42 RICHARD B MELROSE 7 Tempered distributions A good first reference for distributions is 2 4 gives a more exhaustive treatment The complete metric topology on S Rn is described above Next I want to try to convice you that elements of its dual space S 0 Rn have enough of the properties of functions that we can work with them as generalized functions First let me develop some notation A differentiable function n Rn C has partial derivatives which we have denoted x j R C For reasons that will become clear later we put a 1 into the definition and write 1 7 1 Dj i xj We say is once continuously differentiable if each of these Dj is continuous Then we defined k times continuous differentiability inductively by saying that and the Dj are k 1 times continuously differentiable For k 2 this means that Dj Dk are continuous for j k 1 n Now recall that if continuous these second derivatives are symmetric 7 2 D j Dk D k Dj This means we can use a compact notation for higher derivatives Put N0 0 1 we call an element Nn0 a multi index and if is at least k times continuously differentiable we set12 1 1 n whenever 1 2 n k i x1 xn Now we have defined the spaces 7 4 C0k Rn Rn C D C00 Rn k 7 3 D Notice the convention is that D is asserted to exist if it is required to be continuous Using hxi 1 x 2 we defined 7 5 hxi k C0k Rn Rn C hxik C0k Rn and then our space of test functions is S Rn hxi k C0k Rn k 12Periodically there is the possibility of confusion between the two meanings of but it seldom arises LECTURE NOTES FOR 18 155 FALL 2004 43 Thus 7 6 S Rn D hxik C00 Rn k and all k Lemma 7 1 The condition S Rn can be written hxik D C00 Rn k k Proof We first check that C00 Rn Dj hxi C00 Rn j 1 n Since C00 Rn hxiDj C00 Rn j 1 n Dj hxi hxiDj Dj hxi 1 1 x hxi is a bounded continuous function i j and Dj hxi Then consider the same thing for a larger k 7 7 this is clear D hxip C00 Rn p 0 p k hxip D C00 Rn p 0 p k I leave you to check this as Problem 7 1 Corollary 7 2 For any k N the norms X khxik kC k and kx Dx k k k are equivalent Proof Any reasonable proof of 7 2 shows that the norms X khxik D k khxik kC k and k are equivalent Since there are positive constants such that X X C 1 1 x hxik C2 1 x k k the equivalent of the norms follows Proposition 7 3 A linear functional u S Rn C is continuous if and only if there exist C k such that X u C sup x Dx k k Rn 44 RICHARD B MELROSE Proof This is just the equivalence of the norms since we showed that u S 0 Rn if and only if for some k u Ckhxik kC k Lemma 7 4 A linear map T S Rn S Rn is continuous if and only if for each k there exist C and j such that if k and k X 0 0 7 8 sup x D T C sup x D S Rn 0 j 0 j Rn Proof This is Problem 7 2 All this messing about with norms shows that xj S Rn S Rn and Dj S Rn S Rn are continuous So now we have some idea of what u S 0 Rn means Let s notice that u S 0 Rn implies 7 9 7 10 7 11 xj u S 0 Rn j 1 n Dj u S 0 Rn j 1 n u S 0 Rn S Rn where we have to define these things in a reasonable way Remember that u S 0 Rn is supposed to be like an integral against a generalized function Z 7 12 u u x x dx S Rn Rn Since it would be true if u were a function we define 7 13 xj u u xj S Rn Then we check that xj u S 0 Rn xj u u xj X C k k C0 X sup x D xj Rn k 1 k sup x D Rn LECTURE NOTES FOR 18 155 FALL 2004 45 Similarly we can define the partial derivatives by using the standard integration by parts formula Z Z 7 14 Dj u x x dx u x Dj x dx Rn Rn if u C01 Rn Thus if u S 0 Rn again we define Dj u u Dj S Rn Then it is clear that Dj u S 0 Rn Iterating these definition we find that D for any multi index defines a linear map D S 0 Rn S 0 Rn 7 15 In general a linear differential operator with constant coefficients is a sum of such monomials For example Laplace s operator is 2 2 2 D12 D22 Dn2 x21 x22 x2n We will be interested in trying to solve differential equations such as u f S 0 Rn We can also multiply u S 0 Rn by S Rn simply defining u u S Rn 7 16 For this to make sense it suffices to check that X X sup x D 7 17 sup x D C k k Rn k Rn k This follows easily from Leibniz formula Now to start thinking of u S 0 Rn as a generalized function we first define its support Recall that 7 18 supp clos x Rn x 6 0 We can write this in another weak way which is easier to generalize Namely 7 19 p supp u S Rn p 6 0 u 0 In fact this definition makes sense for any u S 0 Rn Lemma 7 5 The set supp u defined by 7 19 is a closed subset of Rn and reduces to 7 18 if u S Rn 46 RICHARD B MELROSE Proof The set defined by 7 19 is closed since 7 20 supp u p Rn S Rn p 6 0 u 0 is clearly open the same works for nearby points If S Rn we define u S 0 Rn which we will again identify with by Z 7 21 u x x dx Obviously u 0 0 simply set in 7 21 Thus the map 7 22 S Rn 3 7 u S 0 Rn is injective We want to show that 7 23 supp u supp on the left given by 7 19 and on the right by 7 18 We show first that supp u supp Thus we need to see that p supp p supp u The first condition is that x 0 in a neighbourhood U of p hence there is a C …