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 function 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 continuous function 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 φ CC1(R 0 ), we can write using| polar| coordinates| | 2 \{ } 1 s n 1 f(φ)= r u(!)φ(r!)r − d!dr n 1 Z0 ZS − 1 s = r+Φ(r)dr, Z0 where rs = 0 on r 0, exp(s log r) on r>0, and + n 1 Φ(r)= u(!)φ(r!)r − d!. n 1 ZS − n n This lets us view f as an element of C1(R 0 ). Now for φ (R ), since u(!) is bounded being a continuous function on a compact set, we\{ can} see that Φ is2 smoothS by di↵erentiating under the integral sign and we have sup r↵dβΦ(r) < x n,↵,β k | | 1 2R because the derivatives of φ(r!) are sums of bounded numbers of of r` terms times derivatives of φ 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 φ is Schwartz, so is Φ. 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 integer (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+/Γ(s + 1) for Γ 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 Γ(s + 1) = sΓ(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 δ(r) is a distribution supported in r 0 − − s n 1≥ with the right homogeneity degree. This means that the distribution ⇢ ( n 1 u(!)φ(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 subset of − corresponding n 1 § exactly in coordinates other than the ith in S − lying within the support 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 sheaf 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 Uβ, u↵(x)=uβ(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 tangent 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 chain rule.