Fourier transform, assorted stuff...

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1 Fourier transform, assorted stuff... M. Carlsson October 9, An example from control theory To get an idea of where real applications are, lets begin with an example from control theory. Example 1.1. Suppose we measure a signal u of time t, i.e. u : R Ñ R. A filter is a devise F that transforms u into another signal v with the following properties: (i) Causality: For any T P R we have supp u Ă rt, 8q ùñ supp F puq Ă rt, 8q, i.e. if you do nothing, nothing happens. (ii) Time-invariance: F pup T qq Dup T q, i.e. if you delay a signal by time T, the response is unchanged but also delayed by T. Let us derive a formula for F. Let δ 0 be the "delta-distribution" at zero. Let h be the impulse response h F δ 0. Given times tt n u 8 n 8 Ă R and amplitudes tu n u 8 n 8 Ă R we consider the "function" u ÿ u n δ tn ÿ u n δ 0 p t n q. npz npz Using piq and piiq we get (1.1) F puq F p ÿ npz u n δ tn q ÿ npz u n hp t n q Taking an intuitive limit of the above expression, we get that for any "nice function" u one has ż (1.2) F puq uptqhp tqdt u h How to study/understand/predict properties of the above operation? R With this cliffhanger we leave the example for a while to review basic properties of the Fourier transform. 2 The Fourier transform Given u P L 1 prq we define the Fourier transform F as the function (2.1) pupξq Fpuqpξq? 1 ż uptqe itξ dt, where ξ P R. R 1

2 F is a continuous operator from L 1 prq onto CpRq. Moreover, by Plancherel s identity (Parseval s formula for F) we have (2.2) }u} 2 }Fu} 2, for all u P L 1 prq X L 2 prq. In other words, F is an isometry with respect to the L 2 -norm. In particular, it is continuous and we can extend F by continuity to an isometric operator from L 2 prq into L 2 prq. It turns out that F is also surjective. Surjective isometric operators are called unitary. An operator T is unitary if and only if it is invertible and T 1 T, where T denotes the adjoint. For F, one has (2.3) quptq F 1 puqptq? 1 ż upξqe itξ dξ. Note that q pu u. Exercise 2.1. Show that (2.4) z u h? pu p h Exercise 2.2. Prove (some or all) of the above statements concerning F. Example 1.1 continued. Thanks to (2.4), the complicated and frequently occuring operation (1.2) transforms into the simplest possible operation, namely multiplication; (2.5) u h? F 1 ppu p hq. This is one (of many) reasons why the Fourier transform belongs to top 10 of the most important mathematical inventions. Due to Example 1.1 piq, we have supp h Ă R`. Moreover, since we can not measure future events, it is natural to have supp u Ă R. To better understand (2.5), we need to study the structure of the Fourier transform of functions with support on a semi-axis. This is the topic of Hardy spaces! R 3 Hardy spaces As u 0 on R` we have (3.1) pupξq 1? ż 0 8 uptqe itξ dt, ξ P R. Set ζ ξ ` iη, ξ, η P R and define the function (3.2) ũpζq? 1 ż 0 8 uptqe itζ dt? 1 ż 0 8 uptqe itξ e tη dt Fpue η q, which exists for all ζ P C` tξ ` iη : η ą 0u. Exercise 3.1. Set L 2 R prq tu P L 2 prq : supp C`, (written ũ P HolpC`q), for all u P L 2 R prq. u Ă R u. Show that ũ is an analytic function in Two questions immediately arise; What is the relationship between ũ P HolpC`q and pu P L 2 prq? What properties of ũ signify this or that concerning (1.2)? 2

3 To answer these questions we have to study the Hardy space H 2 pc`q Ă HolpC`q. If instead we work with sampled signals, i.e. instead of uptq we have the sequence (3.3) pu n q 8 n 0 pup nt 0 qq 8 n 0 where t 0 is some small fixed sampling-interval, then we have to study the Hardy space H 2 pdq Ă HolpDq, where D tz P C : z ă 1u, (3.4) H 2 pdq tfpzq Another question to be investigated is: 8ÿ a k z k : pa k q 8 k 0 P l 2 pnqu. k 0 What is the relationship between H 2 pdq and H 2 pc`q? 4 Hankel operators To introduce the second topic of the course we return again to Example 1.1. Example 1.1 continued. Recall the relationship (1.2): F puq u h. We want to understand how the future signal is affected by past events, i.e. setting v F puq tą0 we want to understand the map u ÞÑ v. Lets discretize u according to (3.3) and v, h according to (4.1) pv n q 8 n 0 pvpnt 0 qq 8 n 0; ph n q 8 n 0 phpnt 0 qq 8 n 0 If we replace the integral (eq12) with a Riemann sum we obtain (4.2) v n ż 8 0 up tqhpnt 0 ` tq dt 8ÿ m 0 h m`n u m The matrix taking pu n q into pv n q is easily seen to have the following structure (Exercise!): h 0 h 1 h 2. h 1 h (4.3) Γ phn q. h Such operators Γ phn q : l 2 pnq Ñ l 2 pnq are called Hankel operators, and ph n q is called the symbol. The use of the symbol Γ for Hankel operators is widespread, and one might wonder why not H became the standard choice. I believe this has the following slightly amusing explanation: the Russians have been very influential in the modern development of Hankel operator theory, and in the Cyrillic alphabet, H does not exist and is usually replaced by G, or rather Γ. As anyone who has ever been to a Mc Donald s in Russia knows, they sell Gamburgers. Ok, to be fair to the Russians one should admit that most peoples have gotten the H wrong. The english knows how to pronounce it but call the symbol something else, whereas neither the french nor the spanish seems to have any clue what to do with the letter. Of course, Swedes got it all right; We call it H and use it as a H. 5 Further exercises Let C k prq denote the set of k times differentiable functions on R. We also define the following important subsets; C0 k prq those that vanish at 8, Cc k prq those that have compact support and Cb k prq those which are bounded. 3

4 Exercise 5.1. Approximate identities: A sequence pϕ k q 8 k 1 Ă C8 prq such that ϕ k p q ě 0 ş ϕ k 1 lim kñ8 ş ϵ ϵ ϕ k 1 for all ϵ ą 0 is called an approximate identity. Show that lim 8 k 1 ϕ k δ 0 as a distribution, i.e. show that ż fpxqϕ k pxq dx δ 0 pfq lim kñ8 for any f P C 8 c, (where δ 0 pfq is just a really silly way of writing fp0q). Also show that x ϕ k Ñ 1? uniformly on compacts. Exercise 5.2. Continuity of F. Show that F : L 1 prq Ñ L 8 prq is continuous. Exercise 5.3. Calculation-rules for F. We use x, y, t for independent variables in the "timedomain" or "space-domain", and ξ, ζ,... for the "Fourier-domain". (These terms is convenient engineering slang. From a mathematical point of view, there is of course no difference between R on one side or the other of the Fourier-transform). Given u P L 1 prq, show that piq Fpupx x 0 qqpξq e ix 0ξ pupξq piiq Fpe ixξ0 uq pupξ ξ 0 q piiiq Fpupaxqq 1 a pup ξ a q pivq Fpu 1 q iξpu for all u P C 1 c pvq d dξ pu ixu, whenever ş p1 ` x q upxq dx ă 8, pviq If u P C 2 c prq then pu P L 1 prq. Exercise 5.4. Calculate F for the characteristic function of an interval χ ra,bs. Then show that FpL 1 q Ă C 0 prq (this is known as the Riemann-Lebesgue lemma). Note that pu R L 1 can happen, which is a source of headache... Exercise 5.5. Smoothness of a function and rapid decay of its Fourier transform go hand in hand. For example, show that if u P C 8 c prq, then ξ n pupξq is bounded for every n P N. Exercise 5.6. Set φpxq e x2 {2. Show that piq ş φ? piiq φ is an eigenvector for F with eigenvalue 1 piiiq Fpe ax2 q 1? 2a e x2 {4a pivq Set φ n nφpn q. Show that pφ n q 8 n 1 is an approximate identity. (Which can be very useful for getting around the headache mentioned in Exercise 5.4). Exercise 5.7. Show that if u P L 1 then ûpξq lim ηñ0` ũpξ ` iζq, where η Ñ 0` signifies η Ñ 0 and η ą 0. (See Section 3 for definition of ũ) Unfortunately, things are not as simple in L 2, as we shall see. 4

5 Hints to exercises Hint 2.2 First do Exercise 5.1 to 5.6. Then show F is given by (2.3) for u P L 1 prq X L 2 prq. Recall xf, gy 2 ş fḡ for C-valued functions. F Fpuq FF puq u for all u P Cc 8 prq. It suffices to show F Fpuq u, since FF only differs by a silly minus. Note that the second integral is well defined due to Exercise 5.3 pviq. However, the problem is that we can not use Fubini s theorem in the following integral: F Fpuqpxq 1 ż ż e ipx tqξ uptqdt dξ One way to deal with this is to insert the functions xφ n pξq (recall Exercise 5.1 and see Exercise 5.6) and take a limit. A slick alternative is to use Exercise 5.3 and 5.6 piiq to show that ˆ ˆ F F a ˆ a φ φ b b for all a P R and b ą 0. This shows that F F is the identity on the linear span of such functions. But by basic integration theory, this linear span is dense in L 2 prq, and hence the conclusion can be lifted to L 2 via continuity arguments, see below. F extends by continuity to a unitary operator on L 2 prq By now we have that F Fpuq u holds for a dense subset of L 2 prq. Thus F is isometric on that set, and hence in particular it is bounded. By basic functional analysis we can now define F on all of L 2 prq by continuity! (It is important to realize that the formula (2.1) no longer is valid.) Several of the things you have already shown now extends by continuity, e.g. the formula F F FF I. Hint 3.1 e it0ζ is clearly analytic since it is defined by a uniformly convergent power series. First assume that u P CpRq has compact support in R. Approximate the integral (3.2) with a Riemannsum and note that the resulting sequence of analytic functions converge uniformly on compacts, and hence the statement is true for such u s. Then pick a sequence pu j q 8 j 1 Ă CpRq converging to a given u P L 2 R prq, and prove that pru j q 8 j 1 converges uniformly on compacts in C` to ru. Alternatively, check that the d dz -derivative of ũ is zero by applying differentiation inside the integral, which can be justified by basic integration theory. Hint 5.3 pvq. First use piiq to show that it suffices to prove the identity at 0. The most novel strategy is of course to show by "bare hands" that ϵ 1 puppϵq pup0qq converges to the desired limit, using Taylor series of sinpxq and cospxq, basic estimates and some real analysis. Alternatively one can use the dominated convergence theorem and the estimate e iϵx 1 ă 2 minp x, ϵ 1 q. A more lazy approach is to apply some theorem stating that interchanging of integration and differentiation is allowed... pviq. by pivq and Exercise 5.2 deduce that ξ 2 pupξq is bounded. Hint 5.4 Use Theorem?? and Exercise 5.2. Hint 5.5 Combine 5.3 pivq with the Riemann-Lebesgue lemma. Hint 5.6 piq calculate ş e px2`y 2 q{2 via Fubini s theorem and via polar coordinates. piiq note that φ satisfies the differential equation (5.1) d φ ` xφ 0. dx 5

6 Moreover, so does pφ and by ordinary differential equations we conclude that pφ cφ for some c P R. To determine c 1, evaluate pφp0q using piq. Alternatively, put z x ` iy and note that φpzq is an analytic function. Moreover φ goes so rapidly to zero that that ˆφpzq is also analytic (exercise). The point is that we can calculate explicitly pφpiyq and conclude that pφpiyq φpiyq, y P R. Since analytic functions have discrete zeroes, we deduce that pφ φ in all of C, (which in particular contains the real line). In more detail, we have Fpe x2 {2 qpiηq? 1 ż e x2 {2 e ixpiηq dx 1 ż? e px ηq2 {2`η 2 {2 dx 1?? e η 2 {2 piiiq use Exercise 5.3 piiiq. 6