Real Analysis Chapter 8 Solutions Jonathan Conder. = lim n. = lim. f(z) dz dy. m(b r (y)) B r(y) f(z + x y) dz. B r(y) τ y x f(z) f(z) dz

Transcription

1 3. (a Note that η ( (t e /t for all t (,, where is a polynomial of degree. Given k N, suppose that η (k (t P k (/te /t for all t (,, where P k (x is some polynomial of degree (k. By the product rule and the chain rule, η (k (t P k (/t( t e /t + P k (/te /t t P k (/te /t for all t (,, where P k (x : x (P k (x P k (x is a polynomial of degree + (k k. It follows by induction that this result holds for all k N {}. (b By definition η ( (. Given k N, suppose that η (k (. Note that (n + e (n+ lim n ne n lim + n n e e <, so that n ne n converges by the ratio test. This implies that lim n ne n, and hence η (k (t η (k ( P k (te /t lim lim t t t t e /t P k ( lim t t The left-hand derivative of η (k at is clearly, so η (k (. By induction, η (k ( for all k N. 4. This question is surprisingly subtle, and one of the few instances where it is important to keep track of null sets explicitly. It can be shown that A r f is uniformly continuous for arbitrary f L and r (,, but the δ constructed depends on r, which will cause issues later.. Instead, let ε (, and take δ (, such that τ y f f < ε for all y B δ(. If r (, and x, y R n satisfy x y < δ, then A r f(x A r f(y f(z dz f(z dz m(b r (x B r(x m(b r (y B r(y f(z + x y dz f(z dz m(b r (y B r(y m(b r (y B r(y τ y x f(z f(z dz m(b r (y B r(y τ y x f f < ε. In particular, A r f is uniformly continuous. If x R n and r, s (, δ, then the above inequality gives A r f(x A s f(x A rf(x A s f(y dy + A s f(y dy A s f(x m(b r (x B r(x m(b r (x B r(x (f(y A s f(y dy m(b r (x B r(x + A s f(y A s f(x dy m(b r (x B r(x ( < f(y f(z dz dy m(b r (x B r(x m(b s (y B s(y + ε m(b r (x B r(x dy f(y f(z dz dy + ε m(b r (xm(b s (y B r(x B s(y f(y τ z f(y dz dy + ε m(b r (xm(b s (y B r(x B s( f(y τ z f(y dy dz + ε m(b r (xm(b s (y B s( B r(x

2 < ε. τ z f f dy dz + ε m(b r (xm(b s (y B s( B r(x Therefore (A /n f n is uniformly Cauchy, so it converges uniformly to a function g which is uniformly continuous (by a standard argument. Theorem 3.8 implies that f g almost everywhere. 7. Choose R (, so that g is supported on B R (. The partial derivatives of g (up to order k are also supported on B R (, and hence they are bounded. If x R n and y B (x then f g(y f(zg(y z dz (fχ BR+ (x g(y, so (f g B (x C k (B (x by Proposition 8.. It follows that f g C k (R n. 8. If x R n, e j is the j th standard basis vector in R n and t R is non-zero then (f g(x + te j (f g(x ( j f g t t (f(x + te j y f(x yg(y dy ( j f(x yg(y dy (t ((τ tej f(x y f(x y ( j f(x yg(y dy t ((τ tej f(x y f(x y ( j f(x y g(y dy ( ( t ((τ tej f(x y f(x y ( j f(x y p dy /p g q t ((τ tej f(y f(y ( j f(y p dy /p g q t (τ tej f f j f p g q by Hölder s inequality, so by the squeeze theorem j (f g ( j f g. 9. uppose that f is absolutely continuous on every bounded interval and that f L p. If t, x R and t then (t (τ t f f f (x t ((τ t f(x f(x f (x t (f(x + t f(x f (x and hence, by Minkowski s inequality for integrals t (τ t f f f p x+t t f (y dy f (x t [t,] [,t] t x t (f (x + y f (x dy [t,] [,t] ( [t,] [,t] t (f (x + y f (x dy /p t (f (x + y f (x dx p dy ( /p (τ y f (x f (x dx p dy

3 t [t,] [,t] τ y f f p dy. Given ε (,, since f L p there exists δ (, such that τ y f f p < ε for all y ( δ, δ. In particular, if t < δ then [t,] [,t] τ yf f p dy [t,] [,t] ε dy ε t, in which case t (τ t f f f p ε. This shows that f is the L p derivative of f. Conversely, suppose that f has an L p derivative h. Define φ C c (X by φ(x : ( x χ [,] (x. Clearly φ 4( + x and φ, so f φ /n n converges to f pointwise on L f. If a, b L f and a < b then h L p ([a, b] L ([a, b], and the previous exercise implies that (f φ /n n h φ /n n. This sequence converges to h in L ([a, b]. Moreover f φ /n is absolutely continuous for all n N, because (f φ /n (x (h φ /n (x h(x yφ /n (y dy /n /n for all x [a, b] (so its derivative is bounded on [a, b]. It follows that f(x f(a lim n ((f φ /n(x (f φ /n (a lim n h(x y nφ(ny dy n h L [a /n,b+/n] x a h φ /n for all x L f [a, b]. By redefining f on a null set, it follows that f AC([a, b] and f h almost everywhere on [a, b]. Given an interval with endpoints not in L f, there is a larger interval whose endpoints do lie in L f, so f is absolutely continuous, with f h, on every bounded interval. This implies that f h almost everywhere, in which case f L p.. Let x R n and t (,. et E : B t ( and for each k N define E k : B k t( \ B k t(. Then f(x yφ t (y dy f(x y t n φ(t y dy E E t n f(x y C( + t y n ε dy E C m(b ( f(x y dy m(b t ( E C m(b ( f(y dy m(b t (x B t(x C m(b ( Hf(x. Moreover, if k N then f(x yφ t (y dy f(x y t n φ(t y dy E k E k t n f(x y C( + t y n ε dy E k C t n E k f(x y ( + k n ε dy C kn m(b ( (k (n+ε m(b k t( C m(b ( (k ε n m(b k t(x 3 B k t ( B k t (x f(x y dy f(y dy x a h

4 It follows that f φ t (x f(x yφ t (y dy C m(b ( C m(b ( (k ε n Hf(x. ( + k n (k ε Hf(x C m(b ( ( + n ε Hf(x and hence M φ f C m(b (( + n Hf. ε 3. (a Clearly f( ( x dx. If κ Z \ {} then we integrate by parts: e πiκx f(κ ( x (b By Plancherel s theorem k k 4π e πiκx dx ( x πiκ f(k π f(κ π f π κ Z k x a e πiκx + πiκ dx + 4πiκ + e πiκx 4π κ 5. (a Let a (,, and note that χ [ a,a] L. If ξ R \ {} then χ [ a,a] (ξ e πiξx χ [ a,a] (x dx e πiξx xa πiξ eπiξa e πiξa πiξ ( x dx π 3 ia sin(πξa πiξa ( 3 x πiκ. π 6. a sinc(ξa, and it is clear that χ [ a,a] ( χ [ a,a] a a sinc(a. It follows that χ [ a,a] (ξ χ [ a,a]( ξ a sinc( ξa a sinc(ξa for all ξ R. (b ince the Fourier transform is linear, it is clear that H a is a subspace of L. If f n n is a sequence in H a which converges to f L, then f n n converges to f in L (because the Fourier transform is unitary, and some subsequence of f n n converges to f pointwise almost everywhere (because it converges in measure. In particular f(ξ for almost all ξ R \ [ a, a], which implies that f H a. This shows that H a is a closed subspace of L, so it is a Hilbert space. Define φ : (a / χ [ a,a] and f : φ. If ψ then F ψ is the reflection of ( ψ ψ about. Therefore F agrees with the reflection operator on a dense subset of L, so the two operators are equal and f is the reflection of φ about. This shows that f φ, so f H a. For each k Z define f k : τ k/a f, so that f k (ξ e πiξk/a f(ξ for all ξ R, and in particular f k H a. If j, k Z then f j, f k f j, f k e πiξj/a πiξk/a f(ξe f(ξ dξ e πiξ(k j/a φ(ξ dξ a e πiξ(k j/a dξ, a a and if j k it follows that f j, f k πi(k j (eπi(k j e πi(j k. Moreover, f k, f k a a a. This shows that {f k } is an orthonormal set in H a. If g H a and g f k for all k Z, then a ĝ(ξe πiξk/a dξ a ĝ(ξe πiξk/a φ(ξ dξ a ĝ, f k a g, f k a for all k Z. This implies that ĝ [ a,a] M, where M is the closed span of the collection of functions of the form ξ e πiξk/a (where k Z. But M L ([ a, a] by the tone-weierstraß theorem and the fact that the inclusion C([ a, a] L ([ a, a] is a bounded linear map with dense range (this is essentially Theorem 8.. Therefore ĝ [ a,a] almost everywhere, so ĝ almost everywhere and hence g almost everywhere. This shows that {f k } is an orthonormal basis for H a. 4

5 (c Given f H a, the series f, f k f k f, f k f k converges to f in H a (thus in L. If k Z then a f, f k f(xe πixk/a e πixk/a φ(x dx f(x dx e πixk/a F f(x f( k/a dx f(k/a. a a a a a Let Z Z. By Hölder s inequality f, f k f k (ξ f, f k f k (ξ f, f k a sinc(aξ k for all ξ R. Parseval s identity shows that the first sum f, f k can be made small by choosing Z appropriately, so the series f, f k f k will be uniformly Cauchy provided that the second sum is uniformly bounded. It suffices to show this for ξ [, a because a sinc(aξ j a sinc(a(ξ + i/a k a sinc(aξ + i k j Z for all ξ R and i Z. Given ξ [, a define Z : Z \ {,, }, so that (since sinc a sin(πaξ πaξ πk a sinc(aξ k 6a + 6a + 6a + 6a + 6a + 6a + 6a + 6a π <. a πaξ πk a (π k πaξ a (π k π a (π k / 8a π k This shows that f(k/a sinc(aξ k is uniformly Cauchy for all ξ R. It is clear that sinc is continuous, and in fact sinc C because sinc(ξ πξ for all ξ R \ {}. Hence, the above series converges uniformly to some g C. A subsequence of its partial sums converges pointwise almost everywhere to f (because the series converges to f in measure. This implies that f g almost everywhere. k k 6. (a If k N and x R then f k (x χ [,] (x yχ [ k,k] (y dy, if x (, k ] [k +,, if x [ k, k ] χ [x,x+] [ k,k] k + x if x [k, k + ] k + + x if x [ k, k]. It clearly follows that f k u. 5

6 (b If k N and ξ R then Therefore f k (ξ f k ( ξ χ [,] ( ξ χ [ k,k] ( ξ sinc( ξ k sinc( ξk f k 4k sin(πξ sin(πkξ. (πξ 4k sin(πξ sin(πkξ 4 sin(πζ/6k sin(πζ/6 48k sin(πζ/6k sin(πζ/6 (πξ dξ (πζ/k dζ π ζ dζ. Note that sin(πζ/6 for all ζ i Z([, 5] + 6i and (provided that k 6 at least m : k of these intervals are contained in [k, 5k]. Indeed, at least one of them begins in [k, k +5], after which there is an available length of 4k 5 6m, which is enough room for m such intervals. Hence, there exists a 6Z such that m i [a i, b i ] [k, 5k], where a i : + a + 6i and b i : a i + 4 for all i {,..., m }. Therefore f k m i bi k m a i π ζ dζ k π which implies that lim k f k. i ( a i b i m i k (b i a i a i b i m i 4k (5k 4m 5 k 5, (c If f L and f then f L and hence f ( f almost everywhere. In particular F L has trivial kernel. Moreover F L is injective (it L(L, C because f(ξ e πiξx f(x dx f for all f L and ξ R, by definition. If F(L C, then F L has an inverse G L(C, L by the open mapping theorem (since L and C are Banach spaces. This contradicts the previous parts, because G Gf k f k u with k 6. k 5 for all k N 7. (a Note that f(x dx e πx x a dx π (π a e πx (πx a dx (π a e t t a dt Γ(a (π a and hence f L. imilarly, if a > then f L because f(x dx e 4πx x a dx (4π (4π a e 4πx (4πx a dx 4π (4π a e t t a dt Γ(a (4π a. (b Given ξ R, define γ : R C by γ(x : π(iξ + x. Note that f(ξ e πiξx f(x dx e π(iξ+x x a π(iξ + dx (π(iξ + a By part (a x e πiξx f(x is in L, so the dominated convergence theorem implies that e π(iξ+x (π(iξ + x a dx. N e π(iξ+x (π(iξ + x a dx lim e π(iξ+x (π(iξ + x a dx N π(iξ + lim e z z a dz. N γ [,N] Given N N, define γ : [, πn] C by γ (x : x and γ : [ : πn] by γ (x : πn + iξx. Clearly γ followed by γ is homotopic to γ [,N], so by Cauchy s theorem e z z a dz γ [,N] e z z a dz + γ e z z a dz. γ 6

7 Note that lim N γ e z z a πn dz lim N e x x a dx Γ(a, whereas e z z a πn dz e (πn+iξx (πn + iξx a iξ dx e πn ξ γ πn πn + iξx a dx. If a then πn + iξx a (πn + ξ x a (πn( + ξ a for all x [, πn]. Otherwise πn + iξx a (4π N + ξ x (a / (πn a for all x [, πn]. In either case, the above integral exhibits polynomial growth as N (which is slower than exponential decay, so lim N e z z a dz lim γ N [,N] e z z a dz + lim γ N e z z a dz Γ(a + Γ(a. γ Combining the above calculations, it follows that f(ξ (π(iξ + a Γ(a. (c Define g : R R in a similar manner to f but with b instead of a. By part (a both f and g are in L, and ( ix a ( + ix b dx (π a+b (π( + ix a (π( + ix b dx (π a+b f(xγ(a ĝ(xγ(b dx (πa+b (ĝ, f Γ(aΓ(b by part (b. ince the Fourier transform is unitary (from L to L, (ĝ, f (g, f f(xg(x dx e πx x a e πx x b dx (e πx x a+b dx h(x dx where h : R R is defined in a similar manner to f but with a+b instead of a. By part (a, it follows that ( ix a ( + ix b dx (πa+b Γ(a + b (4π a+b Γ(aΓ(b a b πγ(a + b. Γ(aΓ(b 9. ince {x R n f(x } has finite measure, f L and hence f(ξ e πiξ x f(x dx for all ξ R n. Given a measurable set E R n, the Minkowski and Hölder inequalities imply that f e πiξ x f(x dx dξ E E e πiξ x f(x χ E (ξ dx dξ ( e πiξ x f(xχ E (ξ dξ dx ( f(x χ E (ξ dξ dx ( m(e f(x dx m(e( f χ f m(m(e. 7

8 . (a Let ξ, ζ R n and suppose that ξ ζ. If ξ then ξ ζ and hence J(ξ J(ζ. Otherwise, set α : ξ, and note that {αξ} and {αζ} can be extended to orthonormal bases for R n. Let R O(n be the change of basis matrix between these bases, and note that R(αξ αζ, implying that Rξ ζ. It follows that J(ζ J(Rξ eix (Rξ dσ(x ei(r x ξ dσ(x det(r eix ξ dσ(x J(ξ. Therefore J(ξ depends only on ξ. By Theorem.49 F (ξ e πiξ x f( x dx e πiξ (rx f( rx r n dσ(x dr e i(πrξ ( x dσ(xf(rr n dr J(πrξf(rr n dr j(πr ξ f(rr n dr, so ( F is radial and g(s j(πrsf(rr n dr for all s [,. (b If k {,..., n} then ξ k e ix ξ ix k e ix ξ and hence e ix ξ x k for all x and ξ R n. Moreover e ix ξ (ix k e ix ξ x k eix ξ, so ξk the dominated convergence theorem implies that ( k J(ξ e ix ξ x k for all x and ξ R n. ince dσ σ( <, e ix ξ dσ(x ξk e ix ξ dσ(x x k eix ξ dσ(x for all ξ R n, in which case n k k J + J. Indeed, if ξ Rn then n ( n ( n ( k J(ξ x k eix ξ dσ(x x k e ix ξ dσ(x x e ix ξ dσ(x J(ξ. k k (c If k {,..., k} and ξ R n is non-zero then and hence This implies that ξ k (ξ ξ / (ξ ξ / (ξ k ξ k ξ, so that ( k J(ξ j( ξ j ( ξ ξ k ξ ( k J(ξ ( j ( ξ ξ k j ( ξ ξ k ξ ξ + j ( ξ ξ ξ k ξ ξ. n k J(ξ + J(ξ j ( ξ k n k j ( ξ + j ( ξ ξk n ξ + ξ j ( ξ ξ j ( ξ k ( n ξ + j( ξ ξ j ( ξ + (n j ( ξ ξ + j( ξ, n k ξ k ξ 3 + j( ξ which is zero by part (b, in which case ξ j ( ξ + (n j ( ξ + ξ j( ξ, for all ξ R n \ {}. Moreover ( J(,..., j ( ( J(,...,, since J(r,,..., j(r J( r,,..., for all r [,, so j ( and hence rj (r + (n j (r + rj(r for all r [,. 8

9 (d Define f : [, C by f(r : rj(r, so that f (r j(r + rj (r and hence f (r j (r + rj (r for all r [,. It follows from part (c that f + f, which is a second order ODE satisfied by the function g : [, R defined by g(x : σ( sin(x. ince f ( j( dσ σ( g ( and f( g(, it follows that f g, in which case j(r r f(r σ(r sin(r 4πr sin(r for all r (,. ince j( σ( 4π, this statement can be interpreted for r by taking limits. 3. (a Define linear operators P, Q on (R by P f(x : f (x and Qf(x : xf(x. If f, g (R then (Qfg xf(xg(x dx f(xxg(x dx f(qg. Moreover, (fg f g + fg, and (by the monotone convergence theorem applied separately to the positive and negative real and imaginary parts of f g, followed by the fundamental theorem of calculus It follows that N (fg lim (fg lim (f(ng(n f( Ng( N. N N N (P fg f g (fg fg f(p g, in which case (T fg (Qf P fg (Qfg (P fg f(qg + f(p g f(qg + P g f(t g. Thus (T fg f(t g. If x R then (P Qf(x (Qf (x f(x + xf (x f(x + (QP f(x, so [T, T ] [Q + P, Q P ] [Q, Q] [Q, P ] + [P, Q] [P, P ] [P, Q] I. This implies that [T, T ] I T. If k N with k > and [T, T k ] (k T k, then T T k T T k T [T, T k ]T + T k T T (k T k + T k [T, T ] + T k T T kt k + T k T, in which case [T, T k ] kt k. By induction, this shows that [T, T k ] kt k. for all k N. (b If k N then T h k (k! / T k+ h k + ((k +! / T k+ h k + h k+, and since (T h (x xh (x + h (x π /4 xe x / + π /4 ( xe x /, part (a implies that T h k (k! / T T k h (k! / ([T, T k ] + T k T h (k! / kt k h kh k. These formulae also hold for k if we let h be an arbitrary function, and it clearly follows that T T h k T ( kh k kt h k k kh k kh k for all k N {}. (c Note that T T + I (Q P (Q + P + I Q + [Q, P ] P + [P, Q] Q P. Hence, if f (R then f(x Q f(x P f(x x f(x f (x for all x R. Moreover if k N {} then h k T T h k + h k kh k + h k (k + h k. (d Note that h h e x π dx π π. If k N and h k, then h k h k h k k (T T h k h k k khk (T h k (T h k k khk h k, 9

10 so by induction h k for all k N {}. If j, k N {} and j > k then (h j, h k h j h k j (T T h j h k j (T h j (T h k j jhj ( kh k so by induction (h j, h k k(k j(j (j k (h j k, h. This shows that {h k } k is orthonormal. (e If k N and k j (h j, h k T k f(x ( k ( k/ e x / dk / dx k (e x f(x, for all x R (which is clearly true when k, the product rule implies that ( T k f(x ( k k/ xe x / dk / dx k (e x f(x xe x / dk / dx k (e x f(x e x / dk / dx k (e x f(x ( k k/ e x / dk dx k (e x / f(x. for all x R. Therefore T k f(x ( k k/ e x / d k dx k (e x / f(x for all x R and k N {}, by induction. If k N {} then h k k / T h k (k! / T k h, so for all x R h k (x (k! / ( k k/ e x / dk dx k (e x / h (x ( k π / k k! ex / dk dx k e x. (f Given k N {}, it is easily shown by induction and the product rule that dk dx k e x P k (xe x for all x R, where P k (x is some polynomial of degree k. The formula for h k from part (e implies that H k (x ( k π / k k! ex P k (xe x ( k π / k k! P k(x, so H k (x is also a polynomial of degree k. In particular H (x is a non-zero constant, so all the constant polynomials are in span{h (x}. If the polynomials of degree less than k are in span{h j (x} j k and c R \ {} is the leading term of H k (x, then x n c H k (x span{h j (x} k j, which implies that xn span{h j (x} k j and hence every polynomial of degree at most k is in span{h j (x} k j. By induction, this shows that span{h j(x} k j is the set of polynomials of degree at most k, for all k N {}. (g Let f L and suppose that f h k for all k N {}. Define g : R C by g(x : f(xe x /, so that g L (by Hölder s inequality. If ξ, x R and N N then N ( πiξx k N g(x k! πξx k f(x e x / e π ξx x / f(x. k! k If ξ R then x e π ξx x / is clearly in L, so x e π ξx x / f(x is in L. Moreover ĝ(ξ e πiξx g(x dx k k ( πiξx k k! g(x dx lim N N k ( πiξx k g(x dx k! by the dominated convergence theorem. If N N then N (πiξx k k k! span C {H k (x} N k, and since H kg fh k for all k N {}, it follows that ĝ(ξ lim N. In particular ĝ L, so by the Fourier inversion theorem g (ĝ almost everywhere. ince e x / > for all x R, this implies that f in L. Therefore {h k } k is an orthonormal basis for L.

11 (h Clearly A is linear and bijective (its inverse is given by A f(x : (π /4 f((π / x. If f L then π f(x Af Af(x dx π dx π f(t dt f, π which shows that A is unitary (by the polarisation identity. Moreover, if ξ R then (assuming f L Âf(ξ e πiξx (π /4 f(x π dx (π /4 e πiξx π f(x π dx (π /4 e πiξt f(t dt, in which case f(ξ A Âf(ξ (π /4 (π /4 e πi(π / ξt f(t dt π e iξt f(t dt. If f then clearly f, and it follows that π T f(ξ e iξt T f(t dt T f(te iξt dt f(t (teiξt + iξe iξt dt f(t(t iξe iξt dt. On the other hand πit ( f(ξ i (ξ e iξt f(t dt d dξ e iξt f(t dt. ince d dξ e iξt f(t ite iξt f(t tf(t for all t R, and t tf(t is in L (as f, it follows that πit ( f(ξ i (ξ i ( ( π T f(ξ. e iξt f(t dt ξe iξt f(t dt + te iξt f(t dt (t iξe iξt f(t dt d dξ e iξt f(t dt ite iξt f(t dt iξe iξt f(t dt In particular T f it ( f. Note that Ah (x (π /4 h ( πx /4 e πx for all x R and hence Âh (ξ /4 e πξ for all ξ R (by Proposition 8.4. It follows that h (ξ A Âh (ξ (π /4 /4 e π(π ξ π /4 e ξ / h (ξ for all ξ R. If k N then h k (k! / T k h, so for all ξ R hk (ξ (k! / T k h (ξ (k! / ( it ( T k h (ξ (k! / ( i k T k h (ξ ( i k h k (ξ. ince {h k } k is an orthonormal basis for L, its unitary image {φ k } k is also an orthonormal basis for L. Moreover, for each k N {} it is clear that φ k Âh k AA Âh k A h k A( i k h k ( i k φ k. 5. (a Let f AC(R be periodic and suppose that f L p (T. If x, y T and x > y then x x f(x f(y f f f p χ [y,x] q f p (x y /q, y y

12 in which case f(x f(y f p x y /q for all x, y R, because f is periodic. Therefore f Λ /q (T. Given α ( q,, choose β ( q, α and define g : [, ] R by g(x : βxβ. Note that g L p ([, ] because (β p > ( q p. Now define f : R R by f(x : x ((τ k gχ [k.k+ (τ k+ gχ [k+,k+, so that, by definition, f is periodic and absolutely continuous on every bounded interval. Moreover f(x f( x α x α x g xβ β x α x β α for all x (,, and lim x x β α, so f / Λ α (T. (b Let α (, and define h : [, ] [, ] by h(t : sin(πt. ince α < the series k k /α converges. Let (, be its sum and for each n N set t n : n k k /α. Now define f : [, ] [, ] by ( χ [tn,t f(t : n+ ](t x tn h. n t n+ t n n If x [, ] then x [t n, t n+ ] for some n N. Given y [t n, t n+ ], the mean value theorem implies that f(x f(y ( ( x n h tn y tn h t n+ t n t n+ t n x t n n y t n t n+ t n t n+ t n x y n t n+ t n x y α t n+ t n α n t n+ t n x y α n t n+ t n α Given y [t n+, t n+ ], set z : t n+ and note that x y α n( n /α α α x y α. f(x f(y /α /α ( f(x f(z /α + f(z f(y /α /α ( x z + z y, so f(x f(y α ( x z + z y α α (z x + y z α α x y α. If y [t n+, ] then α x y α ((y x α ((t n+ t n α n + f(x + f(y n 4 f(x f(y. 4 This shows that f(x f(y 4 α x y α for all x, y [, ], in which case f Λ α (T. However n for all N N, so f is not of bounded variation. N ( f tn + t n+ f(t n N n N n n n

13 3. Define Φ : R n R by Φ(ξ : e π ξ, so that Φ( and Φ Φ L. Note that + ξ dt + ξ n log( + ξ ξ t ξ dt ξ π ξ + n log( + n for all ξ (, n (n,. Moreover n log( + ξ n log( + n π ξ + n log( + n for all ξ [ n, n]. It follows that ( + ξ n Ce π ξ, where C : ( + n n, and hence Φ(ξ Φ(ξ C( + ξ n, for all ξ R. ince L f, Theorem 8.35 implies that lim t f(ξe πt ξ dξ lim t f(ξφ(tξe πiξ dξ f(, in which case, by Fatou s lemma, f f(ξ dξ lim f(ξe πt ξ dξ f( <. t 3