Katznelson Problems. Prakash Balachandran Duke University. June 19, 2009

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1 Katznelson Problems Prakash Balachandran Duke University June 9, 9 Chapter. Compute the Fourier coefficients of the following functions (defined by their values on [ π, π)): f(t) { t < t π (t) g(t) { t t < t π < t < t < t h(t) t π < t < π What relation do you see between f and? What about g and? Solution: ˆ (n) π (t)e int dt π ˆf(n) π f(t)e int dt e int sin n dt π π n. ( t )e int dt [ sin n n ( ) sin n π n ˆf(n). t e int dt n πn sin ĝ(n) π g(t)e int dt [ e int dt π e int dt i sin n πn in ˆ (n). ĥ(n) π h(t)e int dt π π π te int dt i( )n. n

2 . Show that the Fourier series of a function f L ( ) is formally eual to A + (A n cos nt + B n sin nt) where A n ˆf(n) + ˆf( n) and B n i( ˆf(n) ˆf( n)). Euivalently: A n f(t) cos ntdt π n B n π f(t) sin ntdt. Show also that if f is real valued, then A n and B n are all real; if f is even, then B n for all n; and if f is odd, then A n for all n. Solution: B n ˆf() i [ S[f + n ˆf(n)e int ˆf() + ( ˆf(n)e int + ˆf( n)e int ) n ( ˆf(n) + ˆf( n)) eint + e int + i( ˆf(n) ˆf( n)) eint e int i n A n [ A + (A n cos nt + B n sin nt). f(t)(e int e int )dt n Obviously if f is real, then A n and B n are all real. f(t)(e int + e int )dt f(t) cos ntdt π i f(t) i sin ntdt π f(t) sin ntdt. If f is even, then f sin nt is odd for all n. aking the fundamental domain to be [ π, π, this implies that B n for all n. If f is odd, then f cos nt is odd for all n, so that again taking the fundamental domain to be [ π, π implies that A n for all n. 3. Show that if S a j cos jt then S a j sin jt Proof: Given S j a j cos jt ( j a j j > b. By definition: S n isgn(n)b n e int e ijt +e ijt ) ib n e int + ib n e int n n b je ijt where b j a j, b j a j for ( ) e int e int a j i j a j sin jt. j

3 4. Let f L ( ) and let P (t) N n N a ne int. Compute the Fourier coefficients for fp. Solution: Let g(t) f(t)p (t). hen, ĝ(n) g(t)e int dt f(t)p (t)dt 5. Let f L ( ) and let m be a positive integer. Write N m N a n f (m) (t) f(mt). f(t)e i(n m)t dt N m N a n ˆf(n m). Show that f (m) (n) { ˆf ( n m) if m n otherwise Proof: First, assume that m n. Notice that the function f (m) (t) has period. Using this fact and the m substitution u mt: f (m) (n) f (m) (t)e int dt f(mt)e int dt m ˆf ( n m). m f(mt)e int dt f(u)e i n m u du Now suppose that m n. Let p(t) N n N a ne int be a trigonometric polynomial. hen: since m n. p (m) (n) N k N a k e i(mk n)t dt Now, since f L ( ), choose a trigonometric polynomial p(t) N j N a je ijt such that f p < ɛ (proven in the next chapter). hen: f (m) (n) [ f(mt) p(mt) dt m f (m) (t)e int dt f(mt)e int dt (f(mt) p(mt))e int dt + p(mt)e int dt m Since ɛ was arbitrary, we have that f (m) (n). f(mt) p(mt) dt f(t) p(t) dt f p < ɛ. 3

4 6. Show that no trigonometric polynomial of degree n > can have more than n zeros on. Note: he trigonometric polynomial cos nt (eint + e int ) is of degree n and has exactly n zeros on. Proof: Let P (t) n j n a je ijt be a trigonometric polynomial of degree n. hen, we can identify this polynomial with P (z) z n n j n a jz n+j z n p(z) on z, where p(z) n j n a jz n+j. Since p(z) is a polynomial of degree n the fundamental theorem of algebra implies that p has at most n zeros in C, and hence at most n zeros on. Since z n is always nonzero on z, we have that P (z) can have no more than n zeros on. 7. Denote by C the multiplicative group of complex numbers different from zero. Denote by the subgroup of all z C such that z. Prove that if G is a subgroup of C which is compact, then G. Proof: Let z G C. hen, z re iθ for some r > and θ [, ). Consider the seuence {z j } j. Since G is a subgroup of C, {z j } G. Furthermore, since G is compact, this seuence contains a convergent seuence, so that z jm z as m. Since G is a compact subset of C it is closed, so that z G. Now, z jm r jm e ijmθ. If r < then z jm which is certainly not in G. If r > then this seuence diverges, so that we must have r. hus, z e iθ. Since z was arbitrary, we must have G. 8. Let G be a compact proper subgroup of. Prove that G is finite and determine its structure. Proof: Suppose that G is infinite. hen, since G is compact, there exists a convergent seuence {g n} in G. Since G is closed, g n g G. hus, g n g g n e in G. Identifying with [, ) we may WLOG assume g n as n. Now, let t. Given ɛ > there exists N such that g N < ɛ. Consider the grid of [, ) formed by {m g N } M m where M is the largest integer such that M g N <. If (m )g N t < mg N for m {,..., M}, then t (m )g N < ɛ so that since G a subgroup implies (m )g N G, we can get ɛ close to t by elements in G. If M g N t < (M + ) g N then t M g N < ɛ. hus, we can always get within ɛ of t by elements of G, so that there exists {f n } in G such that f n t. Since G is closed, this implies t G. Since t was arbitrary, G, contradicting that G was proper. Any such group G must obviously generated by finitely many elements of the form p for p, N. Remark: his does not hold for n, as can be exhibited by the compact proper subgroup [,. p 9. Let G be an infinite subgroup of. Prove that G is dense in. Proof: G so that one can easily verify that Ḡ. hus, Ḡ is a compact subgroup of with Ḡ. If it were proper, then (8) would imply that G is finite, which is a contradiction. hus, Ḡ so that G is dense in.. Let α be an irrational multiple of. Prove that {nα(mod)} n Z is dense in. Proof: Let G α {nα(mod)} n Z hen, G is an infinite subgroup of, so that the density statement follows by (9). 4

5 . Prove that a continuous homomorphism of into C is necessarily given by an exponential function. Proof: Let φ be the continuous ( ) homomorphism. ( ) Since is compact φ( ) C is compact, so that by (7) ( ) ( ) φ( ). Now, φ φ so that φ is a -th root of unity; that is φ e i. Now, let G be the subgroup generated by rational multiples of. hen, G is an infinite subgroup of so that (9) implies that it is dense in. If r is a an element in, then there exists n r such that n are rational multiples of. Continuity of φ implies that φ(r) lim n φ( n ) lim n e in e ir so that φ(t) e it for all t.. If E is a subset of and τ, define E + τ {t + τ : t E}. Say that E is invariant under translation by τ if E E + τ. Show that, given a set E, the set of τ such that E is invariant under translation by τ is a subgroup of. Hence, prove that if E is a measurable set on and E is invariant under translation by infinitely many τ then E or its complement has measure zero. Proof: Given E measurable, let Γ E {τ : E + τ E}. he subgroup statement is trivial. Let t Γ E. hen, {E ( ɛ, ɛ)} + t E (t ɛ, t + ɛ) µ{e ( ɛ, ɛ)} µ{{e + ( ɛ, ɛ)} + t} µ{e (t ɛ, t + ɛ)}. ( ) Now, if Γ E then Γ E is an infinite subgroup of so that it is dense in by (9). So, given t there exists t n Γ E such that t n t. he bounded convergence theorem guarantees that µ{e (t n ɛ, t n + ɛ)} µ{e (t ɛ, t + ɛ)} so that ( ) implies for all t. hus, for all t. µ{e ( ɛ, ɛ)} µ{e ( ɛ, ɛ)} lim ɛ µ{e (t ɛ, t + ɛ)} lim ɛ µ{e (t ɛ, t + ɛ)} So, if t is a point of density of E, then every point in is a point of density of E. he relation µ{e (t ɛ, t + ɛ)} + µ{e c (t ɛ, t + ɛ)} µ{(t ɛ, t + ɛ)} then implies that no point of is a point of density of E c so that µ{e c }. Switching the roles of E and E c, the result follows. 3. If E and F are subsets of, write E + F {t + τ : t E, τ F } and call E + F the algebraic sum of E and F. We define similarly the sum of any finite number of sets. A set E is called a basis for if there exists an integer N such that E + E + + E (N times) is. Prove that every set E of positive measure on is a basis. Proof: If E contains an interval, then the assertion follows. o see this, suppose I E is an interval, say I (a, b). hen, E + E contains the interval (a, b), and so E + E + + E (N times) contains the interval (Na, Nb) which has length N(b a). So, for N large enough, < N(b a), so that the interval 5

6 (Na, Nb) contains a (closed) interval of form [nπ, (n + )π. Since this is a translated interval of [,, we have that ( N a, N b) (modulo ) covers. hus, for this N, E + E + + E. Now, let E be a set of positive measure. We claim that E + E contains an interval. Since almost every point in E is a point of density of E, by translating E if necessary (which preserves Lebesgue measure), we may assume that E is a point of density. Suppose ɛ >, y E c B(, ɛ) such that y x E c, x E. Let < ɛ <. hen, there exists y E c B(, ɛ ) such that y x E c for all x E. So: ( ɛ) µ(e B(, ɛ ɛ )) (ɛ ɛ ) ɛ ɛ ɛ µ(e B(, ɛ ɛ )) (ɛ ɛ ) µ(e B(, ɛ ɛ )) µ(e B(, ɛ y )) µ(y E B(, ɛ y )) µ(ec B(, ɛ)). Since < ɛ < was arbitrary, and is a point of density of E, taking ɛ we have that so that lim ɛ ( ɛ) µ(e B(, ɛ ɛ )) (ɛ ɛ ) lim ɛ µ(e c B(, ɛ)) µ(e c B(, ɛ)) lim ɛ implying that is a point of density of E c contradicting that is a point of density of E. So, ɛ > such that y E c B(, ɛ) y x E for some x E. For this ɛ, we thus have E c B(, ɛ) E +E; since clearly E B(, ɛ) E +E we have B(, ɛ) E +E. By repeating the first part of the proof, we thus have that E + E + + E N-times. 4. Show that a measurable proper subgroups of have measure zero. Proof: Suppose otherwise. hen, there exists G < that is a measurable proper subgroup with positive measure. (3) implies that G is a basis of so that there exists an integer N such that G + G + + G (N times) is. But, since G is a subgroup, G + G + + G (N times) is exactly G. hus, G contradicting that G is proper. 5. Show that measurable homomorphisms of into C map it into. 6. Let f be a measurable homomorphism from into. Show that for all values of n except possibly one value, ˆf(n). Proof: Fix t. Using the substitution x y+t and the hypothesis that f is a measurable homomorphism: ˆf(n) f(x)e inx dx f(y + t)e in(y+t) dy f(t)e int f(y)e iny dy f(t)e int ˆf(n) ˆf(n)( f(t)e int ). Since t was arbitrary, if there exists t for which f(t)e int then since n was arbitrary, we must have ˆf(n) for all n. If every t satisfies f(t)e int f(t) e int, then obviously ˆf(m) for all values of m except m n in which case ˆf(n). 6