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


cos x is continuous: Let x 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 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 pθ for θ 0, the point on the equator that is intersected by the line through the origin and (cosθ,sinθ,0).

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

Let pθ 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, 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 pθ and pθ+π have the same temperature.

5


Define {an} and {bn} by an = -1∕n and bn = 1∕n. Then of course both sequences converge to zero. Then the function f : defined as
    {
f =   0   x ≤ 0
       1x  otherwise

has the desired property of f(an) 0 and f(bn) 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 an = ∘ ------
n L + 1n - 1, then an 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(an,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 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) = ex


For any x [0,) and h we have that
|ex+h - ex| = |ex(eh - 1)|

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

(c) f(x) = --1-
1+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) = √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 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-+) where δ- is such that |x-y| < δ-⇒|√ --
  x-√--
 y| < ε2 for all x,y [0,1] and δ+ is such that |x - y| < δ+ ⇒|√ --
  x -√ --
  y| < ε2 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| < ε2 and |√--
 y -√-
 1| < ε2. 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 are not homeomorphic.


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

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


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

(c) Prove 2 and {(x,y) 2 : y > 0} are homeomorphic.


Since f(x) = ex is a homeomorphism from (0,) then the function g : 2 →{(x,y) 2 : y > 0} defined by
             y
g((x,y)) = (x,e )

will also be a homeomorphism.

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


The map f : (-1,1) 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 , 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


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 must be f(x) = cx for some c.


Let f : be continuous and have that f(x + y) = f(x) + f(y) for all x,y . 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 has f(q) = qc. With this, we finally have that for x and a rational sequence {an} with an 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 be a set and f : E 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:

  1. when one of x and y is a limit point not in E and the other a point of E
  2. 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 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 . Therefore, the set f(E) is also compact since f is continuous. Thus f(E) is bounded since it’s a subset of . 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.

14