Analysis IV, Assignment 4

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1 Analysis IV, Assignment 4 Prof. John Toth Winter 23 Exercise Let f C () an perioic with f(x+2) f(x). Let a n f(t)e int t an (S N f)(x) N n N then f(x ) lim (S Nf)(x ). N a n e inx. If f is continuously ifferentiable at x o [, ], { f(x t) f(x ) t t an t < Define F (t) f (x ) t By the assumptions on f, F is boune on [, ]. ecall the Dirichlet Kernel D N efine by sin((n+/2)t) sin(t/2). One can verify that (S N f)(x) (f D N )(x) an that 2 D N. Then (S N f)(x ) f(x ) f(x t)d N (t)t f(x ) (f(x t) f(x )) D N (t)t t F (t) sin((n + /2)t)t sin(t/2) From here we write sin((n + /2)t) sin(nt) cos(t/2) + cos(nt) sin(t/2) an take the limit as N. Since the terms in the integran not epening on N are boune this integral converges to by the iemann-lebesgue lemma. x Exercise 2 The Volterra integral operator T : L 2 ([, ]) L 2 ([, ]) given by (T f)(x) f(y)y for x [, ] is compact with spectrum {}.

2 T is a boune operator since T f 2 2 f 2 2 ( x 2 f(y)y) x f 2 x Notice that we can rewrite T as (T f)(x) { k(x, y)f(y)y where k(x, y) y < x y x Since k : [, ] 2 C L 2 ([, ] 2 ), T is Hilbert-Schmit an therefore compact. From the general theory of compact operators, λ anλ σ(t ) λ is an eigenvalue, an since L 2 ([, ]) is infinite imensional, σ(t ). Suppose that f(y)y λf(x). Since f L ([, ]), by the funamental theorem of calculus we may ifferentiate the equation to obtain f(x) λf (x) which has solution f(x) ce x λ for λ. Putting back into the integral equation will give that c, an we conclue that λ is not possible. Exercise 3 The Meellin transform efine as M : L 2 ( +, t/t) L 2 (), (Mf)(x) 2 f(t)tix t is a unitary operator. An alternate efinition, perhaps more stanar, of the Meellin transform is M : L 2 ( + ) L 2 (). ( Mf)(x) 2 f(t)t /2+ix t. Consier the map φ : L 2 ( + ) L 2 () ( φf)(z) e z/2 f(e z ). φ is unitary since φf 2 2 ez f(e z ) 2 z f(u) 2 u f 2 2 an is easily seen to be surjective. Now notice that M F φ, where F enotes the Fourier transform. Inee, 2 ez/2 f(e z )e izx z 2 f(t)t ix t Similarly one checks that M F φ where φ : L 2 ( +, t/t) L 2 () (φf)(z) f(e z ) an F enotes the inverse Fourier transform. φ is unitary since φf 2 2 f(ez ) 2 z f(u) 2 u u f 2 2. In either case, we have the composition of two unitary maps, so the Meellin transform is unitary. Exercise 4 Let C be open an H be the subspace of L 2 () consisting of holomorphic functions on. Then H is close subspace of L 2 () an hence a Hilbert space with inner prouct f, g f(x + iy)g(x + iy)xy 2

3 Obviously the secon part of the statement follows from the first as close subspaces of Hilbert spaces are Hilbert spaces with the same inner prouct. We shall nee the following result from complex analysis: Let {f n (z)} n be a sequence of complex-value functions analytic on an open connecte set D C which converge uniformly to f(z). Then f(z) is analytic on D. We also nee the following inequality for finite measure spaces: If µ() <, then for < p < q <, f p f q µ() /p /q which is prove by using Holer s inequality with conjugate exponents q p an f p p f p f p q/p q f p q p qµ() q p q q q p : Let g be a holomorphic function on. Then by the mean value property of holomorphic functions for any z an any r r(z) > sufficiently small so that B(z, r) we have f(z) r f(ζ)a Then f(z) B(z,r) 2 r f 2,B(z,r) r f 2,B(z,r) r f 2 In particular, if K C is compact, an z K, then f(z) ro f 2 where r o (K, c ). Hence if {f n } is a Cauchy sequence in L 2 (), then {f n } is uniformly Cauchy on all compacts K, an therefore converges uniformly on all compacts K. Combining this with previous result state earlier completes the proof. Exercise 5 Let {ϕ n } be an orthonormal basis for H. Then:. z, ϕ n (z) 2 (z, c ) 2 2. B(z, w) ϕ n (z)ϕ n (w) calle the Bergman kernel converges absolutely for (z, w) an is inepenent of the choice of orthonormal basis. 3. The linear transformation T : L 2 () H, (T f)(x) B(z, w)f(w)w is the orthogonal projection onto H. 4. In the special case where is the unit isc, f H iff f(z) a n z n for some a n satisfying a n 2 n+ <. Also {zn n+ } is an orthonormal basis of H an B(z, w) ( zw) 2. We start with a lemma: Let {a n }, {b n } be sequences of complex numbers. Then b n 2 sup a n b n. an 2 3

4 By Cauchy-Schwarz, we have a (N) n N b j 2 j. For { bn n N n > N b n 2 sup an 2 the supremum attains a n b n. By taking b n 2. a n 2, from the previous exercise we have a n ϕ n (z) (z, c ) a n ϕ n (z) 2 (z, c ), Now taking a n a (N) an applying the lemma yiels the result. n N ϕ j(z) 2 j 2. The fact that B(z, w) converges absolutely follows from applying Cauchy- Schwarz an invoking the previous result, (an remembering that for complex series, convergence implies absolute convergence). 3. Since H is a Hilbert subspace of L 2 () we may complete {ϕ n } with {ψ k } k to an orthonormal basis of L2 (). For f L 2 (), say f a n ϕ n + b k ψ k we have (T f)(z) k ( ) ( ) ϕ n (z)ϕ n (w) a m ϕ m (w) + b k ψ k (w) w m a m ϕ n (z) a n ϕ n (z) m ϕ n (w)ϕ m (w)w + k k { ϕn (z) n N n > N b k ϕ n (z) ϕ n (w)ψ k (w)w The fact that the kernel B(z, w) is inepenent of the choice of basis is now easy to see, because if B(z, w) correspons to {ϕ n } an B(z, w) correspons to { ϕ n }, then for all f L 2 () we have ( B(z, w) B(z, ) w) f(w)w which implies that for all z, B(z, w) B(z, w). 4. If n m, ϕ n, ϕ n n + n + B(,) 2 r z n z n z θ r 2n rrθ If n m, 4

5 (n + )(m + ) ϕ n, ϕ m (n + )(m + ) B(,) 2 r z n z m z θ r n+m e iθ(n m) rrθ If f H, then f(z) a n z n for some a n. Then f an so a n 2 n+ <. Finally B(z, w) n+ zn n+ wn where we use the fact that for ζ <, ζ (n + )ζ n a n n+ ϕ n (n + )(zw) n ( zw) 2 ζ n implies ( ζ) 2 Exercise 6 Let L be a linear partial ifferential operator with constant coefficients. Then the space of solutions u of Lu with u C ( ) is infinite imensional for 2. First we argue that a non-constant polynomial in two variables has uncountably many zeroes. If p(x, y) C[x, y] then we can fin polynomials q o (y),..., q n (y) with q n (y) not ientically zero such that p(x, y) n q i (y)x i. Let n enote the set of zeroes of q n (y) in C an for fixe λ, i Z λ the set of zeroes of p(x, λ) in C. Since q n (y) is not ientically zero, n is a finite collection an so C \ n is uncountable. Now for fixe λ C \ n, the polynomial p(x, λ) is of egree n, an therefore has at least one zero. It follows that Z λ {λ} is an uncountable istinct λ C\ n union of zeroes of p(x, y). This same argument can be one for higher imensions. For simplicity of notation, we continue in 2. Now we nee to show that we can fin infinitely many solutions to P (ξ)û(ξ) with u C ( ). Most of the work is alreay one because we can fin an infinite sequence of istinct roots (r (k), r(k) 2 ) k, i.e. P (r(k), r(k) 2 ). Take then û (k) (ξ, ξ 2 ) e r(k) ξ+r(k) 2 ξ2. It is easy to check that u (k) C ( ) an that the collection {û (k) : k N} is linearly inepenent for all N. 5

6 Exercise 7 Let f L 2 ( ). Then there exists g L 2 ( ) such that ( x) α f(x) g(x) in the weak sense iff (2iξ) α ˆf(ξ) ĝ(ξ) L 2 ( ). We let L ( ) α x so that L ( ) ( ) α alpha. x Notice that Lˆ ψ(ξ) ( ) α (2iξ) α ˆψ(ξ) (2iξ) α ˆψ(ξ) If (2iξ) α ˆf(ξ) ĝ(ξ), then by Plancherel s formula we have for any ψ C : g(x)ψ(x) ĝ(ξ) ˆψ(ξ) (2iξ) α ˆf(ξ) ˆψ(ξ) ˆf(ξ) Lˆ ψ(ξ) f(x)l ψ(x) so g Lf weakly. Conversely, suppose g Lf weakly. Then ĝ(ξ) ˆψ(ξ) g(x)ψ(x) f(x)l ψ(x) ˆf(ξ) Lˆ ψ(ξ) (2iξ) α ˆf(ξ) ˆψ(ξ) which implies that (2iξ) α ˆf(ξ) ĝ(ξ) 6