n 1 @ @2 = r @r @r2

n 2 2 since j=1 xi = r . We now consider f(g). We can extend g to Rn 0 by homogeneity of degree 0, setting P \{ } n 1 g˜(x)=g(x/ x ). Now it remains to show that g˜ restricted to S is !g for some di↵erential | | n operator ! of degree 2 on C1(S ). Sinceg ˜ is homogeneous of degree 0, g˜ is homogeneous of 2 n 1 degree 2, so g˜(x)=r if g˜ restricts to a di↵erential operator on S . ! ! We have that the gr on the sphere of radius r is given related to the Euclidean metric n 1 ij 1 ij n 1 geuc restricted to S by gr = r2 geuc, since the sphere of radius r is a scaling of S by a factor of r Since the principal symbol for the Laplacian on a compact Riemannian manifold with a metric 2 g was calculated in class to be g and this property was proved to characterize the Laplacian on || · || 2 1 2 M, this means that the principal symbols for the Laplacians are related by gr = r2 geuc 1 || · || || · || so we can define the operator ! = geuc such that v = gr v˜ = r2 geuc v, and thus we get that 1 g = r2 !g. Putting this all together, we have

n 1 @ @2 1 (fg)= (f)g + f g r @r @r2 r2 ✓ ◆ so the Euclidean Laplacian has been expressed as desired.

Problem 5 (Eigenfunctions of ! extend to harmonic functions).

3 James Rowan 18.155 Problem Set 9 December 9, 2016

n 1 Solution. Suppose that !v = v, for C and v C1(S ). We can extend v to a smooth 2 2 function on Rn 0 that is homogeneous of degree s by writing u x sv (where by v we mean v(x/ x ), \{ } | | n 1 | | but for the purposes of this problem we think of x as rx = x !, ! S )Wehave,bytheresult of problem 4, that this means | | 2 n 1 1 u = @2(rsv) @ (rsv)+ (rsv) r r r r2 ! s 2 s 2 s 2 = s(s 1)r v (n 1)sr v + r v ! s 2 = s(s + n 2) + r v. (5.1) So if we choose s to be a solution to the quadratic equation s2 +(n 2)s = 0, we can guarantee that there is an s such that the extension x sv of the eigenfunction v to s is harmonic on Rn 0. | | \ Problem 6 (Eigenfunctions of ! are restrictions of polynoials). Note: this solution ends up being a more harmonic-functions flavor than the hinted solution; it seemed like the tempered distribution result was too general for this setting and so my first thought was extending harmonic functions rather than looking at the tempered extension of x sv since the hypothesis in problem 2 was just continuity and not the much stronger condition of| | harmonicity (harmonicness?).

2 Solution. We know that ! =(d + ) (where the operators d and = d⇤ are defined for the n 1 2 2 compact manifold S ), so ! = d + d + d + = d + d is self-adjoint. This means that n 1 all the eigenvalues of ! must be real. Now we also know that C1(S ) has positive definite hermitian inner product given by v, w = vwd!, h i n 1 ZS so we can write

v, v = (dv + d)v, v h ! i h i = dv, v + dv, v h i h i = v,v + dv, dv h i h i = v 2 + dv 2 0. || || || || If v is an eigenfunction of with eigenvalue , we can scale it to somev ˜ with v˜ = 1 so that ! || || v,˜ v˜ = v,˜ v˜ h ! i h i = .

So if is an eigenvalue of ! with eignefunction v, there is a correspondingv ˜ such that !v,˜ v˜ = . Since we know v, v 0 for all v, this tells us that all eigenvalues of must be nonnegative.h i h ! i ! Now this means that for any eigenvalue of !, we have a quadratic equation in s given by

s2 +(n 1)s = 0 (6.1) for > 0. For s far enough away from 0 in either direction, this quadratic takes on positive values, while for s = 0 the value of the quadratic at 0 is equal to 0. The intermediate value theorem tells us that we can always choose the s problem 5 gives us that has x sv homogeneous, smooth, | | and harmonic on Rn 0 to be greater than or equal to 0. Now since s 0 and\{ }v on the sphere is a smooth function on a compact set and thus bounded, n 2 s 1 s we have limx 0 x x v = 0 for n 3 and limx 0(log x ) x v = 0 for n = 2 (for n = 1, we ! | | | | ! | | | | s can choose s =1when = 0 and we don’t need to remove a singularity since limx 0 x v =0 ! | |

4 James Rowan 18.155 Problem Set 9 December 9, 2016 to begin with). By the removable singularity theorem, this condition is enough to guaratee we can smoothly extend x sv, and since ( x sv) = 0 on Rn 0 , the extension satisfies u = 0 on all of | | | | \{ } Rn. We prove on an earlier pset that this implies that u is a . Since u = x sv (extended n 1 | | across 0) restricted to S is v, we have shown that for any v which is an eigenfunction of !,we n 1 can find a polynomial u such that v is the restriction of u to the sphere S . This analysis also shows that s must be an integer since the degree of homogeneity of a polynomial must be an integer. So the allowed values of are the s(s + n 2) for nonnegative integers s.

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