Folland: Real Analysis, Chapter 8 Sébastien Picard

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1 Folland: Real Analysis, Chapter 8 Sébastien Picard Problem 8.3 Let η(t) = e /t for t >, η(t) = for t. a. For k N and t >, η (k) (t) = P k (/t)e /t where P k is a polynomial of degree 2k. b. η (k) () exists and equals zero for all k N. (a) It is clear that for t > all derivatives η (k) (t) exist and can by computed by the usual rules of calculus. We proceed by induction. First, notice η () (t) = d dt e /t = t 2e /t = P (/t)e /t, where P = x 2 is a polynomial of degree 2. Suppose η (k) (t) = P k (/t)e /t for some k, where P k is a polynomial of degree 2k. Then ( ) d η (k+) (t) = dt P k(/t) e /t + P ( k(/t) e /t = t 2 t 2P k(/t) ) t k(/t) e /t. 2P This yields η (k+) (t) = P k+ (/t)e /t where P k+ = x 2 P k x 2 P k. Since P k is a polynomial of degree 2k, x 2 P k has highest order coefficient of degree 2k+2. On the other hand, P k is a polynomial of degree less than or equal to 2k ; therefore x2 P k has degree less than or equal to 2k +. It follows that P k+ = x 2 P k x 2 P k is a polynomial of degree 2(k +). The proof follows by induction. (b) We proceed again by induction. First, we compute η(t) η() e /t lim = lim t + t t + t The left-handed limit is easy since η(t) when t : Therefore, η () exists and is equal to zero. η(t) η() lim =. t t = lim s se s =. Suppose η (k) () exists and equals zero. We compute η (k+) (). Since η(t) for t, we have η (m) (t) = for all t < and all m N. By induction hypothesis, η (k) () =. Therefore we have the following left-handed limit:

2 η (k) (t) η (k) () lim =. t t On the other hand, we can use (a) and the fact that e x decays faster than any polynomial to compute the following right-handed limit: η (k) (t) η (k) () P k (/t)e /t lim = lim t + t t + t = lim s sp k (s)e s =. This shows that η (n+) () =. By induction, η (k) () exists and equals zero for all k N. Problem 8.4 If f L and τ y f f as y, then f agrees a.e. with a uniformly continuous function. (Let A r f be as in Theorem 3.8. Then A r f is uniformly continuous for r > and uniformly Cauchy as r.) Define A r f as in Theorem 3.8: A r f(x) = f(z)dz. m(b(r,x)) B(r,x) The first step is to show that for r >, A r f is uniformly continuous. Choose δ > such that for all c R n such that c < δ, we have τ c f f < ǫ. Then when x y < δ, we have A r f(x) A r f(y) = m(b(r,x)) = m(b(r,x)) m(b(r,x)) τ x yf f m(b(r,x)) < ǫ. B r(x) B r(x) B r(x) f(z)dz f(z)dz B r(y) B r(x) f(z)dz f(z +y x)dz f(z) f(z (x y)) dz B r(x) dz Next, we show that A r f is uniformly Cauchy. Choose δ > such that τ c f f < ǫ/2 when c < δ. Then for all r,s > such that r < δ, s < δ, for any x we have 2

3 A r f(x) A s f(x) A r f(x) f(x) + A s f(x) f(x) = m(b(r,x)) (f(z) f(x))dz + B r(x) m(b(s,x)) τ z x f(z) f(z) dz + m(b(r,x)) m(b(s,x)) < ǫ/2+ǫ/2 = ǫ. B r(x) B s(x) B s(x) (f(z) f(x))dz τ z x f(z) f(z) dz Taking the supremum over all points x, we see that A r f is uniformly Cauchy. At each point x, {A r f(x)} is Cauchy, so we can define f(x) = lim r A r f(x). We show A r f converges to f as r in the uniform norm. Select δ > such that A s f A s2 f u < ǫ for all < s < δ, < s 2 < δ. Then if < r < δ, for each point x we have A r f(x) f(x) = lim s A r f(x) A s f(x) lim s A r f A s f u < ǫ. We now show that f is uniformly continuous. Select r > such that A r f f u < ǫ/3 and δ > such that A r f(x) A r f(y) < ǫ/3 when x y < δ. Then for any points such that x y < δ, f(x) f(y) f(x) A r f(x) + A r f(x) A r f(y) + A r f(y) f(y) f A r f u + A r f(x) A r f(y) + A r f f u < ǫ/3+ǫ/3+ǫ/3 = ǫ. By Theorem 3.8, A r f f as r pointwise a.e. By unicity of the limit, we must have f = f a.e. Problem 8.8 Suppose that f L P (R). If there exists h L p (R) such that lim y y (τ y f f) h p =, we call h the (strong) L p derivative of f. If f L p (R n ), L p partial derivatives of f are defined similarly. Suppose that p and q are conjugate exponents, f L p, g L q, and the L p derivative j f exists. Then j (f g) exists (in the ordinary sense) and equals ( j f) g. First, we denote g(x) = g( x). After a change of variables, we see that g q = g q. Denote the L p derivative of f as h L p. Then at any point x R n, we have 3

4 j (f g)(x) (h g)(x) = lim f(x+tej y)g(y) f(x y)g(y) t dy t lim g(y) t (f(x+te j y) f(x y)) h(x y) dy t = lim t g(x z) t (f(z +te j ) f(z)) h(z) dz lim τ t x g q t ( τ tej f f ) h p = g q lim t ( τ tej f f ) h p =. t h(x y)g(y)dy Problem 8.4 (Wirtinger s Inequality) If f C ([a,b]) and f(a) = f(b) =, then b a ( ) 2 b a b f(x) 2 dx f (x) 2 dx. π (By a change of variable it suffices to assume a =, b = /2. Extend f to [ /2,/2] by setting f( x) = f(x), and then extend f to be periodic on R. Check that f, thus extended, is in C (T) and apply Parseval identity.) The first step is to show that, without loss of generality, we can assume that a =, b = /2. Suppose the inequality holds for this specific case. Then via the change of variables x = 2(b a)z+a, we obtain b /2 f(x) 2 dx = 2(b a) f(2(b a)z +a) 2 dz a ( ) 2 /2 2(b a) d 2π dz f(2(b a)z +a) 2 dz ( ) 2 /2 = (2(b a)) 3 f (2(b a)z +a) 2 dz 2π ( ) 2 b a b = f (x) 2 dx. π a a Next, we extend f to [ /2,/2] by setting f( x) = f(x), and then extend f to be periodic on R in the obvious way. To check that f C (T), it suffices to verify that f is differentiable and its derivative is continuous at ; all other points of the form n/2, n Z are done similarly. By the definition of C ([,/2]) given in Exercise 5.9, f has a one-sided derivative at the endpoint : f(t) f() lim = c. t + t 4

5 The left-handed derivative can be computed using the symmetry of f: f(t) f() f( t) lim = lim t t t + t (f(t) f()) = lim = c. t + t Hence f () exists and is equal to c. Next, we show that f (x) = f ( x). Indeed, d/dx(f( x)) = d/dx(f(x)) = f (x), and on the other hand by the chain rule, d dx (f( x)) = f ( x) ( ). Cancelling ( ) from both sides yields the result. We now show that the derivative is continuous at zero. Since f C ([,/2]), On the other hand, This completes the proof that f C (T). lim (t) = c. t +f lim (t) = lim ( t) = lim (t) = c. t f t +f t +f Now that we have reduced the problem to an easier case, we prove Wirtinger s inequality. Using Parseval s identity and integration by parts, we compute /2 /2 /2 /2 f(x) 2 dx = f 2 L = f(x) e i(2πkx) dx 2 2 k Z /2 /2 = f(x)dx+ /2 f(x) e i(2πkx) dx = k,k Z = k,k Z = k,k Z 4π 2 /2 /2 /2 /2 4π 2 k 2 k,k Z f(x) k,k Z /2 i(2πk) d (e i(2πkx) ) f (x) i(2πk) e i(2πkx) dx /2 /2 /2 /2 2 f (x) e i(2πkx) dx f (x) e i(2πkx) dx 2. Since f = at the endpoints, we see that /2 /2 f =. Hence by Parseval s identity we have /2 /2 f(x) 2 dx 4π 2 k Z /2 /2 f (x) e i(2πkx) dx 2 = 4π /2 /2 2 f (x) 2 dx.

6 Wirtinger s inequality then follows by symmetry: /2 /2 f 2 = 2 /2 f 2 2 ( ) 2 /2 /2 ( ) 2 /2 /2 f 2 = f 2. π π /2 Problem 8.6 Let f k = χ [,] χ [ k,k]. a. Compute f k (x) explicitly and show that f u = 2. b. fk (x) = (πx) 2 sin2πkxsin2πx, and fk as k. (Use Exercise 5a, and substitute y = 2πkx in the integral defining fk.) c. F(L ) is a proper subset of C. (Consider g k = fk and use the open mapping theorem.) (a) By definition of convolution, we have f k (x) = k For k >, k N, unravelling the definitions yields f k (x) = We see that f k C, and f k u = 2. (b) We use Exercise 5a to compute k χ [,] (x y)dy. 2 if (k ) x k, if x (k +), or x k +, (+k) x if k x k +, (+k)+x if (k +) x (k ). (f k ) (x) = (χ [,] χ [ k,k] ) = (χ [,] ) (χ [ k,k] ) = (πx) 2 sin2πkxsin2πx. By substituting y = 2πkx, we then obtain fk = (πx) sin2πkxsin2πx 2 dx = 4 π siny y sin(y/k) y/k dy. We know that for any y, lim sin y/k y/k =. k Also, for any B >, we have siny y sin y/k y/k χ [,B] χ [,B] L (R). 6

7 By the Dominated Convergence Theorem, we obtain 4 lim siny sin(y/k) k π y y/k χ [,B]dy = 4 π This leads to the conclusion that for all B >, we have lim k f k 4 B siny π y dy. siny y χ [,B]dy. However, as shown in Exercise 2.59a (which was on last semester s final exam), we know that y siny dy =. Hence the right hand side can be made arbitrarily large, and f k as k. (c) Suppose F : L C is onto. By Corollary 8.27, we conclude that F : L C is therefore a bijection. By the Open Mapping Theorem, F : C L is a bounded linear operator. In other words, there exists a C > such that for all f C, we have f C f u. However, f k u = 2 for all positive integers k, but f k can be made arbitrarily large as k. This contradiction establishes that F(L ) is a proper subset of C. Problem 8.26 The aim of this exercise is to show that the inverse Fourier transform of e 2π ξ on R n is φ(x) = Γ( 2 (n+)) π (n+)/2 (+ x 2 ) (n+)/2. a. If β, e β = π (+t2 ) e iβt dt. (Use (8.37).) b. If β, e β = (πs) /2 e s e β2 /4s ds. (Use (a), Proposition 8.24, and the formula (+t 2 ) = e (+t2 )s ds. c. Let β = 2π ξ where ξ R n ; then the formula in (b) expresses e 2π ξ as a superposition of dilated Gauss kernels. Use Proposition 8.24 again to derive the asserted formula for φ. (a) By (8.37), we have ( ) e 2π ξ = F = π(+x 2 ) If we let ξ = (2π) β, we obtain π (+x 2 ) e i2πxξ dx. e β = π (+t 2 ) e iβt dt. 7

8 (b) To justify using Fubini s Theorem later on, we first of all compute: e s e st2 e iβt dsdt = e s(+t2) dsdt = dt +t 2 = We now compute from (a) and the formula (+t 2 ) = e (+t2 )s ds, e β = π (+t 2 ) e iβt dt = π By Fubini s Theorem, we can swap the order of the integrals, e β = π We now substitute z = β(2π) t and obtain e β = π e s ( e s e st2 e iβt dtds. e (+t2 )s e iβt dsdt. arctant = π <. ) 2π )z2 e π(4πs/β2 e 2πiz dz ds = 2 e s F(e π(4πs/β2 )z 2 )() ds. β β The Fourier transform of the Gaussian can be computed using Proposition 8.24, which yields e β = 2 β ( ) β e s 2 /2 e β2 /4s ds = 4πs (πs) /2 e s e β2 /4s ds. (c) As done in (b), before lauching into the proof, we compute an integral that will later allow us to use Fubini s Theorem. The Gaussian is integrated using Proposition 2.53: (πs) /2 e s e π2 ξ 2 /s e 2πix ξ dξds = = = (πs) /2 e s e π2 ξ 2 /s dξds ( ) n/2 π (πs) /2 e s ds π 2 /s s n/2+/2 e s ds π (n+)/2 = π (n+)/2γ(n/2+/2) <. We now prove the claim. By part (b), letting β = 2π ξ we obtain e 2π ξ = (πs) /2 e s e π2 ξ 2 /s ds. 8

9 We use this to write (e 2π ξ ) (x) = e 2π ξ e 2πix ξ dξ = Applying Fubini, we exchange the order of integration (πs) /2 e s e π2 ξ 2 /s e 2πix ξ dsdξ. (e 2π ξ ) (x) = (πs) /2 e s e π2 ξ 2 /s e 2πix ξ dξds = Proposition 8.24 gives us the inverse Fourier transform of a Gaussian, hence (e 2π ξ ) (x) = (πs) /2 e s ( s π) n/2e s x 2 ds = We substitute z = s(+ x 2 ) and obtain π (n+)/2 (πs) /2 e s (e π2 ξ 2 /s ) (x) ds. s (n )/2 e s(+ x 2) ds. (e 2π ξ ) (x) = ( ) (n+)/2 z π (n+)/2 + x 2 2 (n+) e z dz = Γ((n+)) 2 (+ x 2 ) (n+)/2. π (n+)/2 Problem 8.3 Suppose a >. Use (8.37) to show that k 2 +a = π +e 2πa 2 a e 2πa. Then subtract a 2 from both sides and let a to show that k 2 = π 2 /6. Using the notation φ from (8.37), we have k 2 +a 2 = π a 2 +( k a )2 = π a 2 φ(k/a). By the Poisson summation formula, we obtain π a 2 Using Theorem 8.22b, we have φ(k/a) = π a 2 (φ(k/a)). π a 2 (φ(k/a)) = π a 2 9 aˆφ(ka).

10 We now apply (8.37) to substitute for ˆφ: k 2 +a 2 = π a = π a ( = π a = π a = π a ˆφ(ka) e 2π ka +2 ( +2 ( ) ) e 2πa k k= +e 2πa e 2πa. ( e 2πa )) Now that we have shown the desired identity, we subtract a 2 from both sides: After simplification, we obtain k 2 +a k= k 2 +a = π +e 2πa 2 a e 2πa a 2. k 2 +a = aπ +e 2πa (+aπ). 2 a 2 ( e 2πa ) Letting a, the left hand side goes to 2 k 2. For the right hand side, we repeatedly apply l Hopital s rule (three times in total): aπ +e 2πa (+aπ) lim a a 2 ( e 2πa ) = lim a π +πe 2πa 2πe 2πa (+aπ) 2a( e 2πa )+2πa 2 e 2πa = lim a = lim a = 4π3 2π. Therefore 2 k 2 = π 2 /3, which completes the proof. 4aπ 3 e 2πa 2 2e 2πa +8aπe 2πa 4π 2 a 2 e 2πa 4π 3 e 2πa 8aπ 4 e 2πa 2πe 2πa 24π 2 ae 2πa +8π 3 a 2 e 2πa Problem 8.35 The purpose of this exercise is to show that the Fourier series of most continuous functions on T do not converge pointwise. a. Define φ m (f) = S m f(). Then φ C(T) and φ = D m.

11 b. The set of all f C(T) such that the sequence {S m f()} converges is meager in C(T). (Use Exercise 34 and the uniform boundedness principle.) c. There exists f C(T) (in fact, a residual set of such f s) such that {S m f(x)} diverges for every x in a dense subset of T. (The result of (b) holds if the point is replaced by any other point in T. Apply Exercise 4 in 5.3. (a) We have φ m (f) = S m f() = f(y)d m (y)dy. T By linearity of the integral, φ m acts linearly on C(T), and φ m (f) f D m dy f u D m. T Therefore, φ m C(T). To show φ = D m, we apply the Riesz Representation Theorem for Radon measures. Indeed, since Lebesgue measure is a Radon measure and D m L, we know dµ = D m dy is a Radon measure. Hence we can rewrite φ m (f) = fdµ. By definition, d µ = D m dy, hence by the Riesz Representation Theorem (7.7): φ m = µ = µ (T) = D m dy = D m. T T (b) Proceed by contradiction. Suppose the set of all f C(T) such that the sequence {φ m (f)} converges is nonmeager inc(t). Then sup m φ m (f) < forall f insome nonmeager subset of C(T). By the uniform boundedness principle, sup m φ m <. However, by Exercise 34, φ m = D m as m. This is a contradiction, and hence the set of all f C(T) such that the sequence {φ m (f)} converges is meager in C(T). (c) Enumerate the rational {r k } inside (,). This yields a countable dense subset of T R/Z. Define T jk (f) = S j f(r k ). Then T jk (f) = f(y)d j (r k y)dy = f(y)τ rk D j ( y)dy. T T Denote D j ( y) = D j. We can repeat the same analysis done in (a) and obtain T jk C(T) and

12 T jk = τ rk Dj = D j. By the same argument of part (b), we conclude that for all r k, the set of all f C(T) such that sup j T jk f < is meager in C(T). We now apply the principle of condensation of singularities (Exercise 5.4). We have shown that for each k there exists a f C(T) (indeed, a residual set of f s) such that sup{ T jk f : j N} =. By the principle of condensation of singularities, there is a residual set of f s in C(T) such that sup{ T jk f : j N} = for all k. Hence we obtain we a residual set of f C(T) such that {S m f(r k )} diverges for each r k. 2