Math 508: Advanced Analysis

Homework 6

Lawrence Tyler Rush

<me@tylerlogic.com>

October 27, 2014

http://coursework.tylerlogic.com/courses/upenn/math508/homework06

1 Prove cos x and sin x are continuous for all x ∈ ℝ

cos x is continuous: Let x ∈ ℝ For any |h| we have

|cos(x+ h)- cosx| = |cosxcosh- sinhsinx - cosx| = |cosx(cosh - 1)+ sinh (- sin x)| ≤ |cosx(cosh - 1)|+ |sin h(- sinx)| ≤ |cosh- 1|+ |sinh|

where the last inequality comes from the fact that -1 ≤-sinx ≤ 1 and -1 ≤ cosx ≤ 1 for all values of x. Now, as h → 0, cosh → 1 and sinh → 0. This implies the last line in the equation above approaches zero as h → 0. Thus, for any ε > 0 we can find a small enough δ so that for |h| < δ |sin(x + h) - sinx| < ε. Hence sinx is continuous at x.

sin x is continuous: Let x ∈ ℝ For any |h| we have

|sin(x+ h) - sinx| = |sin xcosh+ sinhcosx - sinx| = |sin x(cosh - 1)+ sinh cosx| ≤ |sin x(cosh - 1)|+ |sin hcosx| ≤ |cosh- 1|+ |sin h|

where the last inequality comes from the fact that -1 ≤ sinx ≤ 1 and -1 ≤ cosx ≤ 1 for all values of x. Like we saw in the previous paragraph, the last line of the above equation can be made arbitrarily close to zero as h → 0. Hence, similar to what we saw in the previous paragraph, cosx is continuous at x.

2

Let f(x) = x² + 4x and 0 < ε < 4. Define δ = ε
5

. Then 0 < δ < 1, given the constraints on ε. Assume that |x| < δ, so that 0 < |x| < δ < 1, in particular 0 < |x|² < |x| < 1. With this and the triangle inequality, we then have

2 2 2 |f(x)| = |x +4x| ≤ |x |+ |4x| = |x| + 4|x| < |x|+ 4|x | = 5|x| < 5δ = ε

as desired.

3 Prove the existence of an x ∈ [1, 2] such that x⁵ + 2x + 5 = x⁵ + 10

Put f(x) = x⁵ + 2x + 5 and g(x) = x⁵ + 10. Being polynomials, f and g are both continuous. Therefore the intermediate value theorem implies that on the interval [1,2] f will take on every value between f(1) = 8 and f(2) = 41 and g will take on every value between g(1) = 11 and g(2) = 26. Since f(1) = 8 < 11 = g(1) and g(2) = 26 < 41 = f(2), f and g must therefore intersect on the interval [1,2]. Hence there’s a point x on [1,2] with f(x) = g(x), i.e. where x⁵ + 2x + 5 = x⁵ + 10.

4 Show there exists two diametrically opposite points on the Eath’s equator that have the same temperature.

We’ll use the 3-dimensional cartesian coordinate system and orient the earth so that the z-axis contains the earth’s axis of rotation, the north pole has a positive z coordinate, and the center of the earth is at the orgin. Then the equator will be contained in the xy-plane. Finally, we will denote, by p_θ for θ ∈ ℝ_≥0, the point on the equator that is intersected by the line through the origin and (cosθ,sinθ,0).

Now let T : ℝ_≥0 → ℝ be the function whose value at θ ∈ ℝ_≥0 is the temperature of p_θ. We will assume that this temperature function is continuous. Define the function D : ℝ_≥0 → ℝ by D(θ) = T(θ) -T(θ + π), i.e. it is the difference in temperature of p_θ and its diametrically opposite point, p_θ+π. Since D is made up of the composition and difference of continuous functions, then it too is continuous.

Let p_θ be a point on the equator that doesn’t have the same temperature as its diametric opposite (there must be such a point, otherwise there would be nothing to prove as all points diametrically opposite would have the same temperature). Then D(θ) = -D(θ + π) and neither values are zero. Without loss of generality, assume that D(θ) < 0 < D(θ + π). Since D is continuous and it’s domain, ℝ_≥0, is connected, then this and the intermediate value theorem implies that there is a θ′∈ (θ,θ + π) such that D(θ′) = 0. But this implies that T(θ′) = T(θ′ + π), i.e. the diametrically opposed points p_θ′ and p_θ′+π have the same temperature.

5

Define {a_n} and {b_n} by a_n = -1∕n and b_n = 1∕n. Then of course both sequences converge to zero. Then the function f : ℝ → ℝ defined as

{ f = 0 x ≤ 0 1x otherwise

has the desired property of f(a_n) → 0 and f(b_n) being unbounded.

Does there exist such a function that’s continuous at x = 0 This is not possible as it would contradict Rudin’s Theorem 4.2.

6

Let f(a,n) = (1 + a)ⁿ where a and n are positive.

(a) Behavior of f(a,n) for constant a or n

For constant a how does f(a,n) behave as n →∞? When a is constant, 1 + a > 1 since a was assumed positive. So f(a,n) = (1 + a)ⁿ →∞ as n →∞.

For constant n how does f(a,n) behave as a → 0? As a → 0, 1 + a → 1. So for constant n, f(a,n) = (1 + a)ⁿ → 1 as a → 0.

(b)

Defining a_n = ∘ ------
n L + 1n

- 1, then a_n is a sequence of positive values, and furthermore

( (∘ ------ ))n ( ∘ -----)n n n 1- n 1- 1- f(an,n) = (1 +an) = 1 + L + n - 1 = L + n = L + n

so that f(a_n,n) → L as n →∞.

7 Which of the following functions are uniformly continuous on [0,∞)

(a) f(x) = x sin x

For any x ∈ [0,∞) and h ∈ ℝ we have that

|(x+ h)sin(x + h)- xsinx| = |h sin(x+ h) + xsin(x +h )- xsin x| = |h sin(x+ h) + x(sinxcosh + sinh cosx)- xsin x| = |h sin(x+ h) + x(sinxcosh + sinh cosx- sin x)|

For a fixed h, hsin(x + h) is bounded, but x(sinxcosh + sinhcosx - sinx) gets arbitrarily large as x gets large. Thus no matter how small h is, we can find x large enough so that |hsin(x + h) + x(sinxcosh + sinhcosx - sinx)|, and therefore |(x + h)sin(x + h) - xsinx|, can be made arbitrarily large. Hence there is no way for f(x) = xsinx to be uniformly continuous on [0,∞).

(b) f(x) = e^x

For any x ∈ [0,∞) and h ∈ ℝ we have that

|ex+h - ex| = |ex(eh - 1)|

so no matter the value of h, |e^x(e^h - 1)| can be made arbitrarily large, and therefore so can |e^x+h -e^x|. Hence e^x cannot be uniformly continuous on [0,∞)

(c) f(x) =

For any ε > 0, set δ = ε. Thus for any x,y ∈ [0,∞ when |y - x| < δ we have

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

Since x,y ∈ [0,∞) then 1 + y ≥ 1 and 1 + x ≥ 1 which implies

| | ||-1-- -1--|| ---|y--x|---- |1 + y - 1+ x| = |(1+ y)(1+ x)| ≤ |y - x| < δ = ε

so that f(x) is uniformly continuous on [0,∞)

8 Show that f(x) = is continuous ∀x ≥ 0

Let ε > 0 and x ∈ [0,∞). Define δ = ε √x--

so that in particular ε = δ√--
x

. Then for any y ≥ 0 such that |x - y| < δ we have

√-- √-- -|x--y|-- |x--y| -δ- |f (x) - f (y)| = | x - y| = |√x-+ √y-| ≤ √x- < √x-= ε

indicating that f is continuous at x.

Is the function uniformly continuous? Yes. First note that because f is continous, as per above, then because [0,1] is compact in ℝ f is uniformly continuous on [0,1]. Furthermore, for any x,y ∈ [1,∞), √y-- + √x-- ≥ 1. Therefore for any ε > 0, setting δ = ε means that when |y - x| < δ we have

√ -- √ -- --|x---y|- | x- y| = |√x-+ √y| ≤ |x - y| < δ = ε

so that f is uniformly continuous on [1,∞).

Finally for any ε > 0, define δ = min(δ^-,δ⁺) where δ^- is such that |x-y| < δ^-⇒| √ --
x - √--
y | < for all x,y ∈ [0,1] and δ⁺ is such that |x - y| < δ⁺ ⇒| - √ --
y | < for all x,y ∈ [1,∞). Thus with this definition, whether x,y ∈ [0,1] or x,y ∈ [1,∞), |x - y| < δ will imply | √--
x - √--
y | < ϵ.

It remains to be seen that |x-y| < δ implies | √--
x - √ --
y | < ϵ when x ∈ [0,1] and y ∈ [1,∞). This nevertheless holds, since for x ∈ [0,1] and y ∈ [1,∞), |x - y| < δ implies both |x - 1| < δ < δ^- and |y - 1| < δ < δ⁺. These two equations in turn imply | √ --
x - √ -
1 | < and | √--
y - √-
1 | < . Hence we obtain

√-- √-- √-- √ -- √-- √- √-- √- ε ε | x - y| = | x - 1|+ | y- 1| = | x - 1|+ | y - 1| <- + - = ε 2 2

Therefore f is uniformly continous on all of [0,∞).

9

(a) Prove [0, 1] and ℝ are not homeomorphic.

Since [0,1] is compact but ℝ is not, there exists no continuous function from [0,1] to ℝ. Thus there exists no homeomorphism between them either.

(b) Prove ℝ and (0,∞) are homeomorphic.

The function f : ℝ → (0,∞) defined by f(x) = e^x is both bijective and continuous, and thus is a homeomorphism between the two sets.

(c) Prove ℝ² and {(x,y) ∈ ℝ² : y > 0} are homeomorphic.

Since f(x) = e^x is a homeomorphism from ℝ → (0,∞) then the function g : ℝ² →{(x,y) ∈ ℝ² : y > 0} defined by

y g((x,y)) = (x,e )

will also be a homeomorphism.

(d) Prove ℝ and (-1, 1) are homeomorphic.

The map f : (-1,1) → ℝ is continuous since its composed of the quotient, product, and difference of continuous functions. It’s injective, because

--x--- --y--- x2 - 1 = y2 - 1 x(y2 - 1) = y(x2 - 1) 2 2 xy 2- x = yx2- y xy + y = yx + x y(xy+ 1) = x(yx+ 1) y = x

It’s surjective because for any y ∈ ℝ, if it were to be true that

y = --x--- x2 - 1

then we’d have

2 yx - x - y = 0

and the quadratic formula yields

∘ ------- 1+ 1+ 4y2 x = ----2y------

as one of the roots. Thus because

1 + ∘1-+-4y2 ∘1-+-4y2 ∘4y2- 2y ------------< --------< ----- = -- = 1 2y 2y 2y 2y

then we have a solution for x < 1. Thus f is a homeomorphism.

10

Let

{ f(x ) = x sin(1∕x) x ⁄= 0 0 otherwise

(a) Show that f is continuous on ℝ

The function f is the composition and product of functions x, sinx, and 1∕x, which are all continuous at nonzero reals. Therefore f is continuous at all nonzero reals. It remains to be shown that f is continuous at zero.

Let ε > 0 and set δ = ε. The for |x| < δ we have

|x sin(1∕x)- f(0)| = |xsin(1∕x) - 0| = |xsin(1∕x)| ≤ |x| < δ = ε

so that f is continuous at zero.

(b) Is f uniformly continous on [0, 2π]?

Yes, f is continous on [0,2π] and [0,2π] is compact.

(c)

11

(a)

(b)

12 Show a continuous real-valued function f that has f(x + y) = f(x) + f(y) for all x,y ∈ ℝ must be f(x) = cx for some c.

Let f : ℝ → ℝ be continuous and have that f(x + y) = f(x) + f(y) for all x,y ∈ ℝ. Put c = f(1). Then for integer n, f(n) = f(1) + f(n- 1) = f(1) +

+ f(1) = nf(1) = nc. Thus for any rational n∕m we have

mf (n ∕m) = f (n∕m )+ ⋅⋅⋅+ f(n∕m) = f(m(n∕m )) = f(n) = nc

so that f(n∕m) = (n∕m)c. Thus any rational q ∈ ℚ has f(q) = qc. With this, we finally have that for x ∈ ℝ and a rational sequence {a_n} with a_n → r, the continuity of f yields

f(r) = lim f(an) = lim (can ) = c lim an = cr n→ ∞ n→∞ n→∞

as desired.

13

Let E ⊂ ℝ be a set and f : E → ℝ be uniformly continuous.

(a) Show that if E is bounded, then so is f(E)

Lemma 13.1. If E ⊂ X is a bounded set, then so is its closure, E.

Proof. Let E be a bounded subset of a metric space X, and x,y two points of its closure, E. If x,y ∈ E, then there’s nothing to prove since E is already bounded. We must address two remaining case:

when one of x and y is a limit point not in E and the other a point of E
when both x and y are limit points not in E

Assume the first, and without loss of generality, let x ∈ E and y be a limit point of E. Then any open ball centered at y, say B₁(y), contains some point of x ∈ E. Thus d(x,y) ≤ d(x,x) + d(x,y) < M + 1 where M is a bound for the set E. Hence the distance between any point of E and a limit point is bounded.

Now assume the second case above. By the previous paragraph we have d(x,y) ≤ d(x,z) + d(z,y) < 2(M + 1) for any z ∈ E. So in this case, the distance is bounded.

Hence we have that the closure of E is bounded. __

Let E ⊂ ℝ be a bounded set. Then the closure, E, is also bounded by the above lemma. Hence it’s closed and bounded, implying that it’s compact as a subset of ℝ. Therefore, the set f(E) is also compact since f is continuous. Thus f(E) is bounded since it’s a subset of ℝ. But because E ⊂E, then f(E) ⊂ f(E), and so f(E) must also be bounded.

(b) If E is not bounded, give an example showing f(E) might not be bounded.

Let E = (0,∞) and f(x) = x. Then f is uniformly continuous, E is not bounded, and f(E) = (0,∞) which is also not bounded.