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ACM 105: Applied Real and Functional Analysis. Solutions to Homework # 2. Andy Greenberg, Alexei Novikov. Problem 1. Riemann-Lebesgue Theorem.
ACM 105: Applied Real and Functional Analysis. Solutions to Homework # 2. Andy Greenberg, Alexei Novikov Problem 1. Riemann-Lebesgue Theorem. Theorem (G.F.B. Riemann, H.L. Lebesgue). If f is an integrable function on (−∞, +∞), then +∞ Z

lim

f (x) cos

→0

x

−∞

dx = 0



Proof. We first note that if the interval of integration were finite [a, b] with f (x) ≡ 1 on it, then Zb

f (x) cos

x

a



dx =

Zb

cos

a

x



dx = 

Zb/

cos(y)dy = O() → 0

a/

as  → 0, since the integral of the cosine is bounded. We shall reduce the general case to this simple case. First off, due to the integrability of f on (−∞, +∞) and the following estimates, +∞ Z Zb x  x  f (x) cos dx − f (x) cos dx ≤   a

−∞

+∞ +∞ Za Z Za Z  x   x  ≤ |f |(x) dx + |f |(x) dx f (x) cos  dx + f (x) cos  dx ≤ −∞

−∞

b

b

we can always find a finite interval [a, b], such that +∞ Z Zb x  x  ε f (x) cos dx − f (x) cos dx <   3 a

−∞

(by bounding the integrals along the “tails”). Furthermore, since f is integrable, there exists a sequence of integrable simple functions fn (x), such that Zb

f (x) dx = lim

n→∞

a

Zb

fn (x) dx

a

implying that for N large enough, b Z Zb fN (x) dx − f (x) dx < ε 3 a

a

1

We thus have b Z Zb x  x  Zb  x  f (x) cos dx − dx f (x) cos ≤ |f (x) − f (x)| · cos dx ≤ N N    a

a

a



Zb

|f (x) − fN (x)|dx
0, and a step function g(x), such that g(x) = fN (x) on A with b Z Zb fN (x) dx − g(x) dx < ε 3 a

a

But g(x) is a step function, so there exists a finite number of points xj , j = 0, . . . , m with a = x0 < x1 < . . . < xm = b, such that g(x) ≡ const = cj on [xj−1 , xj ]. Therefore Zb a

g(x) cos

x



dx =

x m Zj X

cj cos

j=1 xj−1

x



dx = 

m  X

j=1

 x  xj cj sin 

→0

xj−1

as  → 0. Since the above manipulations guarantee that +∞ Z Zb  x  x dx − g(x) cos dx < ε f (x) cos   a

−∞

for an arbitrarily small ε, we are done.

Problem 2. Lax Equivalence Theorem: Solution Operator. For the initial value problem   du(t)

dt

= Lu(t), 0 ≤ t ≤ T

(1)

 u(0) = u , 0

and the norm k · kV in the space V 3 u(t), we denote S(t) the solution operator: if u(t) is the solution of (1), then u(t) = S(t)u0 , u0 ∈ V0 . We have



1

lim (u(t + ∆t) − u(t) − Lu(t))

=0 ∆t→0 ∆t

(2)

a. From the continuous dependence property we have







sup S(t)(u10 − u20 ) ≤ C0 (u10 − u20 )

0≤t≤T

V

V

and sup kS(t)u0 kV ≤ C0 ku0 kV

0≤t≤T

∀ u0 ∈ V0

This implies that kS(t)k ≤ C0 , so S(t) is bounded on V0 . Since V0 is dense in V , the required extension of S(t) on the whole space V exists and is unique by the extension theorem for operators. 2

b. (continuity of generalized solutions). Choose a sequence {u0,n } ⊆ V0 which converges to u0 in V : ku0,n − u0 kV → 0 as n → ∞ Fix t0 ∈ [0, T ] and let t ∈ [0, T ]. We write u(t) − u(t0 ) = S(t)u0 − S(t0 )u0 = S(t)(u0 − u0,n ) + (S(t) − S(t0 ))u0,n − S(t0 )(u0 − u0,n ) In estimating the norm of this difference, we use the fact that kS(t)k ≤ C0 in the first and the last terms of the right-hand side: ku(t) − u(t0 )kV ≤ 2C0 ku0,n − u0 kV + k(S(t) − S(t0 ))u0,n kV Given an ε > 0 we choose a sufficiently large n, such that 2C0 ku0,n − u0 kV
0, such that for |t − t0 | < δ, ε k(S(t) − S(t0 ))u0,n kV < 2 Then for t ∈ [0, T ] with |t − t0 | < δ,we have ku(t) − u(t0 )kV < ε, so that u(t) is continuous. c. (semi-group property). The solution to the problem (1) is u(t) = S(t)u0 . We have u(s) = = S(s)u0 and S(t)u(s) is the solution of the differential equation on [s, T ] with the initial condition u(s) imposed at time s. By the uniqueness of solution, S(t)u(s) = u(t + s) i.e., S(t)S(s)u0 = S(t + s)u0 , and therefore S(t + s) = S(t)S(s), as required.

Problem 3. Lax Equivalence Theorem: Proof. Recall the definitions. Definition 1. A difference method is a one-parameter family of operators C(∆t) : V → V , and there exists a ∆t0 > 0, such that ∀ ∆t ∈ (0, ∆t0 ] kC(∆t)k ≤ c (C is uniformly bounded). The approximate solution is u∆t (m∆t) = C(∆t)m , m = 1, 2, . . . Definition 2. A difference method is consistent if ∃ Vc ⊆ V , a dense subspace of V , such that ∀ u0 ∈ Vc for the corresponding solution of the initial value problem (1), we have ku(t + ∆t) − C(∆t)u(t)k =0 ∆t→0 ∆t lim

uniformly on [0, T ]. Definition 3. A difference method is convergent if for any fixed t ∈ [0, T ] and any u0 ∈ V we have lim k(C(∆ti )mi − S(t))u0 k = 0 ∆ti →0

3

where the two sequences: {mi }, of integers and {∆ti }, of step sizes are such that lim mi ∆ti = t. i→∞

Definition 4. A difference method is stable if the operators

{C(∆t)m 0 < ∆t ≤ ∆t0 , m∆t ≤ T } are uniformly bounded, i.e., there exists a constant M0 > 0 such that k C(∆t)m kV →V ≤ M0

∀ m : m∆t ≤ T, ∀ ∆t ≤ ∆t0

Theorem (P.D. Lax). Suppose the initial value problem (1) is well-posed. Then for a consistent difference method, stability is equivalent to cnvergence. Proof. (Stability =⇒ Convergence). Consider the error C(∆t)m u0 − u(t) =

m−1 X

C(∆t)j [C(∆t)u((m − 1 − j)∆t) − u((m − j)∆t)] + u(m∆t) − u(t)

j=1

We use the hint: first consider u0 ∈ Vc . Then since the method is stable,

C(∆t)u(t) − u(t + ∆t)

+ ku(m∆t) − u(t)k

kC(∆t) u0 − u(t)k ≤ M0 m∆t sup

∆t m

t

The sup goes to 0 by consistency, while continuity implies ku(m∆t) − u(t)k → 0, so convergence is established in this case. Now consider the general case when u ∈ V and not necessarily in Vc . Since Vc is dense in V , we have a sequence {u0,n } ⊆ V0 such that u0,n → u0 in V . Writing (recall the construction in Problem 2) C(∆t)m u0 − u(t) = C(∆t)m (u0 − u0,n ) + [C(∆t)m − S(t)]u0,n − S(t)(u0 − u0,n ), we obtain, by triangle inequality, kC(∆t)m u0 − u(t)k ≤ kC(∆t)m (u0 − u0,n )k + k[C(∆t)m − S(t)]u0,n k + kS(t)(u0 − u0,n )k Since the initial value problem (1) is well-posed and the method is stable, kC(∆t)m u0 − u(t)k ≤ cku0 − u0,n k + k[C(∆t)m − S(t)]u0,n k Given any ε > 0, we can find a sufficiently large n, so that cku0 − u0,n k
0 ∃ N = N (ε) such that for any n, m > N we have kvn (x) − vm (x)k = max |vn (x) − vm (x)| < ε 0≤x≤π

By the uniform convergence criterion, the condition above implies that the Cauchy sequence converges uniformly on [0, π] to a function v(x). But a uniform limit of continuous functions is continuous, so v ∈ C[0, π]. By continuity, since vn (0) = vn (π) = 0 ∀ n, it follows that also v(0) = v(π) = 0, so v ∈ V , and V is a complete linear normed space, i.e., a Banach space. Now take V0 , the space of linear combinations of sines. Obviously any v ∈ V0 is continuous and vanishes at the boundaries (0 and π). Therefore, V0 ⊂ V . A well-known result from Fourier analysis says that any continuous function on [0, π], vanishing at the boundaries, can be represented as a Fourier sine series, ∞ X

v(x) =

ak sin(kx)

k=1

Therefore for any function v ∈ V we can find a sequence {vn } ⊂ V0 , vn (x) =

n X

ak sin(kx)

k=1

so that vn → v. Thus the closure of V0 , V¯0 ⊃ V , so V0 is dense in V . 5

b. Define C(∆t, r)v(x) = (1 − 2r)v(x) + r(v(x + ∆x) + v(x − ∆x)) p

Here ht = ∆t, hx = ∆x = ∆t/r. If x ± ∆x ∈ / [0, π], then we take the odd 2π-periodic extension of v. Obviously, C(∆t) : V → V is a linear operator. Moreover, kC(∆t)vkV = k(1 − 2r)v(x) + r(v(x + ∆x) + v(x − ∆x))kV ≤ ≤ k(1 − 2r)vkV + krvkV + krvkV = (|1 − 2r| + 2r)kvkV so kC(∆t)k ≤ |1 − 2r| + 2r,

(4)

and the family {C(∆t)} is uniformly bounded. For consistency analysis, take Vc = V0 . For any initial condition u0 (x) ∈ Vc we then obtain a smooth solution as a Fourier sine series u(t, x) =

n X

2

bk e−k t sin(kx) +

Pn

k=1 bk

Fk (t) sin(kx)

k=1

k=1

Here u0 (x) = Fk looks like

n X

sin(kx) and Fk (t) are due to the source term. (It can be shown that Fk (t) =

Zt

fk (τ )ek

2 (τ −t)



0

where f (t, x) = mainder to write

Pn

k=1 fk (t) sin(kx).)

Now use Taylor series expansion in (x, t) with re-

C(∆t)u(x, t) − u(x, t + ∆t) = (1 − 2r)u(x, t) + r(u(x + ∆x, t) + u(x − ∆x, t)) − −u(x, t + ∆t) = (1 − 2r)u(x, t) + r(2u(x, t) + uxx (x, t)(∆x)2 ) + r + (∆x)4 (uxxxx (x + θ1 ∆x, t) + uxxxx (x − θ2 ∆x, t)) − 4! h 1 (∆x)2 −u(x, t) − ut (x, t)∆t − utt (x, t + θ3 ∆t)(∆t)2 = ∆t −ut (x, t) + r uxx (x, t) + 2 ∆t i 1 (∆x)2 (∆x)2 +r (uxxxx (x + θ1 ∆x, t) + uxxxx (x − θ2 ∆x, t)) − utt (x, t + θ3 ∆t)∆t ∆t 24 2 Since ∆x = ∆t/r, (r/∆t)(∆x)2 = 1, and the first two terms combine to f (x, t)∆t. For consistency, the remaining terms must go to zero faster than ∆t. We have p

r

1 (∆x)2 (∆x)2 (uxxxx (x + θ1 ∆x, t) + uxxxx (x − θ2 ∆x, t)) − utt (x, t + θ3 ∆t)∆t = ∆t 24 2 h ∆x ∆x i 1 = ∆t (uxxxx (x + θ1 ∆x, t) + uxxxx (x − θ2 ∆x, t)) − utt (x, t + θ3 ∆t) ∆t 24 2

Now it is indeed clear that the |hx /ht | = |∆x/∆t| ≤ c is suffiecient to bound the term in the square brackets, so that

1



∆t (C(∆t)u(x, t) − u(x, t + ∆t) − f (x, t)) ≤ c˜∆t → 0

uniformly, i.e., the difference method based on C(∆t) is consistent. 6

c. We prove that r ≤ 1/2 is the necessary and sufficient condition for stability and then invoke Lax’s theorem to get the same result for convergence. We have established earlier (4) that k C(∆t)k ≤ |1−2r|+2r. It immediately follows that if r ≤ 1/2, then k C(∆t)k ≤ 1, so that k C(∆t)m k ≤ 1, i.e., uniformly bounded, so the method is stable. For the other direction (necessity), we prove that for r > 0 (4) is actually an equality, i.e., that max k C(∆t)vkV = = (|1 − 2r| + 2r)kvmax kV is achieved. We just need to present the function vmax at which this maximum is attained. For r ≥ 1/2, the maximum is obviously attained on vmax ≡ 1. For 0 < r < 1/2, let xj = (j/π)∆x be the spatial discretization. Define a piecewise linear function by vmax (xj ) =



0, j = 0 or j = n (−1)j , otherwise

with straight line segments connecting the vertices. (This is no other than the familiar “sawtooth” shape.) Obviously, the supremum norm of this function is 1, and since its values on the lattice are alternating +1’s and -1’s, the norm kC(∆t)vmax = |(1 − 2r)(±1) + +2r(∓1)| = 4r − 1 = (|1 − 2r| + 2r)kvkV . Thus the inequality in (4) is in fact an equality. It is now easy to establish the required result. If the method is stable, then C(∆t)m is uniformly bounded. Obviously, if k C(∆t)k > 1, then k C(∆t)m k → ∞, so the uniform bound can exist only if k C(∆t)k ≤ 1, i.e., |1 − 2r| + 2r ≤ 1 ⇔ r ≤ 1/2. Thus r ≤ 1/2 is necessary and sufficient for stability of the difference method. Since the method is consistent, by the Lax equivalence theorem it is also necessary and sufficient for convergence.

Problem 5. Geometric Series Theorem. a. Geometric Series Theorem. Let V be a Banach space and L : V → V with kLk < 1. Then the operator I − L is a bijection on V , its inverse is a bounded operator and k(I − L)−1 k ≤

1 . 1 − kLk

(5)

Proof. Define the following sequence of operators on V , Mn = p ≥ 1,

n+p

X

kMn+p − Mn k =

Since kLk < 1, we have

k=n+1



Lk



n+p X

kLk k ≤

k=n+1

kMn+p − Mn k ≤

n+p X

Pn

k k=0 L ,

n ≥ 0. For

kLkk

k=n+1

kLkn+1 1 − kLk

so sup kMn+p − Mn k → 0 as n → ∞, and {Mn } is a Cauchy sequence in L(V ), the space of p≥1

linear operators on V . But this space is complete, so there exists an operator M : V → V , such that kMn − M k → 0 as n → ∞. Now, (I − L)Mn = Mn − LMn =

n X

k=0

7

k

L −

n X

k=0

Lk+1 = I − Ln+1

Similarly, Mn (I − L) = I − Ln+1 , so (I − L)Mn = Mn (I − L) = I − Ln+1 . Letting n → ∞, we get (I − L)M = M (I − L) = I proving that I − L is a bijection and M = (I − L)−1 = lim

n→∞

Furthermore, kMn k ≤

n X

kLkk =

k=0

n X

Lk =

∞ X

Ln

n=0

k=0

1 1 − kLkn+1 ≤ 1 − kLk 1 − kLk

Taking the limit as n → ∞, we obtain kM k = k(I − L)−1 k ≤

1 1 − kLk

proving (5) and concluding the proof of the geometric series theorem. We can pause for a second here to recognize that what we have just obtained is just another generalization of the geometric series formula: if 0 < q < 1, then the sum of the geometric series 1 1 + q + q2 + . . . + qn + . . . = 1−q We also know from complex analysis that if z ∈ C is also inside the unit disc (i.e., |z| < 1), then the series ∞ X

zn

n=0

converges uniformly to the function (1 − z)−1 , which is bounded on the unit disc. Now we also know that the same holds for linear operators acting on a Banach space! b. The linear integral equation of the second kind, λu(x) −

Zb

k(x, y)u(y) dy = f (x)

(6)

a

with λ 6= 0, k(x, y) ∈ C([a, b] × [a, b]) and f ∈ C[a, b], can be written in the form (λI − K)u = f , where K is the integral operator generated by the kernel k(·, ·). Equivalently, 1 1 (I − L)u = f, L = K λ λ By the geometric series theorem, if kLk =

1 kKk < 1 |λ|

then the inverse operator (I − L)−1 exists and k(I − L)−1 k ≤

8

1 1 − kLk

Since kKk = max

a≤x≤b

Zb

|k(x, y)| dy

a

we thus have that, as long as kKk < |λ|, (λI − K)−1 exists and k(λI − K)−1 k ≤

1 |λ| − kKk

Therefore, if only max

a≤x≤b

Zb

|k(x, y)| dy < |λ|

a

for any f ∈ C[a, b], there is a unique solution u ∈ C[a, b] to the integral equation of the second kind (6).

Problem 6. Different Banach Spaces with the Same Norm. a. Consider the space l∞ of bounded real sequences with the k·k∞ (i.e., supremum) norm. All axioms of the norm are trivially verified. Also, linear combinations preserve boundedness, so l∞ is a linear normed space. Consider a Cauchy sequence {xn } ⊂ l∞ : ∀ ε > 0 ∃N = N (ε) such that ∀ n, m > N sup |xn,k − xm,k | < ε k

By the uniform convergence criterion xn → x as n → ∞, so for n sufficiently large, sup |xn,k − xk | < ε k

so supk |xk | ≤ supk |xn,k | + ε for an arbitrary ε > 0, so x is also bounded, proving that l∞ is complete, and therefore a Banach space. b. Following the same line of arguments as above, we establish that c is a normed linear space, and if {xn } ⊂ c is a Cauchy sequence, then xn → x with sup |xn,k − xk | < k

ε 2

for an arbitrary ε > 0 and all n > N1 (ε). Since {xn } ⊂ c, it consists of converging sequences, so for all k > N2 (ε), we have |xn,k −C| < ε/2. Therefore, for all k > max{N1 , N2 }, we have |xk − C| ≤ ε so x ∈ c and c is a Banach space. The exact same argument applies to c0 . c. First consider c0 , the space of sequences converging to 0, with the supremum norm. We show that the space of bounded linear functionals on c0 is isomorphic to l1 , the space of P absolutely summable sequences (i.e., those with ∞ k=1 |xk | < ∞), with the norm kf k =

∞ X

k=1

9

|fk |

To prove this, first note that for any element f = (f1 , f2 , . . . , fn , . . .) ∈ l1 , the formula f˜(x) =

∞ X

xk fk

(7)

k=1

defines a functional f˜ on the space c0 , which is clearly linear. Moreover, |f˜(x)| ≤ kxk

∞ X

|fk |

k=1

so kf˜k ≤ kf k

(8)

Consider the following vectors in c0 : e1 = (1, 0, . . . , 0, . . .), e2 = (0, 1, . . . , 0, . . .), etc., and let n X fk ek x(n) = |f | k k=1 (if fk = 0, then we set fk /|fk | = 0). Then x(n) ∈ c0 and kx(n) k ≤ 1. Moreover, f˜(x(n) ) =

n n ∞ X X X fk ˜ f (ek ) = |fk | → |fk | = kf k

k=1

|fk |

k=1

k=1

as n → ∞. This implies kf k ≤ kf˜k lim kx(n) k ≤ kf˜k. Together with (8), the opposite n inequality established earlier, this implies kf k = kf˜k. Therefore the mapping carrying f into f˜ is a norm-preserving mapping from the space of linear functionals on c0 to l1 . It remains to prove that every functional has a unique representation like (7), where f ∈ l1 . Let x = (x1 , . . . , xk , . . .) ∈ c0 . Then x=

∞ X

xk ek

k=1

where the series on the right converges in c0 to the element x since

n

X

xk = sup |xk | → 0

x −

k>n k=1

as n → ∞. Since the functional f˜ is continuous, f˜(x) = f˜

lim

n→∞

n X

xk ek

!

= lim

n→∞

k=1

n X

f˜(xk ek ) =

k=1

∞ X

xk f˜(ek )

k=1

so f˜ has a unique representation of the form (7). We only need to establish that ∞ X

|f˜(ek )| < ∞

(9)

k=1

For any n, let (n)

x ˜

=

n X f˜(ek )

k=1

10

|f˜(ek )|

ek

Then x ˜(n) ∈ c0 and k˜ x(n) k ≤ 1, so n X

|f˜(ek )| =

k=1

n X f˜(ek )

f˜(ek ) = f˜(˜ x(n) ) ≤ kf˜k < ∞

˜ k=1 |f (ek )|

Since n can be made arbitrarily large, this proves (9). Thus the space of bounded linear functionals on c0 is isomorphic to l1 : ∀ x ∈ c0 f (x) =

∞ X

fk xk

k=1

with fk = f (ek ) (ek defined above) and

P

|fk | < ∞.

Now consider the more general space c of all bounded convergent sequences. The obvious idea is to notice that for any x ∈ c, converging to x0 , x ˜ = x − x0 ∈ c0 , so we should be able to use the result for functionals on c0 for the case of c. We make this idea precise in the following way. Consider the vectors e1 = (1, 0, 0, . . . , 0, . . .), e2 = (0, 1, 0, . . . , 0, . . .), etc., for all k ≥ 1 and also e0 = (1, 1, 1, . . .). Then for {en }n≥0 ⊂ c, and for any linear functional f on c we can define fn = f (en ), n = 0, 1, . . . We use these values to represent all bounded linear functionals on c. We first prove that the set x(n) =

n X

λi ei ,

λi ∈ Q

i=0

is a dense subset of c. Indeed, if a = (a1 , a2 , . . . , ak , . . .) ∈ c, then ∃ a0 : lim ak = a0 < ∞. k→∞

Then for any ε > 0 we can find a sufficiently large n, so that n X ε (ai − a0 )ei < a − a0 e0 − 2 i=0

since the sequence (a − a0 )k → 0 as k → ∞. Now, since ai ∈ R, there exist rational numbers λi ∈ Q, such that |λi − (ai − a0 )| < ε/2i+1 , so that n n X X ε λi ei − (ai − a0 )ei < 2 i=1 i=1

and therefore (n)

|a − x

n X | = a − a0 e0 − λi ei < ε i=1

so the x(n) ’s are indeed dense in c. Moreover, since λi ∈ Q, {x(n) } is a countable dense subset of c. For any x ∈ {x(n) } and any linear functional f we can represent f (x) = f0 λ0 +

n X

λi fi

i=1

where fi = f (ei ). This functional will be bounded as long as |fi | < ∞. Since x(n) is dense in c, then by the Hahn-Banach theorem we can extend the linear functional f to a P

11

bounded linear functional on the whole space c. Since this extension is unique, it has to look like ∞ f (x) = f (x1 , x2 , . . . , xk , . . .) = f0 x0 +

X

fi xi

i=1

where, as above, fi = f (ei ). In other words, we have proved that c = c0 ⊕ R, so that the space of bounded linear functionals on c is isomorphic to l1 ⊕ R, just as we had expected.

References [1] Royden H.L., Real Analysis, 3rd ed., Prentice-Hall, 1988. [2] Kolmogorov A.N., Fomin S.V., Introductory Real Analysis, Dover, 1975. [3] Atkinson K., Han W., Theoretical Numerical Analysis: a Functional Analysis Framework, Springer, 2001. [4] Zorich V.A., Mathematical Analysis (Calculus), 2nd ed., MCNMO, 1998 (in Russian).

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