Fourier Analysis, Stein and Shakarchi The Fourier Transform on R

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1 Fourier Analysis, Stein and Shakarchi Chapter 5 The Fourier Transform on Yung-Hsiang Huang Abstract After Problem 6, we contain a proof for Widder s uniqueness theorem in the class of nonnegative solutions of heat equation in the strip S T := {(x, t) d (, T )}. It s taken from [3]. Exercises. Corollary.3 in Chapter leads to the following simplified version of the Fourier inversion formula. Suppose f is a continuous function supported on [ M, M], whose Fourier transform ˆf is of moderate decrease. (a) Fix L with L/ > M, and show that f(x) = a n (L)e πinx/l where a n (L) = L L/ L/ f(x)e πinx/l dx = L ˆf(n/L). Alternatively, we may write f(x) = δ ˆf(nδ)e πnδx with δ = /L. (b) Prove that if F is continuous and of moderate decrease, then F (ξ) dξ = lim δ + δ F (δn). (c) Conclude that f(x) = ˆf(ξ)e πxξ dξ. Proof. (c) is proved by taking F (ξ) = f(ξ)e πiξx in (b) and then apply (a). Department of Math., National Taiwan University. d4@ntu.edu.tw

2 For (a), we note that the Fourier series of f on [ L, L ] is a n (L)e πinx/l. which converges absolutely and hence uniformly since we have, by the moderate decreasing property of f, a n (L) = L ˆf(n/L) L L + n. The uniqueness theorem of Fourier series and continuity of f imply the desired equality at every point x. (b) Given ɛ >, we have N = N(ɛ) such that n n F (ξ) dξ F (ξ) dξ < ɛ whenever n N. Note that there is a L ɛ > such that for each L L ɛ and for each < δ <, δ n > L+ δ F (δn) Aδ n > L δ + + δ n A Therefore, for M =: max{n, L ɛ + } we have F (ξ) dξ δ M F (δn) F (ξ) dξ F (ξ) dξ + n Z M + δ F (δn) δ F (δn) n Z L δ δ + δ x dx = A(π arctan L) < ɛ. M M F (ξ) dξ δ n M δ F (δn) n M δ M ɛ + F (ξ) dξ δ M n M δ F (δn). Then by the uniform continuity of F on [ M, M], there is a δ ɛ > such that the last term is less than ɛ whenever δ (, δ ɛ ).. Let f and g be the functions defined by f(x) = χ [,] (x) and g(x) = χ [,] (x) ( x ). Although f is not continuous, the integral defining its Fourier transform still make sense. It s easy to see that f(ξ) = sin πξ πξ and ĝ(ξ) = with the understanding that f() = and ĝ() =. ( sin πξ ) πξ convolution of cf δ with itself, for some c, δ and f δ (x) := f(δx). Note that one can interpret g as a 3. The following exercise illustrates the principle that the decay of f is related to the continuity properties of f. (a) Suppose that f is a function of moderate decrease on whose Fourier transform f is continuous and satisfies ( ) f(ξ) = O as ξ ξ +α

3 for some < α <. Prove that f satisfies a Hólder condition of order α. (b) Let f be a continuous function on which vanishes for x, with f() =, and which is equal to / log(/ x ) for all x in a neighborhood of the origin. Prove that f is not of moderate decrease. In fact, there is no ɛ > so that f(ξ) = O(/ ξ +ɛ ) as ξ. Proof. (a) Since f L (), we have, for each h > f(ξ)[e πi(x+h)ξ e πixξ ] dξ + f(x + h) f(x) = f(ξ)[e πi(x+h)ξ e πixξ ] dξ f(ξ)[e πi(x+h)ξ e πixξ ] dξ ξ > h ξ h C ξ α dξ + h f(ξ) sin(π hξ ) dξ. ξ h By the hypothesis, there is for some M > such that f M/( + ξ +α ) for all ξ where K is independent of h. h f(x + h) f(x) 4Cα h α sin(π hξ ) + M dξ + ξ +α 4Cα h α + M h α sin(πu) u du +α 4Cα h α + M h α M u α du = K h α, (b) The continuity of f is a general fact. Suppose f satisfies the decay condition for some > ɛ >, then f is ɛ-hölder, especially near the origin. However one find that as h, This is a contradiction. f(h) f() = h ɛ. h ɛ log h 4. Examples of compactly supported functions in S() are very handy in many applications in analysis. Some examples are: (a) Suppose a < b, and f is the function such that f(x) = if x a or x b and f(x) = e /(x a) e /(b x) if a < x < b. Show that f is indefinitely differentiable on. (b) Prove that there exists an indefinitely differentiable function F on such that F (x) = if x a, F (x) = if x b, and F is strictly increasing on [a, b]. 3

4 (c) Let δ > be so small that a + δ < b δ. Show that there exists an indefinitely differentiable function g such that g is if x a or x b, g is on [a + δ, b δ], and g is strictly monotonic on [a, a + δ] and [b δ, b]. Proof. We may assume a > (why?). (a) is easy (by mathematical induction). For (b) consider F (x) = c x f(t) dt where c is the reciprocal of f(t) dt. For (c) we first observe that h(x) = F (x ) is a function that equals to on [ a, a], decreasing on { a x b} and vanishes outside { x b}. After a suitable scaling and translation, we have a desired one. 5. Suppose f is continuous on and of moderate decrease. (a) Prove that f is uniformly continuous and vanishes at infinity. (b) Show that if f(ξ) = for all ξ, then f is identically. Proof. (a) (st proof, without LDCT) Given ɛ >, pick large N such that A x >N ɛ/ where A > is the moderate constant. Then f(ξ + h) f(ξ) f(x) e πixh dx A ( N A N provided h is small enough such that sin πhx < sin πhx + + x x >N ɛ 8AN smallness for h is independent of ξ, so the continuity is uniform. sin πhx + x dx ) + x dx < ɛ (nd proof, with LDCT.) The continuity is directly from LDCT. Note that So f(ξ) = f(ξ) = f(x)e πixξ dx = dx < +x for all x [ π, π]. Note that the f(y ξ )e πiyξ dy. [ f(x) f(x ξ )] e πixξ dx and hence the LDCT implies the decay at. (b) Given g S(), then by Fubini s theorem f(x)ĝ(x) dx = f(x) g(y)e πiyx dy dx = g(y) f(x)e πiyx dx dy = g(y) f(y) dy =. In particular, for each y, we take g y (ξ) = e πξδ e πiξy whose Fourier transform is ĝ y (x) = e π(x y) /δ δ /. Letting δ, we have f(y) =. 4

5 emark. See Exercise. (iemann-lebesgue lemma) and.5 (the decay rate of f may be slow as ξ ɛ for any ɛ) in Book III. Also consult [3, Section I.4.] or [, Exercise.6] for showing the Fourier transform map from L ( d ) to C ( d ) is not surjective. For L (T) case, see Section 3. of Chapter 4 in Book IV. emark. For functions on T, one can construct functions with arbitrary slow decay for their Fourier transform, see [8, Section 3.3.]. 6. The function e πx is its own Fourier transform. Generate other functions that (up to a constant multiple) are their own Fourier transforms. What must the constant multiples be? To decide this, prove that F 4 = I. Here F(f) = f is the Fourier transform, F 4 = F F F F, and I is the identity operator (If)(x) = f(x) (see also Problem 7). Proof. If F(g) = cg, then c 4 = since F 4 = I. So c {,, i, i}. The case c = is solved in the statement with g(x) = e πx. Note that g = πxg(x) and hence ĝ (x) = πixĝ(x) = πixg(x) = ig (x), that is, the case c = i has an eigenfunction g (x) = πxe πx. Furthermore, we differentiate g again and see that g (x) = πg(x) + 4π x g(x) and hence ĝ (x) = 4π x ĝ(x) = 4π x g(x) = πg(x) g (x), that is, the case c = has an eigenfunction since g + πg(x) = (πg(x) + g (x)). Finally, we differentiate the g once again and see that g (x) = πg (x) + 8π xg(x) + 4π x g (x) = 6πg + 4π x g (x). Taking Fourier transform to both sides, we have g + 3πg (ξ) = 3πĝ (ξ) d dξ ĝ = i(3πg + g )(ξ), that is, g + 3πg is an eigenfunction for c = i. emark 3. Note that these four eigenspaces E, E, E i, E i span all of L ( d ), that is, given f L ( d ) we are searching f k E k for k = ±, ±i such that f = f + f + f i + f i. To find these f k, one can solve this equation with the following three equations through linear algebra (matrix): Ff = f f + if i if i, F f = f + f f i f i, F 3 f = f f if i + if i. This decomposition of L can be used to give new examples of eigenfunctions for F, e.g. the previous f should be (f+ff+f f+f 3 f)/4 and now we take f(x) = e π x (so Ff(x) = π and F f = f. Therefore f (x) = e π x + π +x is also an eigenfunction for c = different +x from the one given in the exercise. I learned this perspective from [, page ]. 7. Prove that the convolution of two functions of moderate decrease is a function of moderate decrease. 5

6 Proof. Let f, g be functions of moderate decrease. Then the continuity of f g can be proved by LDCT if one realize the integral in Lebesgue s sense, or using an standard decomposition of the integral as local part and far-away part like the following proof for moderate decrease if one realize the integral in iemann s sense, we leave the details to the readers. On the other hand, the moderate decreasing property can be proved as follows: f g(x) f(t) dt t x c g + x 4 f(t) g(x t) dt + + g(x t) dt t > x c f + x 4 f(t) g(x t) dt M + x. 8. Prove that if f is continuous, of moderate decrease, and f(y)e y e xy dy = for all x, then f. Proof. We know [f e x ](z) = e z f(y)e y e zy dy = for all z. So f since f(ξ)ê x (ξ) = f e x (ξ) = for all ξ. Thus f by Fourier inversion theorem. emark 4. This is true for all L ( d ) or L ( d ) functions. 9. If f is of moderate decrease, then ( ξ ) f(ξ)e πixξ dξ = (f F )(x), ( ) sin πt where the Fe jer kernel on the real line is defined by F (t) = πt if t and F () =. Show that {F } is a family of good kernels as, and therefore (f F ) tends uniformly to f(x) as. This is the analogue of Fe jer s theorem for Fourier series in the context of the Fourier transform. Proof. Note that for all > and hence for all δ > F (t) dt = F (t) dt = π [ δ,δ] c F (t) dt = π πδ sin x x sin x x dx = π sin z z dx as dz = 6

7 . Below is an outline of a different proof of the Weierstrass approximation theorem. Dfine the Landau kernels by L n (x) = ( x ) n c n χ x. where c n is chosen so that inf L n (x) dx =. Prove that {L n } n is a family of good kernels as n. As a result, show that if f is a continuous function supported in [ /, /], then (f L n )(x) is a sequence of polynomials on [ /, /] which converges uniformly to f. Proof. Note that c n = ( x ) n dx = ( x ) dx x( x ) n dx =. So for n+ each δ >, δ x L n(x) dx ( δ ) n (n + ) as n.. Suppose that u is the solution to the heat equation given by u = f H t where f S(). If we also set u(x, ) = f(x), prove that u is continuous on the closure of the upper half-plane, and vanishes at infinity, that is, u(x, t) as x + t. emark 5. We should not only consider the tangential limit, i.e. (x, t) (x, ), but also the nontangential limit (x, t) (x, ). However, the proof is almost the same idea. Also, the condition on f can be weaken to f C() and of moderate decreasing. Proof. By the even weaker hypothesis stated in the remark, f is uniformly continuous. So for every ɛ >, there is δ > such that f(y) f(x ) < ɛ whenever y x < δ. Note that for each x with x x < δ, we have {y : x y < δ } { x 4 y < δ}, and hence [ ] u(x, t) f(x ) = f(y) f(x ) H t (x y) dy ɛh t (x y) dy + f H t (x y) dy. x y < δ 4 x y δ 4 And hence there is there is t ɛ > independent of x such that u(x, t) f(x ) 3ɛ whenever t (, t ɛ ) and x x < δ 4. Note that u(x, t) f L ()(4πt) for all x, t >. So it tends to zero as x + t and x t. For the case x > t, we note that u(x, t) f(y) (4πt) e x y /4t dy = f(y) (4πt) e x y /4t dy + x y > x (4πt) e x /6t x y > x (4πt) e x /6t f L () + f(y) dy + M + x 4 7. x y x x y x f(y) (4πt) e x y /4t dy M + y (4πt) e x y /4t dy

8 If x + t with t ɛ for some ɛ >, then x (since we assume x > t) and u(x, t) (4πɛ) e x /6 + If x + t with t +, then x and u(x, t) (4πt) e /6t + 4M as x. 4 + x 4M 4 + x as x, t +.. Show that the function u(x, t) = x t (4πt) / e x /4t () satisfies the -d heat equation for t >, () lim t u(x, t) = for every x, (3) lim (x,t) (x,) u(x, t) = for every x and (4) u is not continuous at the origin. emark 6. Hence, it can not serve as an example to the non-uniqueness phenomenon of the heat equation in (, T ). An exact counterexample is given in Problem 4. Proof. ()() are trivial. (3) is a consequence of Lopital s rule and standard squeeze theorem. (4) u(x, x ) = 4πx e /4 as x. 3. Prove the following uniqueness theorem for harmonic functions in the strip S = {(x, y) : < y <, < x < }: if u is harmonic in S, continuous on S with u(x, ) = u(x, ) = for all x, and u vanishes at infinity, then u =. Proof. For each ɛ >, one can use maximum modulus principle (Mean-Value theorem is proved in the textbook) to conclude u ɛ on S. So u. We sketch another way which is similar to the proof for Theorem.7. Suppose not, we may assume M = sup u >. Hence u(x, y ) = M for some (x, y ) S. By the uniform continuity of u and mean-value property with ball B r min{ y, y }(x, y ), we have a contradiction that M M δ for some δ > if r is close to. 4. Prove that the periodization of the Fejér kernel F N on the real line (Exercise 9) is equal to the Fejér kernel for periodic functions of period. In other words, F N (x + n) = F N (x), when N is an integer, and where N ( F N (x) = n= N n N ) e πinx = sin (Nπx) N sin (πx). 8

9 Proof. Since F (ξ) = ( ξ )χ [,](ξ) for all >, ξ. The desired result follows from Poisson summation formula. 5. This exercise provides another example of periodization. (a) Apply the Poisson summation formula to the function g in Exercise to obtain (n + α) = π (sin πα) whenever α is real, but not equal to an integer. (b) Let α \ Z. Prove as a consequence that n + α = π tan πα where the series is defined through the symmetric partial sum about α, the largest integer α. (Note that this series is not absolutely convergent, so we need to assign the order of summation here.) emark 7. Other proofs are given in Exercise 3.9 and Book II s Exercise 3.. Proof. (a) Poisson summation formula applied to f = ĝ and x = α implies that sin πα π (n + α) = sin (π(n + α)) π (n + α) = g( n)e πinα =. (b) Assume that < α < first. Note that for n N, n + α + α n + α = (n + x) + ( n + x) dx Note that n (n+x) converges uniformly on (, ) by M-test with sup x (,) (n + x) = n and sup x (,) ( n + x) = (n ) for all n. Therefore α + n + α + n + α = α n= α = ( α n (n + x) dx = α α lim ɛ + ɛ ) π tan πα + α ( + lim ɛ + π π tan πɛ + ɛ sin πx x dx ) = π tan πα. Finally, for arbitrary α, α = α + α, where < α <. So by the definition of this series and previous result n + α = n + α = π tan πα = π tan πα 9

10 6. The Dirichlet kernel on the real line is defined by f(ξ)e πixξ dξ = (f D )(x) so that D (x) = χ [,] (x) = sin(πx). πx Also, the modified Dirichlet kernel for periodic functions of period is defined by D N(x) = n N Show that the result in Exercise 5 gives e πinx + (e πinx + e πinx ). D N (x + n) = D N(x), where N is an integer, and the infinite series must be summed symmetrically. In other words, the periodization of D N show that if + \ N, then where [] is the largest integer. is the modified Dirichlet kernel D N. Also D (x + n) = D [] (x), emark 8. Corresponding to Problem 3., one can define modified conjugate Dirichlet kernel D N similarly,that is, D N(x) = n N sign(n)e πinx + ( e πinx + e πinx ). Proof. Note that we can t apply Poisson summation formula since D is not in L () (so the inversion formula breaks down). So we go back to the symmetric partial sum: L k= L D (x + k) = L k= L = m [] + e πi[]x e πiξ(x+k) dξ = e πiξm +[] L e πiξx k= L e πiξk dξ = e πiξx D L (ξ) dξ [] f x ( ξ)d L (ξ) dξ + e πi[]x f x ( ξ)d L (ξ) dξ f x ( ξ)d L (ξ) dξ where f x (s) := e πiξs. So the Tauberian theorem for Cesaro or Abel sum (see Exercise.4 and Problem.3) of Fourier series of f x (s), f x (s)χ (,δ)(s) and f x(s)χ ( δ, )(s) (δ ) implies that as L, δ f x ( ξ)d L (ξ) dξ f() = if δ > and Another case is similar, so we complete the proof. f x ( ξ)d L (ξ) dξ f(+) + f( ) =.

11 7. The gamma function is defined for s > by Γ(s) = e x x s dx. (a) One can easily show that for s > the above integral makes sense, that is, the following two limits exist: lim δ + δ e x x s dx and A lim A e x x s dx. (b) One then can use integration by parts to prove Γ(s + ) = sγ(s) whenever s >, and conclude that () for every integer n we have Γ(n + ) = n!; () Γ( ) = π and Γ( 3 ) = π easily. 8. The zeta function is defined for s > by ζ(s) = n=. Verify the identity n s π s/ Γ(s/)ζ(s) = t s (ϑ(t) ) dt whenever s > where Γ and ϑ are the gamma and theta functions, respectively. ϑ(s) := e πns. More about the zeta function and its relation to the prime number theorem can be found in Book II. Proof. For s > t s (ϑ(t) ) dt = = t s ( π s n s e πnt ) dt = n= n= e πnt t s dt e z z s dz = π s/ Γ(s/)ζ(s), where the interchange of the order of sum and integration is permitted by Monotone convergence theorem (if one use Lebesgue integral) or an careful argument to uniform convergence and improper iemann integral (interchange twice, one is for integral over [, M] and sum, another one is for lim M and sum) 9. The following is a variant of the calculation of ζ(m) = n= /nm found in Problem 4, Chapter 3. (a) Applying the Poisson summation formula to f(x) = t/(π(x +t )) and f(ξ) = e πt ξ where t >, we get π t t + n = e πt n.

12 (b) Prove the following identity valid for < t < : π t t + n = πt + π Also note the following trivial fact (c) Use the fact that e πt n = ( ) m+ ζ(m)t m. m=. e πt z e z = z + B m (m)! zm, where B k are the Bernoulli numbers to deduce from the above formula, m= m+ (π)m ζ(m) = ( ) (m)! B m. Proof. (b) For < t <, t t + n = t n (t/n) + = t ( ) m ( t n n )m = n= n= n= m= = m=( ) m+ t m n = ( ) m+ t m ζ(m), m n= m= n= m= m+ tm ( ) n m where the Fubini s theorem, that is, the interchange of the order of the sums n and m is permitted by the absolute convergence of the double series tm n,m ( )m+ n m proved easily. (c) Note that for z = πt, ( < t < ) B m (π) m t m = πt πt = + πt e πt n (m)! e πt m= = + πt t π t + n = + + t ( ) m+ ζ(m)t m = m= By the uniqueness of power series (see Baby udin s Theorem 8.5), which can be ( ) m+ ζ(m)t m. m= B m (π) m (m)! = ( ) m+ ζ(m).. The following results are relevant in information theory when one tries to recover a signal from its samples.

13 Suppose f is of moderate decrease and that its Fourier transform f is supported in I = [ /, /]. Then, f is entirely determined by its restriction to Z. This means that if g is another function of moderate decrease whose Fourier transform is supported in I and f(n) = g(n) for all n Z, then f = g. More precisely: (a) Prove that the following reconstruction formula holds: sin πy f(x) = f(n)k(x n) where K(y) = πy Note that K(y) = O(/ y ) as y. (b) If λ >, then f(x) = λ f(n λ )K λ(x n cos πy cosπλy ) where K(y) =. λ π (λ )y Thus, if one samples f more often, the series in the reconstruction formula converges faster since K λ (y) = O(/ y ) as y. λ. (c) Prove that f(x) dx = f(n). Note that K λ (y) K(y) as emark 9. This is the well-known Shannon sampling theorem. See the survey article []. Its relation to continuous wavelet transform can be found in [, Chapter ] Proof. (a) Using Poisson summation formula for g(x) = f(x), we have Hence f(x) = f(ξ) = χ [, ](ξ) n Z f(ξ)e πiξx dξ = f(ξ + n) = χ [, ](ξ) f(n)e πi( n)ξ. n Z n Z f(n)e πiξ(x n) dξ = n Z f(n) sin πiξ(x n), π(x n) where the interchange of sum and integration is promised by the uniform convergence with upper bound f(n) M from the moderate decreasing. +n (b) It s the same idea: Using Poisson summation formula with g(x) = f(λx) and the characteristic function χ [, ] there is replaced by the function φ(ξ) describe in the Figure. We omit the brutal computation for λ φ(ξ)e πiξ(x n λ λ ) dξ. (c) By Planchel s theorem and Poisson summation formula, we have f(x) dx = f(ξ) dξ = f(m)e πimξ dξ = f(n)e πi( n)ξ n Z m Z f(n)f(m)e πi(m n)ξ dξ = f(n)f(m) (m,n) Z (m,n) Z e πi(m n)ξ dξ = n Z f(n), 3

14 where the third and fourth equalities are due to the absolute (and hence uniform) convergence of the series with bound f(n)f(m) M (+n )(+m ).. Suppose that f is continuous on. Show that f and f cannot both be compactly supported unless f =. This can be viewed in the same spirit as the uncertainty principle. [Hint: Assume f is supported in [, /]. Expand f in a Fourier series in the interval [, ], and note that as a result, f is a trigonometric polynomial.] Proof. Suppose f is compactly supported in [, ] and f is also compactly supported (say, in [ M, M] for some M >. As hint, f(x) n Z f(n)e πinx = n M f(n)e πinx since f(n) = f(t)e πint dt = f(t)e πint. So the series is an trigonometric polynomial that converges on. The uniqueness of Fourier series implies f equals to this polynomial and hence is not compactly supported unless f.. The heuristic assertion stated before Theorem 4. can be made precise as follows. If F is a function on, then we say that the preponderance of its mass is contained in an interval I (centered at the origin) if I x F (x) dx x F (x) dx () Now suppose f S, and () holds with F = f and I = I ; also with F = f and I = I. Then if L j denotes the length of I j, we have L L π. A similar conclusion holds if the intervals are not necessarily centered at the origin. Proof. We may assume f = and hence f = by Planchel s theorem. So Theorem.4 implies the desired result as follows: ( (L L ) L ) ( L ) = 6 f(x) dx 4 x f(x) dx ξ f(ξ) 4π. f(ξ) dξ 6 x f(x) dx I ξ f(ξ) dξ I 4

15 3. The Heisenberg uncertainty principle can be formulated in terms of the operator L = d dx + x, which acts on Schwartz functions by the formula L(f) = d f dx + x f. This operator, sometimes called the Hermite operator, is the quantum analogue of the harmonic oscillator. Consider the usual inner product on S given by (f, g) = f(x)g(x) dx whenever f, g S. (a) Prove that the Heisenberg uncertainty principle implies This is usually denoted by L I. (Lf, f) (f, f) f S. (b) Consider the operators A and A defined on S by A(f) = df dx + xf and A (f) = df dx + xf. The operators A and A are sometimes called the annihilation and creation operators, respectively. Prove that for all f, g S we have (i) (Af, g) = (f, A g), (ii) (A Af, f) = (Af, Af), (iii) A A = L I. In particular, this again shows that L I. (c) Now for t, let A t (f) = df dx + txf and A t (f) = df dx + txf. Use the fact that (A t A t f, f) to give another proof of the Heisenberg uncertainty principle which says that whenever f(x) dx = then ( )( x f(x) dx df ) dx dx 4. emark. See Problem 7 and [5, Chapter 6]. Proof. (a) By Theorem.4 and arithmetic-geometric mean inequality (Lf, f) = df + x f(x) = 4π x dx f(x) + x f(x) ( 4π x f(x) ) /( ) / x f(x) 4π (f, f). 4π 5

16 (b) (i)(ii) Use the integration by parts, (iii) is obvious. (c) Note that (A t f, A t f) = (A t A t f, f) = t x f(x) dx t f(x) dx + df dx (x) dx. Then we complete the proof through the basic algebra that ( ) ( )( f(x) dx 4 x f(x) dx df ) dx (x). Problems. The equation x u x + ax u x = u t with u(x, ) = f(x) for < x < and t > is a variant of the heat equation which occurs in a number of applications. To solve this equation, make the change of variables x = e y so that < y <. Set U(y, t) = u(e y, t) and F (y) = f(e y ). Then the problem reduces to the equation U + ( a) U y y = U t, with U(y, ) = F (y). This can be solved like the usual heat equation (the case a = ) by taking the Fourier transform in the y variable. One must then compute the integral e( 4π ξ +( a)πiξ)t e πiξv dξ. Show that the solution of the original problem is then given by u(x, t) = (4πt) / e (log(v/x)+( a)t) /(4t) f(v) dv v. Proof. We omit the easy proof for equivalence of these two equation. Taking Fourier transform in y to the equation, we have ( 4π ξ + ( a)πiξ)û(ξ, t) = tû(ξ, t) So Û(ξ, t) = e( 4π ξ +πi( a)ξ)t F (ξ) = where G(y, t) is determined by the inversion formula, that is, G(y, t) = e ( 4π ξ +πi( a)ξ)t e πiξy dξ = e 4π ξ t e πiξ(t( a)+y) dξ = 4πt e (t( a)+y) /(4t). 6

17 Hence U(y, t) = F (s) 4πt e (t( a)+y s)/(4t) ds and then (with s = log v and y = log x) u(x, t) = (4πt) / e (log(v/x)+( a)t) /(4t) f(v) dv v.. The Black-Scholes equation from finance theory is V t + rs V s + σ s V rv =, < t < T, s subject to the final boundary condition V (s, T ) = F (s). An appropriate change of variables reduces this to the equation in Problem. Alternatively, the substitution V (s, t) = e ax+bτ U(x, τ) where x = log s, τ = σ (T t), a = r σ, and b = ( + r σ ) reduces the equation to the one-dimensional heat equation with the initial condition U(x, ) = e ax F (e x ). Thus a solution to the Black-Scholes equation is V (s, t) = r(t t) e πσ (T t) e (log(s/s )+(r σ /)(T t)) σ (T t) F (s ) ds. 3. The Dirichlet problem in a strip. Consider the equation u = in the horizontal strip {(x, y) : < y < ; < x < } with boundary conditions u(x, ) = f (x) and u(x, ) = f (x), where f and f are both in the Schwartz space. (a) Show (formally) that if u is a solution to this problem, then û(ξ, y) = A(ξ)e πξy + B(ξ)e πξy. Express A and B in terms of f and f, and show that û(ξ, y) = sinh(π( y)ξ) sinh(πξ) f(ξ) + sinh(πyξ) sinh(πξ) f (ξ). (b) Prove as a result that and u(x, y) f (x) dx as y u(x, y) f (x) dx as y. (c) If Φ(ξ) = (sinh πaξ)/(sinh πξ), with a <, then Φ is the Fourier transform of ϕ where ϕ(x) = sin πa 7 cosh πx + cos πa.

18 This can be shown, for instance, by using contour integration and the residue formula from complex analysis (see Book II, Chapter 3). (d) Use this result to express u in terms of Poisson-like integrals involving f and f as follows: u(x, y) = sin πy ( f (x t) cosh πt cos πy dt + f (x t) ) cosh πt + cos πy dt. (e) Finally, one can check that the function u(x, y) defined by the above expression is harmonic in the strip, and converges uniformly to f (x) as y, and to f (x) as y. Moreover, one sees that u(x, y) vanishes at infinity, that is, lim x u(x, y) =, uniformly in y. Proof. 4. (Tychonoff s counterexample) For a >, consider the function defined by if t g(t) = e t a if t > () (a) Show that there exists < θ < depending on a so that for t >, g (k) (t) k! (θt) k e (b) Apply (a) to show that for each x and t >, the series u(x, t) := n= t a g (n) (t) xn (n)! converges. Moreover, u solves the heat equation, vanishes for t =, continuous up to the initial t = and satisfies the estimate u(x, t) c e c x a/(a ) for some constants c, c >. This example shows that the growth condition u(x, t) Ae a x we meet in the class is almost optimal, since for each ɛ >, we can find a > such that the growth exponent a/(a ) = + /(a ) = + ɛ. (Also see Exercise and Problem 6.) Proof. 8

19 4 Prove that the function u(x, t) = [ ] e xy cos(xy + ty ) + e xy cos(xy ty ) ye y 4 3 cos y 4 3 dy is another non-trivial solution of the Cauchy Problem in + with vanishing initial data. This was observed by osenbloom and Widder []. I saw it in [3, page 8]. Proof. 5. (Weak Maximum Principle for Heat Equation) Theorem. Suppose that u(x, t) is a real-valued solution of the heat equation in the upper half-plane, which is continuous on its closure. Let denote the rectangle = {(x, y) : a x b, t c} and be the part of the boundary of which consists of the two vertical sides and its base on the line t =. Then min u(x, t) = min u(x, t) and max (x,t) (x,t) u(x, t) = max u(x, t). (x,t) (x,t) The steps leading to a proof of this result are outlined below. (a) Show that it suffices to prove that if u on, then u in. (b) For ɛ >, let v(x, t) = u(x, t) + ɛt. Then, v has a minimum on, say at (x, t ). Show that x = a or b, or else t =. To do so, suppose on the contrary that a < x < b and < t c, and prove that v xx (x, t ) v t (x, t ) ɛ. However, show also that the left-hand side must be non-negative. (c) Deduce from (b) that u(x, t) ɛ(t t) for any (x, t) and let ɛ. Proof. 6. The examples in Problem 4 are optimal in the sense of the following uniqueness theorem due to Tychonoff. Theorem. Suppose u(x, t) satisfies the following conditions: (i) it is continuous on d [, T ] and solves the heat equation in d (, T ) with zero initial condition. (ii) u(x, t) Me a x for some M, a, and all x d, t < T. Then u. 9

20 Proof. One can find some discussions and proof in [3, Section 5.4], [4, Section.3] or [7, Section 7.(b)]. 6. There is another well-known uniqueness theorem for heat equation in d + due to David Widder: Let T > and S T := d (, T ]. Suppose u C, (S T ) := {u t, u xi,x j C(S T )} is a nonnegative solution to heat equation in S T and u(, t) in L loc (d ) as t +. Then u in S T. emark 3. I learned the proof from [3, Section 5.4] which is suitable for arbitrary dimension in contrast to Widder s original proof [7, Section 7.(d)]. Also note that Tychonoff s function defined in Problem 4 is sign-changing. Proof. Widder s theorem is proved if one prove that (x, t ) S T First, there is a more general result: d u(y, s)h t (x y, t s) dy u(x, t ) s (, t ). Theorem 4. Let u be a solution for heat equation and satisfies all the regularity hypothesis and initial data describe above. If u satisfies then u in S T. d u(y, s) H t (x y, t s) dy <. We give a proof for this theorem in the end. Now for fixed (x, t ) S T and s (, t ) Lemma 5. u v in S T. Proof of Lemma 5. proof of Theorem (Combined with Problem 4. in Book III) The Hermite function h k (x) are defined by the generating identity h k (x) tk k! = e (x / xt+t )

21 (a) Show that an alternate definition is given by the odrigues formula ( h k (x) = ( ) k e x / d ) ke x. dx They also satisfy the creation and annihilation identities (x d dx )h k(x) = h k+ and (x + d dx )h k(x) = h k (x) for k whereh =. Conclude from the above expression that each h k is of the form P k (x)e x /, where P k is a polynomial of degree k. In particular, the Hermite functions belong to the Schwartz space and h (x) = e x /, h (x) = xe x /. By (d), one can also show the linear span of P,, P m is the set of all polynomials of degree m. (Folland [6, Exercise 8.3(f)]). (b) Prove that the family {h k } is complete in L, that is if f L (), and for all k, then f. (f, h k ) = f(x)h k (x) =, (An equivalent formulation of this fact is that the set of Hermite polynomials {P k } forms an orthogonal basis for L (, e x )). (c) Define h k (x) = h k((π) / x). Then ĥ k (ξ) = ( i)k h k(x). Therefore, each h k is an eigenfunction for the Fourier transform. (d) Show that h k is an eigenfunction for the operator L = d /dx + x, and in fact, prove that Lh k = (k + )h k In particular, we conclude that the functions h k are mutually orthogonal in L (). (f) Show that [h k(x)] dx = π / k k!. (g) Suppose that K(x, y) = H k (x)h k (y) λ k, and also F (x) = T (f)(x) = K(x, y)f(y) dy. Then T is a symmetric Hilbert-Schmidt operator, and if f a kh k, then F a k λ k H k. (h) One can show on the basis of (a) and (b) that whenever f L (), not only is F L (), but also x F (x) L (). Moreover, F can be corrected on a set of

22 measure zero, so is C, F is absolutely continuous, and F L (). Finally, the operator T is the inverse of L in the sense that for every f L (), LT (f) = LF = F + x F = f. Proof. (a) Note that e (x / xt+t ) = e x / e (t x) and one can prove by induction that ( d ) ke z [ ](x) = P k (x)e x dz where p k is a polynomial of degree k which is even if k is even and is odd if k is odd. From the definition, we know for each x, ( h k (x) = e x / d ) ke (t x) t= = e x / P k ( x)e x = ( ) k e x / P k (x)e x dt ( d ) ke = ( ) k e x / z [ ](x). dz The identities can be proved by this formula easily, we omit it. To show P,, P m spans the set of polynomials of degree m, it suffices to check P,, P m are linearly independent. This can be proved by the orthogonality of h k easily, we omit it. (Another easy proof is by noting that each P k is of exact degree k, cf. Folland [5, Lemma 6.].) (b) There are two equivalent ways (Folland and Stein) to prove this result, their principal idea both rely on the Fourier transform. (We present both of them for completeness). (Method I) Given t, note that K (xt) k converges pointwisely to e xt as K, k! K (xt) k K xt k k! e xt for all K N and f(x)e x / e xt L k! x(). Hence by LDCT and (a) f(x)e x /+xt dx = = lim K x / f(x)e lim K K The desired result follows from Exercise 8. K (xt) k k! dx = lim f(x)e x / K K f(x)e x / c k (t)p k (x) dx = lim K K (xt) k dx k! c k (t) f(x)h k (x) dx = (Method II) Follow the hint of Folland [6, Exercise 8.3(g)], it is enough to show ĝ, where g(x) = f(x)e x/. Given ξ, note that K ( πixξ) k converges pointwisely to e πixξ as k! K, K ( πixξ) k K πixξ k k! e π xξ for all K N and f(x)e x / e π xξ L k! x().

23 Hence by LDCT and (a), ĝ(ξ) = = lim K = lim K g(x)e πixξ dx = f(x)e x / K x / f(x)e lim K ( πixξ) k k! c k (ξ) f(x)h k (x) dx =. K K ( πixξ) k dx k! K c k (ξ) f(x)e x / P k (x) dx dx = lim K 8. To refine the results in Chapter 4, and to prove that f α (x) = nα e πin x n= is nowhere dierentiable even in the case α =, we need to consider a variant of the delayed means N, which in turn will be analyzed by the Poisson summation formula. (A simplified proof can be found in Jones [, Section 6.H]). (a) Fix an indefinitely differentiable function Φ satisfying Φ on B () and vanishes outside B (). By the Fourier inversion formula, there exists ϕ S so that ϕ(ξ) = Φ(ξ). Let ϕ N (x) = Nϕ(Nx) so that ϕ N (ξ) = Φ(ξ/N). Finally, set N (x) = ϕ N (x + n). Observe by the Poisson summation formula that N (x) = Φ(n/N)eπinx, thus N is a trigonometric polynomial of degree N, with terms whose coefficients are when n N. Let N (f) = f N Note that S N (f α ) = N (f α ) where N is the largest integer of the form k with N N. (b) If we set N (x) = ϕ N (x) + E N (x) where E N (x) = ϕ N (x + n), n then one sees that: (i) sup x / E N (x) as N. (ii) N (x) cn. 3

24 (iii) N (x) c/(n x 3 ) for x /. Moreover, x / N (x) dx = and x x / N (x) dx as N. (c) The above estimates imply that if f (x ) exists, then (f N)(x + h N ) f (x ) as N, whenever h N C/N. Then, conclude that both the real and imaginary parts of f are nowhere differentiable, as in the proof given in Section 3, Chapter 4. (d) Let a >, b <. Also prove the Weierstrauss function W (x) = n= bn cos(a n x) is nowhere differentiable if a b. emark 6. One can also check another characterization in [, Section 6.H] that is different from the methods of delay means. Proof. (a) (b) (c) (d) emark 7. For the differentiablity of iemann function n= [9, Chapter 3]. sin πn x, see Jarnicki and Pflug n eferences [] Butzer, P. L., G. Schmeisser, and. L. Stens: An introduction to sampling analysis. Nonuniform sampling. Springer, Boston, MA,. 7-. [] Daubechies, Ingrid: Ten lectures on wavelets. Vol. 6. SIAM, 99. [3] DiBenedetto, E: Partial Differential Equations. nd ed., Springer,. [4] Evans, L. C.: Partial Differential Equations, nd Edition. AMS, Providence, I,. [5] Folland, Gerald B: Fourier analysis and its applications. Vol. 4. American Mathematical Soc., 99. [6] Folland, Gerald B: eal analysis: modern techniques and their applications. nd ed., John Wiley and Sons,

25 [7] John, Fritz: Partial Differential Equations. 4th ed., Springer, 99. [8] Grafakos, Loukas: Classical Fourier Analysis. 3rd ed., Springer, 4. [9] Jarnicki, Marek, and Peter Pflug: Continuous Nowhere Differentiable Functions. Springer Monographs in Mathematics, New York (5). [] Jones, Frank: Lebesgue Integration on Euclidean Space. Jones & Bartlett Learning,. [] Linares, Felipe, and Gustavo Ponce: Introduction to nonlinear dispersive equations. nd ed., Springer, 5. [] osenbloom, P. C., and Widder, D.: A temperature function which vanishes initially. The American Mathematical Monthly 65.8 (958): [3] Stein, Elias M., and Guido L. Weiss: Introduction to Fourier Analysis on Euclidean Spaces. Vol.. Princeton University Press, 97. 5