Math 508: Advanced Analysis

Homework 10
Lawrence Tyler Rush
<me@tylerlogic.com>
November 21, 2014
http://coursework.tylerlogic.com/courses/upenn/math508/homework10

1

Let L : S → T be a linear map from vector spaces S to T. Let V 1 and V 2 be distinct solutions of the equation LX = Y 1. Futhermore let W be a solution to the equation LX = Y 2.

(a) Find a solution to LX = 2Y 1 - 7Y 2

Put V = 2V 1 - 7W. Then
LV = 2LV1 - 7LW = 2Y1 - 7Y2

so that V is a solution to LX = 2Y 1 - 7Y 2.

(b) Find a solution (other than W) to LX = Y 2

Let V be as above and set V = V + 6W - 2V 1 + 6W so that
-- L V = LV - 2LV1 + 6LW = 2Y1 - 7Y2 - 2LY1 + 6Y2 = Y2

Hence V is another solution to LX = Y 2.

2

Let f(x) ∈ C([a,b]) and put
∫ -- --1-- b f (x) = b- a a f(x)dx

Because eu ≥ 1 + u for all u, then in particular we have ef-f ≥ 1 + f -f. Thus we have the following sequence

ef(x) 1 ∫ b -b1-a∫ba f(x)dx ≥ 1+ f(x)- b--a- f(x)dx ∫ b e ∫ b( a ∫ b ) ---ef∫(y)----dy ≥ 1+ f(y)- --1-- f(x)dx dy a eb1-a ba f(x)dx a b- a a ∫b f(y) ∫ b ∫ b ∫ b( ∫ b ) --a e∫b-dy- ≥ dy+ f(y)dy- --1-- f(x)dx dy eb1-a a f(x)dx a a a b- a a ∫b f(y) ∫ b ∫ b --a1 e∫b-dy- ≥ (b- a)+ f(y)dy- b--a- f(x)dx eb-a a f(x)dx a b- a a ∫ba ef(y)dy -b1-a∫ba f(x)dxdy ≥ b- a e ∫ b ∫ --1-- ef(y)dy ≥ eb1-a ba f(x)dx b- a a
giving us the desired result.

3

4 Determine which of the following are pointwise and uniformly convergent

Since uniform convergence implies pointwise convergence, the proofs below omit any mention of pointwise convergence if the given sequence is proven to be uniformly convergent.

(a) fn(x) = sinx- n on ℝ

Let ε > 0 be given. Choose integer N so that Nε > 1. Since 0 ≤|sinx|≤ 1 for all x, then
| | ||sinx-|| 1- 0 ≤ | n | ≤ n < ε

for all n ≥ N. Hence fn → 0 uniformly.

(b) fn(x) = -1--- 1+nx on [0, 1]

This sequence converges pointwise to
{ 1 x = 0 f(x) = 0 otherwise

since for x = 0 fn(x) = 1 for all n. And furthermore, for any ε > 0, at a fixed x0 ∈ (0,1] we can choose integer N such that N > 1- x0(1 ) ε - 1 so that

|| 1 || ||1+-nx-|| < ε 0

for all n ≥ N

This sequence of functions, however, does not converge uniformly. Because each fn is continuous on [0,1], converging uniformly would imply that f above is continuous, but it’s not, due to the simple discontinuity at x = 0.

(c) fn(x) = --x--- 1+nx2 on ℝ

For each n,
1 2nx2 f′n(x ) = 1+-nx2-- (1+-nx2)2

which is zero only at x = ± ---- ∘ 1∕n. Since fn(x) is positive on (0,∞) and negative on (-∞,0), we have that fn has a minimum at -∘1-∕n- and a maximum at ∘1-∕n-. Because the minimum and maximum for each n are fn(-∘1-∕n--) = -1∕2∘1-∕n- and fn(∘1-∕n-) = 1∕2∘1∕n-, respectively, we have

1∘ -1 0 ≤ |fn(x)| ≤- -< 1∕n 2 n

for all n ∈ ℕ. Hence, given any ε > 0, choosing integer N so that Nε > 1 gives us |fn(x)| < ε for all n ≥ N and all x ∈ ℝ. Thus fn → 0 uniformly.

5

Let {fn} and {gn} be sequences of functions in C([0,1]). fn → f and gn → g.

(a) If both fn → f and gn → g pointwise, does fngn → fg pointwise fg?

Yes. For a fixed x0 ∈ [0,1], {fn(x0)} and {gn(x0)} are just normal sequences, and so fn(x0)gn(x0) → f(x0)g(x0) since fn(x) → f(x) and gn(x) → g(x).

(b) If both fn → f and gn → g uniformly, does fngn → fg uniformly fg?

Let ε > 0 be given. Because all {fn} and {gn} are continuous, then their uniform convergence implies that both f and g are also continuous. Hence f and g are bounded on [0,1]. Due to this, we can find an M such that |f(x)|≤ M and |g(x)|≤ M. Furthermore, the uniform convergence of fn → f and gn → g we can find integer N such that both
(∘ -ε ε ) |fn(x)- f(x)| < min 3,3M--

and

(∘ -- ) ε -ε-- |gn(x)- g(x)| < min 3,3M

for all x ∈ [0,1] and n ≥ N. Hence we have

|fngn(x) - f g(x)| = |(fn(x)- f(x))(gn(x)- g(x))+ f(x)(gn(x)- g(x))+ g(x)(fn(x)- f(x))| ≤ |fn(x)- f(x)||gn(x)- g(x)|+ |f(x)||gn(x) - g(x)|+ |g(x)||fn(x)- f(x)| (∘ --)2 < -ε + M -ε-+ M -ε-- 3 3M 3M = ε
for all n ≥ N and all x ∈ [0,1], which implies that fngn → fg uniformly.

6

This problem is identical to a problem on the previous homework.

7 Explain which conditions of the Contracting Map Theorem fail for the following

(a) x↦→x + 1 x on [0,∞)

Denote this map by f. This map is not a contraction map because there is no α < 1 such that
d(f(x),f(y)) ≤ αd(x,y)
(7.1)

for all x,y ∈ [0,∞). This is because the difference between d(x,y) and d(f(x),f(y)) is

| ( )| | | | | || 1- 1- || ||1- 1|| ||x---y|| |(x- y)- x+ x - y+ y | = |x - y| = | xy |

which can be made arbitrarily close to zero by making x and y large enough. This is problematic because the contraction condition in 7.1 implies that

(1- α )d(x,y) ≤ d(f(x),f(y))- d(x,y)

must be true; which is not the case when the difference on the righthand side above can be made arbitrarily close to zero.

(b) x↦→x2 on (0, 1]

Denote this mapping by f. The failure here is that the metric space (0,1] is not complete. In particular, the method of successive approximations fails for this mapping since the sequence {xn} defined by xn = f(xn-1) for fixed, arbitrary x0 ∈ (0,1] converges to zero, which is not in (0,1].

8

Let
∑∞ f(x) = sinkx-- k=0 1+ k4

(a) For which real x is f continuous?

Define {fn} to be the partial sums of the summation f. Since |sin(kx)|≤ 1 for all k and all x and 1 + k4 is increasing, then this sum converges for all x and therefore fn → f. Moreover, this means each fn has a bound Mn and ∑ Mn converges. This then implies that fn → f uniformly. Hence, because each fn is continuous, f must also be continuous.

(b) Is f differentiable? Why?

Since f(x) is a summation that converges for all x, then f(x) < ∞, and thus f(x) is continuous since it is the composition of continuous functions. Furthermore, because
∫ ∑∞ k cos(kx) ∞∑ k ∫ ∞∑ k 1 ∑∞ cos(kx) ------4- = ----4- cos(kx) = ----4--cos(kx) = -----4-= f(x) i=1 1+ k i=1 1+ k i=1 1+ k k i=1 1+ k

then f′(x) = kcos(kx) 1+k4 by the Fundamental Theorem of Calculus.

9

For any complex number z = x + iy and integer n we have nz = nx+iy = nxniy. Furthermore we have
| | | | ||niy|| = ||eln(niy)|| = ||eiyln(n)|| = 1

Thus for any sequence bounded by M and all z ∈ ℂ in the set {z = x + iy | x ≥ c} where c > 1, we have

∑∞ a ∑∞ M ∞∑ M ∞∑ M -nz ≤ -z = -x-≤ -c-< ∞ n=1n n=1 n n=1 n n=1n

so that ∑ n=1∞an nz converges.

10 Show that fn(x) = n3xn(1 - x) does not converge uniformly on [0,1].

For xn = 1 - 1∕n (which is always in [0,1]) then
f(x ) = n3(1- 1∕n)n(1- (1- 1∕n)) = n2 (1 - 1∕n )n n n

Therefore since (1 - 1∕n)n → 1∕e as n →∞ then fn can be made arbitrarily large and so cannot converge uniformly.

11 Give an example of a sequence of continuous functions for each of the below.

(a) A sequence of continous functions that converges to zero on [0, 1] but not uniformly

Problem ten has already proven this.

(b)

PIC

(c)

PIC