Mathematical Tripos Part III Michaelmas 2017 Distribution Theory & Applications, Example sheet 1 (answers) Dr A.C.L. Ashton

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1 Mathematical Tripos Part III Michaelmas 7 Distribution Theory & Applications, Eample sheet (answers) Dr A.C.L. Ashton Comments and corrections to acla@damtp.cam.ac.uk.. Construct a non-zero element of D () that vanishes outside (, ). Construct a non-zero element of D ( n ) that vanishes outside the ball B ɛ { n : < ɛ}. Answer: Consider the function ϕ() { ep ( ) ɛ You will find ϕ D( n ) and has the desired properties.. Given ϕ D(X), Taylor s theorem gives ϕ( + h), < ɛ, ɛ. α N h α α! α ϕ() + N (, h). Prove that supp ( N is contained in some fied compact K X for h sufficiently small. Show also that α N o ( h ) N uniformly in for each multi-inde α, i.e. prove [ sup α ] N (, h) h h N for each multi-inde α. [Hint: you may find it convenient to use the following form of remainder N (, h) α N+ h α (N + ) α! and note that (N + )! α N+ hα /α! (h + + h n ) N+. ] Answer: The remainder is N (, h) α N+ h α (N + ) α! ( t) N ( α ϕ)( + th) dt, ( t) N ( α ϕ)( + th) dt, For h n fied, it is clear that N (, h) vanishes unless + th supp(ϕ) for some t [, ]. Hence supp( N ) supp(ϕ) + B h. Choose h small enough so that the left hand side is contained in some compact subset of X. Also sup α N (, h) h β h n βn sup β! β N+ h β h n βn β! β N+ ( ) N+ h + + h n [ n /( ) ] / N+ h + + h n h N+ α+β ϕ( + th) dt where in the second to last line we used Cauchy-Schwarz. The implied constant in this final estimate will depend on {α, N, n, ϕ}. We see α N (, h) o( h N ) uniformly in. Consider the map d : supp(ϕ) [, ) defined by d() dist(, X), where d(, Y ) inf y Y y. Then d is continuous and d() > since X and supp(ϕ) are disjoint. Compactness of supp(ϕ) implies d() δ for some δ >, so taking h < δ would do.

2 3. Which elements of D(X) can be represented as a power series on X? Answer: None other than ϕ. For this problem define the set N { X : ϕ() } so that supp(ϕ) cl(n ). Take y on the boundary of the support of supp(ϕ), i.e. y N. By our hypothesis there eists a δ > such that that ϕ() coincides with with its Taylor series on B δ (y), i.e. ϕ() α ( y) α α ϕ(y), α! B δ (y). By continuity we have α ϕ(y) for each multi-inde α, meaning ϕ() on B δ (y) supp(ϕ). Since y N, there is a sequence n N with n y, so in particular there eists an N such that n B δ (y) for n > N. This means n B δ (y) supp(ϕ) for n > N, contradicting the fact ϕ( n ). 4. Prove the C Urysohn lemma: if K is a compact subset of X n, show that one can find a ϕ D(X) such that ϕ and ϕ on a neighbourhood of K. [Hint: Let χ be a characteristic (indicator) function on a neighbourhood of K and smooth it off by taking the convolution with a test function that approimates δ.] Answer: Choose ɛ > sufficiently small so that y 4ɛ when K and y X c n \ X. Following the hint, we introduce an indicator function on a neighbourhood of K: let χ on the set K ɛ {y : y ɛ for some K} and χ otherwise. Now fi ϕ D(X) with ϕ d and supp(ϕ) B. It follows that ϕ ɛ () ɛ n ϕ(/ɛ) is an approimation to the identity as ɛ and supp(ϕ ɛ ) B ɛ. We consider the function φ() (χ ϕ ɛ )() χ(y)ϕ ɛ ( y) dy. Certainly φ C (X) and it is clear that supp(φ) supp(χ) + supp(ϕ ɛ ) K ɛ + B ɛ K 3ɛ so φ D(X). Since ϕ ɛ d we have φ() ( ) ϕ ɛ (y) χ( y) dy which vanishes if y K ɛ for all y B ɛ. So φ in K ɛ and we are done. 5. Given T D (X), the derivative α T is defined by α T, ϕ ( ) α T, α ϕ ϕ D(X). Show that that α T D (X). If ord(t ) m what can you say about ord( α T )? Answer: By definition, for each compact K X there eist constants such that α T, ϕ def ( ) α T, α ϕ C sup α+β ϕ C sup γ ϕ β N γ N+ α for each ϕ D(X) with supp(ϕ) K. So α T D (X) and if ord(t ) m then ord( α T ) m + α. To construct a distribution for which this inequality is strict for a particular multi-inde, consider the linear map T, ϕ α ϕ(,,..., ) d and calculate T/ i for i and i. 6. Let T D (X) and f C (X). Prove that for each multi-inde α α (T f) ( ) α β f α β T β β α in D (X) (take the case X if you re still getting used to multi-indices).

3 Answer: Using the definitions we have (ft ), ϕ ft, ϕ T, fϕ T, (fϕ) + T, f ϕ (ft + f T ), ϕ i.e. (ft ) f T + ft in D (X). The general case is similar. 7. Let { k } k be a sequence in X with no it point in X. Consider the family of linear maps u α : D(X) C defined by u α, ϕ α ϕ( k ) for each multi-inde α. For which α is u α D (X)? What is ord(u α )? k Answer: If ϕ D(X) then there are only a finite number of the k contained in supp ϕ (if there were infinitely many inside the compact set supp ϕ, we would be able to go to a subsequence which converges, contradicting the fact that the sequence has no it points). Hence u α, ϕ k α ϕ( k ) k I α ϕ( k ) where I {k : k supp ϕ} is a finite set that depends only on supp(ϕ). So the linear map is welldefined. To see that it s a distribution: fi a compact K X, and take ϕ D(X) with supp ϕ K. Then u α, ϕ α ϕ( k ) sup α ϕ I sup β ϕ. k I k I β α Since I is dependent only on K, we deduce that u α D (X) for all multi-indices α and clearly ord(u α ) α. To show that the order is equal to α, set ϕ m () ( ) α ψ(m( )) for some ψ D( n ) with sufficiently small support and ψ(). Then sup β ϕ m m β α, since β ϕ m () ( ) β β γ ( ) α m γ ( γ ψ)(m( )), γ γ β and the right hand side vanishes unless /m. So sup β ϕ m as m if β < α. But u α, ϕ m α! so the ord(u α ) cannot be less than α. For intuition: our sequence of functions looks eactly like ( ) α localised closer and closer around the point. 8. Find the most general solution to the equations in D (). (a) u, (b) u δ, (c) ( e πi ) u Answer: (a) If u then (u ). In lectures we showed that v in D () implied v const in D (). Hence the general solution to u is u + const. (b) Try u cδ for a particular solution. Then u, ϕ u, (ϕ) u, ϕ u, ϕ c δ, ϕ. So we find u δ by setting u δ. Now we need to classify all solutions to the homogeneous problem. Consider equation v in D (). Introducing ρ D() with ρ in a neighbourhood of {} find v, ϕ v, ρϕ + v, ( ρ)ϕ. By Taylor s theorem ϕ() ϕ() + ϕ(), where ϕ C () and φ D(). And so v, ϕ ϕ() v, ρ + v, ρ ϕ + ( ρ())ϕ() φ() v, φ c δ, ϕ 3

4 where c v, ρ. We conclude that the most general solution to v in D () is v c δ. It follows that the most general solution to u is u c H + c, where H is the Heaviside function which satisfies H δ. So the general solution to u δ in D () is u c H + c δ. (c) It is clear that supp(u ) Z. Indeed, if supp(ϕ) Z then u, ϕ (e πi )u, (e πi ) ϕ since (e πi ) is smooth on supp(ϕ). Introduce the functions ρ n D() {, n ɛ ρ n (), n > ɛ where ɛ < / say (so supp(ρ n ) supp(ρ m ) for n m). Then we have (e πi )u, ϕ n (e πi )u n, ϕ n v n, ϕ where u n ρ n u and v n is defined accordingly. Now since the supports of the v n are disjoint and v n it must be the case that v n for each n. Indeed, we have v n, ϕ v n, ρ n ϕ m n v m, ρ n ϕ since supp(ρ n ϕ) supp(v m ) for m n. So our problem is reduced to solving (e πi )u n. We may concentrate on m since the other cases are just translations of this. We have u, ϕ u, ρ ϕ and by Taylor s theorem ϕ() ϕ() + ϕ() for some ϕ C (). Note that we can write where f C () and f > on supp(ρ ). Hence (e πi ) f() u, ϕ ϕ() u, ρ m + u, ρ ϕ c δ, ϕ + We deduce u c H + const. The general solution is then (e πi )u, ρ ϕ c δ, ϕ. f u const + n c n (τ n H). This is formal the series doesn t necessarily converge in D () for any old choice of {c n }. 9. Define the distribution u D ( ) by the locally integrable function {, y u(, y), otherwise. Show that u yu in D ( ). Can you give a physical interpretation of this result? Answer: We are required to compute u yu, ϕ u(, y) (ϕ ϕ yy ) d dy 4

5 We do the first term first: since u when < y we have ( ) u(, y)ϕ (, y) d dy ϕ (, y) d dy y ϕ (y, y) dy. Similarly for the second term ( u(, y)ϕ yy (, y) d dy ) ϕ yy (, y) dy d ϕ y (, ) d. Summing up these contributions and relabelling the dummy variables to τ we have u yu, ϕ ( d ϕ (τ, τ) + ϕ y (τ, τ)) dτ ϕ(τ, τ) dτ dτ i.e. u yu in D ( ). If y is a time-like coordinate then this corresponds to a step wave.. Compute ( n ) in D ( n ) for n 3, i.e. compute ( n ), ϕ n, ϕ ϕ d n for arbitrary ϕ D ( n ). Note that ( + + n) / and i ( / i). [Hint: use d ɛ d + d and treat each integral separately.] >ɛ Answer: Since n is locally integrable on n it follows that ϕ d ɛ n B ɛ (this can be checked using polar coordinates with d r n dr dσ n ). To look at the contribution from Bɛ, c first choose > sufficiently large so that supp(ϕ) B. Now apply Green s theorem to Ω ɛ B \ B ɛ and observe that ( n ) on Bɛ c for any ɛ >. ecall Green s theorem says ( (u v v u) d u v Ω ɛ n v u ) da n where da is the surface element on Ω ɛ. Setting u n, v ϕ and using polar coordinates with r we find ( ϕ n ϕ d r n rɛ r ϕ ) r (r n ) r n dσ n Ω ɛ where dσ n is the surface measure on S n. We have chosen > sufficiently large so that there are no contributions from the outer surface B. The it of the integral on the right is just ( ɛ ϕ ) ( n)ϕ rɛ dσ n (n )ω n ϕ() ɛ r rɛ S n where ω n S n dσ n. We deduce that ( n ) c n δ in D( n ), where c n ( n)ω n.. Let {f k } k be the sequence of smooth functions defined by f k () π Prove that f k δ in D (). Compute the its Ω ɛ k (k) +. in D (). (a) k k e k, (b) k k3 e ik sin(k), (c) k π 5

6 Answer: First few are straightforward, e.g. f k, ϕ k k π kϕ() + k d k π ϕ(/k) d ϕ() + where in the final step we used the dominated convergence theorem to interchange the it and the integral. Using the same substitution trick we find (a) f k. Integrating by parts we find (b) f k. For (c) proceed as follows: given ϕ D() fi > so that supp(ϕ) [, ]. Then f k, ϕ Using integration by parts on the first term we see sin(k) [ ] sin(k) ϕ() ϕ() d + ϕ() d π π [ ϕ() ϕ() sin(k) π ] d O ( ) k where the implied constant depends on ϕ. By changing variables in the second integral we deduce so that f k δ in D (). k f sin k, ϕ ϕ() d ϕ() k k k π. Compute the it in D (, ). k [ + k m cos(πm) ] Answer: Any ϕ D(, ) has an absolutely convergent Fourier series ϕ() n e iπn ˆϕ n, ˆϕ n e iπn ϕ() d. We have u k, ϕ m k m k e iπm ϕ() d ˆϕ n From this we deduce i.e. u k δ in D (, ). k u k, ϕ n ˆϕ n ϕ() 3. We define the principal value of /, written p.v.(/), by ( ) ϕ() p.v., ϕ ɛ >ɛ d for all ϕ D (). Prove that p.v.(/) D () and ord(p.v.(/)). Show that ( ) ( ) p.v. + iπδ in D (). ɛ iɛ 6

7 Answer: For a given ϕ D() fi > so that supp(ϕ) [, ]. Then [ ϕ() ɛ ] >ɛ d ϕ() + ɛ d ϕ() ϕ( ) d ɛ It follows that ( ) ϕ() ϕ( ) p.v., ϕ d. For each [, ] with supp(ϕ) [, ] we have ( ) p.v., ϕ ϕ() ϕ( ) d ϕ (y) dy d hence ( ) p.v., ϕ C sup ϕ C sup α ϕ. α ϕ (τ) dτ d, So p.v.(/) D () and ord(p.v.(/)). If ord(p.v.(/)) then p.v.(/), ϕ C K ϕ for all ϕ D() with supp ϕ K. But this can t be true: consider ϕ n with supp ϕ n [, ] and ϕ n on [/n, ]. Then since ϕ n () for each n (by continuity) we have ( ) ϕ n () ϕ n () p.v., ϕ n d d log n. /n So p.v.(/), ϕ n but ϕ n and supp ϕ n K [, ], so ord(p.v.(/)) >, and by previous it must be. A note for intuition: we picked a collection of test functions that were bounded, but whose derivatives necessarily got bigger and bigger near. For the net part ϕ() iɛ d ϕ() ϕ() iɛ ϕ() ϕ() iɛ ϕ() ϕ() iɛ d + ϕ() d + iϕ() d + iϕ() /ɛ /ɛ d iɛ ɛd + ɛ + ϕ() d + where we discarded one integral since the integrand was odd. Taking the it we get ɛ iɛ, ϕ ϕ() ϕ() d + iπϕ() Since the integrand in the first term is integrable, we can write it as ϕ() ϕ() + ɛ d ϕ() ϕ() ϕ() ϕ() d d d δ δ< < δ δ< < δ δ< < d The second of these vanishes and we deduce ɛ iɛ, ϕ which yields the desired result. 4. Show that (log ) p.v.(/) in D (). p.v. ( ), ϕ + iπδ, ϕ 7

8 Answer: We have (log ), ϕ log ϕ () d ɛ ɛ >ɛ p.v. log ϕ () d [ log(ɛ)ϕ( ɛ) log(ɛ)ϕ(ɛ) + ( ), ϕ where we used the fact ϕ(ɛ) ϕ( ɛ) O(ɛ) and ɛ log(ɛ) o(). 5. Let f f(z) be comple analytic on X C. Suppose f has zeros at {z i } in X with multiplicities {m i }. Prove that >ɛ ϕ() d ] log f π i m i δ zi in D (X), where + y is the Laplacian on. Answer: First we split the integral into neighbourhoods of the zeros of f and their complement log f, ϕ log f ϕ d log f ϕ d + log f ϕ d. X i B ɛ(z i) X\ B ɛ(z i) The first collection of integrals can be made as small as we please by making ɛ small. Indeed, in a neighbourhood of any z i we can write f(z) (z z i ) mi g(z) for some analytic g with g(z i ) so log f m i log z z i + log g in a neighbourhood of z i. So log f is clearly locally integrable. By Greens theorem ( ) log f ϕ ϕ log f d ( log f ϕ X\ B ɛ(z i) i B ɛ(z i) n ϕ ) log f ds n where s is the arc-length. Since log f e log(f) we know that log f is harmonic in X \ B ɛ (z i ), i.e. log f above. By translation invariance it is enough to understand the problem about z i. ( log f ϕ d log f ϕ X\B ɛ() B ɛ() n ϕ ) log f ds n π ( ϕ r (m i log r + log g ) (m i log r + log g ) ϕ ) r dθ r rɛ π m i ϕ(ɛ cos θ, ɛ sin θ) dθ + o() πm i ϕ() + o() where we used / n / r since on the boundary of X \ B ɛ () the normal points into B ɛ (). Taking the it ɛ we deduce log f, ϕ π i m i δ zi, ϕ. 6. Define the distribution u E ( 3 ) the locally integrable function {,, u(), >. Prove that i i( u/ i ) dσ in E ( 3 ), where dσ is the surface element on the sphere S 3. Answer: For ϕ E( 3 ) the divergence theorem gives u i, ϕ ( i ϕ) d (ϕ) d (ϕ) dσ ϕ dσ i i B i B S S since for S. 8