LECTURE NOTES FOR , FALL 2004

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1 LECTURE NOTES FOR , FALL Cone support and wavefront set In disussing the singular support of a tempered distibution above, notie that singsupp(u) = only implies that u C (R n ), not as one might want, that u S(R n ). We an however refine the onept of singular support a little to get this. Let us think of the sphere S n 1 as the set of asymptoti diretions in n R. That is, we identify a point in S n 1 with a half-line {a x; a (0, )} for 0 = x R n. Sine two points give the same half-line if and only if they are positive multiples of eah other, this means we think of the sphere as the quotient (12.1) S n 1 = (R n \ {0})/R +. Of ourse if we have a metri on R n, for instane the usual Eulidean metri, then we an identify S n 1 with the unit sphere. However (12.1) does not require a hoie of metri. Now, suppose we onsider funtions on R n \{0} whih are (positively) homogeneous of degree 0. That is f (a x) = f ( x), for all a > 0, and they are just funtions on S n 1. Smooth funtions on S n 1 orrespond (if you like by definition) with smooth funtions on R n \ {0} whih are homogeneous of degree 0. Let us take suh a funtion υ C (R n \{0}), υ(ax) = υ(x) for all a > 0. Now, to make this smooth on R n we need to ut it off near 0. So hoose a utoff funtion α C (R n ), with α(x) = 1 in x < 1. Then (12.2) υ R (x) = υ(x)(1 α(x/r)) C (R n ), for any R > 0. This funtion is supported in x R. Now, if υ has support near some point φ S n 1 then for R large the orresponding funtion υ R will loalize near φ as a point at infinity of R n. Rather than try to understand this diretly, let us onsider a orresponding analyti onstrution. First of all, a funtion of the form υ R is a multiplier on S(R n ). That is, (12.3) υ R : S(R n ) S(R n ). To see this, the main problem is to estimate the derivatives at infinity, sine the produt of smooth funtions is smooth. This in turn amounts to estimating the deriviatives of υ in x 1. This we an do using the homogeneity.

2 84 RICHARD B. MELROSE Lemma If υ C (R n \ {0}) is homogeneous of degree 0 then (12.4) D α υ C α α x. Proof. I should not have even alled this a lemma. By the hain rule, the derivative of order α is a homogeneous funtion of degree α from whih (12.4) follows. For the smoothed versio, υ R, of υ this gives the estimates (12.5) D α υ R (x) C α x. γα α α This allows us to estimate the derivatives of the produt of a Shwartz funtion and υ R : (12.6) x β D (υ R f) = ( α D α γ υ R x β D γ f = sup x β D (υ R f) C sup f k π x 1 for some seminorm on S(R n ). Thus the map (12.3) is atually ontinuous. This ontinuity means that υ R is a multiplier on S (R n ), defined as usual by duality: (12.7) υ R u(f) = u(υ R f) f S(R n ). Definition The one-support and one-singular-support of a tempered distribution are the subsets Csp(u) R n S n 1 and Css(u) n R S n 1 defined by the onditions (12.8) Csp(u) R n = supp(u) (Csp(u)) S n 1 ={φ S n 1 ; (Css(u)) S n 1 ={φ S n 1 ; (S n 1 R > 0, υ C ), υ(φ) 0, υ R u = 0}, Css(u) R n = singsupp(u) (S n 1 R > 0, υ C ), υ(φ) 0, υ R u S(R n )}. That is, on the R n part these are the same sets as before but at infinity they are defined by oni loalization on S n 1. In onsidering Csp(u) and Css(u) it is onvenient to ombine R n and S n 1 into a ompatifiation of R n. To do so (topologially) let us identify R n with the interior of the unit ball with respet to the Eulidean metri using the map x (12.9) R n x {y R n ; y 1} = B n. x α

3 LECTURE NOTES FOR , FALL Clearly x < x and for 0 a < 1, x = a x has only the solution 1 x = a/(1 a 2 ) 2. Thus if we ombine (12.9) with the identifiation of n S with the unit sphere we get an identifiation (12.10) R n S n 1 B n. Using this identifiation we an, and will, regard Csp(u) and Css(u) as subsets of B n. 21 Lemma For any u S (R n ), Csp(u) and Css(u) are losed subsets of B n and if υ C (S n ) has supp( υ ) Css(u) = then for R suffiiently large υ Ru S(R n ). Proof. Diretly from the definition we know that Csp(u) R n is losed, as is Css(u) R n. Thus, in eah ase, we need to show that if φ S n 1 and φ / Csp(u) then Csp(u) is disjoint from some neighbourhood of φ in B n. However, by definition, U = {x R n ; υ R (x) = 0} {φ S n 1 ; υ(φ ) = 0} is suh a neighbourhood. Thus the fat that Csp(u) is losed follows diretly from the definition. The argument for Css(u) is essentially the same. The seond result follows by the use of a partition of unity on S n 1. Thus, for eah point in supp(υ) S n 1 there exists a oni loalizer for whih υ R u S(R n ). By ompatness we may hoose a finite number of these funtions υ j suh that the open sets {υ j (φ) > 0} over supp( υ ). By assumption (υ j ) Rj u S(R n ) for some R j > 0. However this will remain true if R j is inreased, so we may suppose that R j = R is independent of j. Then for funtion µ = υ j 2 C (S n 1 ) j we have µ R u S(R n ). Sine υ = υ µ for some µ C (S n 1 ) it follows that υ R+1u S(R n ) as laimed. Corollary If u S (R n ) then Css(u) = if and only if u S(R n ). Proof. Certainly Css(u) = if u S(R n ). If u S (R n ) and Css(u) = then from Lemma 12.3, υ R u S(R n ) where υ = 1. Thus v = (1 υ R )u C (R n ) has singsupp(v) = so v C (R n ) and hene u S(R n ). 21 In fat while the topology here is orret the smooth struture on B n is not the right one see Problem?? For our purposes here this issue is irrelevant.

4 86 RICHARD B. MELROSE Of ourse the analogous result for Csp(u), that Csp(u) = if and only if u = 0 follows from the fat that this is true if supp(u) =. I will treat a few other properties as self-evident. For instane (12.11) Csp(χu) Csp(u), Css(χu) Css(u) u S (R n ), χ S(R n ) and (12.12) Csp( 1 u u 2 ) Csp(u 1 ) Csp(u 2 ), Css( 1 u u 2 ) Css(u 1 ) Css(u 2 ) u 1, u 2 S (R n ), 1, 2 C. One useful onsequene of having the one support at our disposal is that we an disuss suffiient onditions to allow us to multiply distributions; we will get better onditions below using the same idea but applied to the wavefront set but this preliminary disussion is used there. In general the produt of two distributions is not defined, and indeed not definable, as a distribution. However, we an always multiply an element of S (R n ) and an element of S(R n ). To try to understand multipliation look at the question of pairing between two distributions. Lemma If K i B n, i = 1, 2, are two disjoint losed (hene ompat) subsets then we an define an unambiguous pairing (12.13) {u S (R n ); Css(u) K 1 } {u S (R n ); Css(u) K 2 } (u 1, u 2 ) u 1 (u 2 ) C. Proof. To define the pairing, hoose a funtion υ C (S n 1 ) whih is identially equal to 1 in a neighbourhood of K 1 S n 1 and with support disjoint from K 2 S n 1. Then extend it to be homogeneous, as above, and ut off to get υ R. If R is large enough Csp(υ R ) is disjoint from K 2. Then υ R + (1 υ) R = 1 + ν where ν C (R n ). We an find another funtion µ C (R n ) suh that υ 1 = υ R + µ = 1 in a neighbourhood of K 1 and with Csp(υ 1 ) disjoint from K 2. One we have this, for u 1 and u2 as in (12.13), (12.14) υ 1 u 2 S(R n ) and (1 υ 1 )u 1 S(R n ) sine in both ases Css is empty from the definition. Thus we an define the desired pairing between u 1 and u 2 by (12.15) u 1 (u 2 ) = u 1 (υ 1 u 2 ) + u 2 ((1 υ 1 )u 1 ).

5 LECTURE NOTES FOR , FALL Of ourse we should hek that this definition is independent of the ut-off funtion used in it. However, if we go through the definition and hoose a different funtion υ to start with, extend it homogeneoulsy and ut off (probably at a different R) and then find a orretion term µ then the 1-parameter linear homotopy between them (12.16) υ 1 (t) = tυ 1 + (1 t)υ 1, t [0, 1] satisfies all the onditions required of υ 1 in formula (12.14). Thus in fat we get a smooth family of pairings, whih we an write for the moment as (12.17) (u 1, u 2 ) t = u 1 (υ 1 (t)u 2 ) + u 2 ((1 υ 1 (t))u 1 ). By inspetion, this is an affine-linear funtion of t with derivative (12.18) u 1 ((υ 1 υ 1 )u 2 ) + u 2 ((υ υ 1 ))u 1 ). 1 Now, we just have to justify moving the smooth funtion in (12.18) to see that this gives zero. This should be possible sine Csp(υ1 υ 1 ) is disjoint from both K 1 and K 2. In fat, to be very areful for one, we should onstrut another funtion α in the same way as we onstruted υ 1 to be homogenous near infinity and smooth and suh that Csp(α) is also disjoint from both K1 and K 2 but α = 1 on Csp(υ1 υ 1 ). Then α(υ1 υ 1 ) = υ1 υ 1 so we an insert it in (12.18) and justify (12.19) u 1 ((υ 1 υ 1 )u 2 ) = u 1 (α 2 (υ 1 υ 1 )u 2 ) = (αu 1 )((υ 1 υ 1 )αu 2 ) = (αu 2 )(υ 1 υ 1 )αu 1 ) = u 2 (υ 1 υ 1 )αu 1 ). Here the seond equality is just the identity for α as a (multipliative) linear map on S(R n ) and hene S (R n ) and the operation to give the ruial, third, equality is permissible beause both elements are in S(R n ). One we have defined the pairing between tempered distibutions with disjoint oni singular supports, in the sense of (12.14), (12.15), we an define the produt under the same onditions. Namely to define the produt of say u 1 and u 2 we simply set (12.20) u 1 u 2 (χ) = u 1 (χu 2 ) = u 2 (χu 1 ) χ S(R n ), provided Css(u 1 ) Css(u 2 ) =. Indeed, this would be true if one of u 1 or u 2 was itself in S(R n ) and makes sense in general. I leave it to you to hek the ontinuity statement required to prove that the produt is atually a tempered distibution (Problem 78).

6 88 RICHARD B. MELROSE One an also give a similar disussion of the onvolution of two tempered distributions. One again we do not have a definition of u v as a tempered distribution for all u, v S (R n ). We do know how to define the onvolution if either u or v is ompatly supported, or if either is in S(R n ). This leads diretly to Lemma If Css(u) S n 1 = then u v is defined unambiguously by x (12.21) u v = u 1 v + u 2 v, u 1 = (1 α( ))u, u 2 = u u 1 r where α C (R n ) has α(x) = 1 in x 1 and R is suffiiently large; there is a similar definition if Css(v) S n 1 =. Proof. Sine Css(u) S n 1 =, we know that Css(u 1 ) = if R is large enough, so then both terms on the right in (12.21) are well-defined. To see that the result is independent of R just observe that the differene of the right-hand side for two values of R is of the form w v w v with w ompatly supported. Now, we an go even further using a slightly more sophistiated deomposition based on Lemma If u S (R n ) and Css(u) Γ = where Γ S n 1 is a losed set, then u = u 1 + u 2 where Csp(u 1 ) Γ = and u 2 S(R n ); in fat (12.22) u = u 1 + u 1 + u 2 where u 1 C (R n ) and 0 / supp(u 1 ), x R n \ {0}, x/ x Γ = x / supp(u 1 ). Proof. A overing argument whih you should provide. Let Γ i R n, i = 1, 2, be losed ones. That is they are losed sets suh that if x Γ i and a > 0 then ax Γ i. Suppose in addition that (12.23) Γ 1 ( Γ 2 ) = {0}. That is, if x Γ 1 and x Γ 2 then x = 0. Then it follows that for some > 0, (12.24) x Γ 1, y Γ 2 = x + y ( x + y ). To see this onsider x + y where x Γ 1, y Γ 2 and y x. We an assume that x 0, otherwise the estimate is trivially true with = 1, and then Y = y/ x Γ 1 and X = x/ x Γ 2 have Y 1 and X = 1. However X + Y 0, sine X = 1, so by the ontinuity of the sum, X + Y 2 > 0 for some > 0. Thus X + Y ( X + Y ) and the result follows by saling bak. The other ase, of x y

7 LECTURE NOTES FOR , FALL follows by the same argument with x and y interhanged, so (12.24) is a onsequene of (12.23). Lemma For any u S (R n ) and χ S(R n ), (12.25) Css(χ u) Css(u) S n 1. Proof. We already know that χ u is smooth, so Css(χ u) S n 1. Thus, we need to show that if φ S n 1 and φ / Css(u) then φ / Css(χ u). Fix suh a point φ S n 1 \ Css(u) and take a losed set Γ S n 1 whih is a neighbourhood of φ but whih is still disjoint from Css(u) and then apply Lemma The two terms χ u 2, where u 2 S(R n ) and χ u 1 where u 1 C (R n ) are both in S(R n ) so we an assume that u has the support properties of u 1. In partiular there is a smaller losed subset Γ 1 S n 1 whih is still a neighbourhood of φ but whih does not meet Γ 2, whih is the losure of the omplement of Γ. If we replae these Γ i by the losed ones of whih they are the ross-setions then we are in the situation of (12.23) and (12.23), exept for the signs. That is, there is a onstant > 0 suh that (12.26) x y ( x + y ). Now, we an assume that there is a utoff funtion υ R whih has support in Γ 2 and is suh that u = υ R u. For any oni utoff, υ R, with support in Γ 1 (12.27) υr (χ u) = υ Ru, χ(x ) = u(y), υ R (y)υr (x)χ(x y). The ontinuity of u means that this is estimated by some Shwartz seminorm (12.28) sup Dy α (υ R (y)υr(x)χ(x y)) (1 + y ) k y, α k C N χ sup(1 + x + y ) N (1 + y ) k C N χ (1 + x ) N+k y for some Shwartz seminorm on χ. Here we have used the estimate (12.24), in the form (12.26), using the properties of the supports of υr and υ R. Sine this is true for any N and similar estimates hold for the derivatives, it follows that υr (u χ) S(Rn ) and hene that φ / Css(u χ). Corollary Under the onditions of Lemma 12.6 (12.29) Css(u v) (singsupp(u) + singsupp(v)) (Css(v) S n 1 ). Proof. We an apply Lemma 12.8 to the first term in (12.21) to onlude that it has oni singular support ontained in the seond term in (12.29). Thus it is enough to show that (12.29) holds when u

8 90 RICHARD B. MELROSE C (R n ). In that ase we know that the singular support of the on- volution is ontained in the first term in (12.29), so it is enough to onsider the oni singular support in the sphere at infinity. Thus, if φ / Css(v) we need to show that φ / Css(u v). Using Lemma 12.7 we an deompose v = v 1 + v 2 + v 3 as a sum of a Shwartz term, a ompat supported term and a term whih does not have φ in its oni support. Then u v 1 is Shwartz, u v 2 has ompat support and satisfies (12.29) and φ is not in the one support of u v 3. Thus (12.29) holds in general. Lemma If u, v S (R n ) and φ Css(u) S n 1 = φ / Css(v) then their onvolution is defined unambiguously, using the pairing in Lemma 12.5, by (12.30) u v(χ) = u(ˇv χ) χ S(R n ). Proof. Sine v(x) ˇ = v( x), Css(ˇ v) = Css(v) so applying Lemma 12.8 we know that (12.31) Css(ˇv χ) Css(v) S n 1. Thus, Css(v) Css(ˇv χ) = and the pairing on the right in (12.30) is well-defined by Lemma Continuity follows from your work in Problem 78. In Problem 79 I ask you to get a bound on Css(u v) S n 1 under the onditions in Lemma Let me do what is atually a fundamental omputation. Lemma For a oni utoff, υ R, where υ C (S n 1 ), (12.32) Css( υ R ) {0}. Proof. This is atually muh easier than it seems. Namely we already know that D α (υ R ) is smooth and homogeneous of degree α near infinity. From the same argument it follows that (12.33) D α (x β υ R ) L 2 (R n ) if α > β + n/2 sine this is a smooth funtion homogeneous of degree less than n/2 near infinity, hene square-integrable. Now, taking the Fourier transform gives (12.34) γ α D β ( υ R ) L 2 (R n ) α > β + n/2. If we loalize in a one near infinity, using a (ompletely unrelated) utoff υr (γ) then we must get a Shwartz funtion sine (12.35) γ α υr (γ)dβ ( υ R (γ) R ) L 2 (R n ) α > β + n/2 = υ υ R S(R n ).

9 LECTURE NOTES FOR , FALL Indeed this argument applies anywhere that γ = 0 and so shows that (12.32) holds. Now, we have obtained some reasonable looking onditions under whih the produt uv or the onvolution u v of two elements of S (R n ) is defined. However, reasonable as they might be there is learly a flaw, or at least a defiieny, in the disussion. We know that in the simplest of ases, (12.36) u v = u v. Thus, it is very natural to expet a relationship between the onditions under whih the produt of the Fourier transforms is defined and the onditions under whih the onvolution is defined. Is there? Well, not muh it would seem, sine on the one hand we are onsidering the relationship between Css(û) and Css( v) and on the other the relationship between Css(u) S n 1 and Css(v) S n 1. If these are to be related, we would have to find a relationship of some sort between Css(u) and Css(û). As we shall see, there is one but it is not very strong as an be guessed from Lemma This is not so muh a bad thing as a sign that we should look for another notion whih ombines aspets of both Css(u) and Css(û). This we will do through the notion of wavefront set. In fat we define two related objets. The first is the more onventional, the seond is more natural in our present disussion. Definition If u S (R n ) we define the wavefront set of u to be (12.37) WF(u) = {(x, φ) R n S n 1 ; χ C (R n ), χ(x) = 0, φ / Css( χu)} and more generally the sattering wavefront set by (12.38) WF s (u) = WF(u) {(φ, p) S n 1 B n ; υ C (S n ), υ(φ) = 0, R > 0 suh that p / Css( υ R u)}. So, the definition is really always the same. To show that (p, q) / WF s (u) we need to find a utoff Φ near p depending on whether p R n or p S n 1 this is either Φ = χ C (R n ) with F = χ(p) 0 or a υ R where υ C (S n 1 ) has υ(p) = 0 suh that q / Css( Φu). One ruial property is Lemma If (p, q) / WF s (u) then if p R n there exists a neighbourhood U R n of p and a neighbourhood U B n of q suh that for all χ C (R n ) with support in U, U Css( χu) = ; similarly

10 92 RICHARD B. MELROSE if p S n 1 then there exists a neigbourhood U B n of p suh that U Css( υ R u) = if Csp(φ R ) U. Proof. First suppose p R n. From the definition of oni singular sup- (S n 1 port, (12.37) means preisely that there exists υ C ), υ(φ) = 0 and R suh that (12.39) υ R ( χu) S(R n ). Sine we know that χu C (R n ), this is atually true for all R > 0 as soon as it is true for one value. Furthermore, if χ C (R n ) has supp(χ ) {χ = 0} then φ / Css( χ u) follows from φ / Css( χu). Indeed we an then write χ = µχ where µ C (R n ) so it suffies to show that if v C (R n ) has φ / Css(v ) then φ / Css( µv) if µ C (R n ). Sine µv = (2ν) n θ u where θˇ = µ S(R n ), applying Lemma 12.8 we see that Css(θ v) Css( v), so indeed φ / Css( χ u). The ase that p S n 1 is similar. Namely we have one ut-off υ R with υ(p) = 0 and q / Css( φ R u). We an take U = {υ R+10 = 0} sine if υr (S n 1 has oni support in U then υ R = υ R υ R for some υ C ). Thus (12.40) υ φ υr u, vˇ = R u = v R. From Lemma and Corollary12.9 we dedue that (12.41) Css( υr u) Css( φ R u) and hene the result follows with U a small neighourhood of q. Proposition For any u S (R n ), (12.42) WF s (u) ω(b n B n ) = (B n S n 1 ) (S n 1 B n ) = (R n S n 1 ) (S n 1 S n 1 ) (S n 1 R n ) and WF(u) R n are losed sets and under projetion onto the first variable (12.43) ν 1 (WF(u)) = singsupp(u) R n, ν 1 (WF s (u)) = Css(u) B n. Proof. To prove the first part of (12.43) we need to show that if ( x, φ) / WF(u) for all φ S n 1 with x R n fixed, then x / singsupp(u). The n 1 definition (12.37) means that for eah φ S there exists χ α C (R n ) with χ α ( x) = 0 suh that φ / Css( χ α u). Sine Css(χu) is losed and S n 1 is ompat, a finite number of these utoffs, χ j C (R n ), an be hosen so that χ j ( x) = 0 with the S n 1 \ Css( χ j u) overing S n 1. Now applying Lemma above, we an find one

11 LECTURE NOTES FOR , FALL χ C (R n ), with support in j {χ j (x) = 0} and χ( x) = 0, suh that Css( χu) Css( χ j u) for eah j and hene χu S(R n ) (sine it is already smooth). Thus indeed it follows that x / singsupp(u). The onverse, that x / singsupp(u) implies ( x, φ) / WF(u) for all φ S n 1 is immediate. The argument to prove the seond part of (12.43) is similar. Sine, by definition, WF s (u) (R n B n ) = WF(u) and Css(u) R n = singsupp(u) we only need onsider points in Css(u) S n 1. Now, we first hek that if β / Css(u) then {β} B n WF s (u) =. By definition of Css(u) there is a ut-off υ R, where υ C (S n 1 ) and υ(β) 0, suh that υ R u S(R n ). From (12.38) this implies that (β, p) / WF s (u) for all p B n. Now, Lemma allows us to apply the same argument as used above for WF. Namely we are given that (β, p) / WF s (u) for all p B n. Thus, for eah p we may find υ R, depending on p, suh that υ(β) = 0 and p / Css( υ R u). Sine B n is ompat, we may hoose a finite subset of these oni loalizers, υ (j) R j suh that the intersetion of the orresponding sets Css( υ (j) R j u), is empty, i.e. their omplements over B n. Now, using Lemma we may hoose one υ with support in the intersetion of the sets {υ (j) = 0} with υ(β) = 0 and one R suh that Css( υ R u) =, but this just means that υ R u S(R n ) and so β / Css(u) as desired. The fat that these sets are losed (in the appropriate sets) follows diretly from Lemma Corollary For u S (R n ), (12.44) WF s (u) = u S(R n ). Let me return to the definition of WF s (u) and rewrite it, using what we have learned so far, in terms of a deomposition of u. Proposition For any u S (R n ) and (p, q) ω(b n B n ), (12.45) (p, q) / WF s (u) u = u 1 + u 2, u 1, u 2 S (R n ), p / Css(u 1 ), q / Css( û 2 ). Proof. For given (p, q) / WF s (u), take Φ = χ C (R n ) with χ 1 near p, if p R n or Φ = υ R with υ C (S n 1 ) and υ 1 near p, if p S n 1. In either ase p / Css(u 1 ) if u 1 = (1 Φ)u diretly from the definition. So u 2 = u u 1 = Φu. If the support of Φ is small enough it follows as in the disussion in the proof of Proposition that (12.46) q / Css( û 2 ).

12 94 RICHARD B. MELROSE Thus we have (12.45) in the forward diretion. For reverse impliation it follows diretly that (p, q) / WF s (u 1 ) and that (p, q) / WF s (u 2 ). This restatement of the definition makes it lear that there a high degree of symmetry under the Fourier transform Corollary For any u S (R n ), (12.47) (p, q) WF s (u)) (q, p) WF s (û). Proof. I suppose a orollary should not need a proof, but still.... The statement (12.47) is equivalent to (12.48) (p, q) / WF s (u)) = (q, p) / WF s (ˆ u) sine the reverse is the same by Fourier inversion. By (12.45) the ondition on the left is equivalent to u = u 1 + u 2 with p / Css(u 1 ), q / Css( û 2 ). Hene equivalent to (12.49) u = v 1 + v 2, v 1 = u 2, v 2 = (2ν) n ǔ 1 so q / Css(v 1 ), p / Css( v 2) whih proves (12.47). Now, we an exploit these notions to refine our onditions under whih pairing, the produt and onvolution an be defined. Theorem For u, v S (R n ) (12.50) uv S (R n ) is unambiguously defined provided and (p, φ) WF s (u) (B n S n 1 ) = (p, φ) / WF s (v) (12.51) u v S (R n ) is unambiguously defined provided (β, q) WF s (u) (S n 1 B n ) = ( β, q) / WF s (v). Proof. Let us onsider onvolution first. The hypothesis, (12.51) means that for eah β S n 1 (12.52) {q B n 1 ; (β, q) WF s (u)} {q B n 1 ; ( β, q) WF s (v)} =. Now, the fat that WF s is always a losed set means that (12.52) remains true near β in the sense that if U S n 1 is a suffiiently small neighbourhood of β then (12.53) {q B n 1 ; β U, (β, q) WF s (u)} {q B n 1 ; β U, ( β, q) WF s (v)} =.

13 LECTURE NOTES FOR , FALL The ompatness of S n 1 means that there is a finite over of S n 1 by suh sets U j. Now selet a partition of unity υ i of S n 1 whih is not only subordinate to this open over, so eah υ i is supported in one of the U j but satisfies the additional ondition that (12.54) supp(υ i ) ( supp(υ i )) = = (12.56) u i v i = u i v î supp(υ i ) ( supp(υ i )) U j for some j. Now, if we set u i = (υ i ) R u, and v i = (υ i ) R v, we know that u u i i has ompat support and similarly for v. Sine onvolution is already known to be possible if (at least) one fator has ompat support, it suffies to define u i v i for every i, i. So, first suppose that supp(υ i ) ( supp(υ i )) =. In this ase we onlude from (12.54) that (12.55) Css( û i ) Css( v î ) =. Thus we may define using (12.20). On the other hand if supp υ i ( supp(υ i )) = then (12.57) Css(u i ) ( Css(v i )) S n 1 = and in this ase we an define u i v i using Lemma Thus with suh a deomposition of u and v all terms in the onvolution are well-defined. Of ourse we should hek that this definition is independent of hoies made in the deomposition. I leave this to you. That the produt is well-defined under ondition (12.50) now follows if we define it using onvolution, i.e. as (12.58) uv = f g, f = u, ǧ = v. Indeed, using (12.47), (12.50) for u and v beomes (12.51) for f and g.