<Home>Teaching><MATH 131A (Winter 2021)>Week 9 Discussion Notes

Week 9 Discussion Notes

Table of Contents

Homework Comments

11.11

I saw a lot of people write something like this:

By Theorem 11.4, every sequence has an increasing subsequence.

Here is the statement of Theorem 11.4:

Every sequence {sn}n\set{s_n}_n has a monotonic subsequence.

Notice that the theorem only says that every sequence has a monotonic sequence, which is why the statement is false. It's possible for a sequence to have no increasing subsequences, e.g., if sn=1ns_n = \frac{1}{n}, then every subsequence of {sn}n\set{s_n}_n is decreasing. However, in this problem, you can show that the monotone sequence you get does actually increase.

12.2

I saw the following (incorrect) statements pop up a bunch of times:

lim sup⁡n→∞sn=0  ⟹  lim sup⁡n→∞ ∣sn∣=0\displaystyle\limsup_{n\to\infty} s_n = 0 \implies \limsup_{n\to\infty} \,\abs{s_n} = 0

Basically, this boils down to the fact that it's possible for

∣sup⁡ {x∣x∈S}∣≠sup⁡ {∣x∣∣x∈S}.\abs{\sup\,\set{x \mid x \in S}} \neq \sup\,\set{\abs{x} \mid x \in S}.

For example, if S=[−1,0]S = \br{-1, 0}, then sup⁡S=0\sup{S} = 0, but {∣x∣∣x∈S}=[0,1]\set{\abs{x} \mid x \in S} = \br{0, 1} has supremum 11. As for a counterexample for this specific statement, the sequence sn=−1+(−1)ns_n = -1 + \p{-1}^n works. This is the sequence that alternates between 00 and −2-2, so

lim sup⁡n→∞sn=0butlim sup⁡n→∞ ∣sn∣=2.\limsup_{n\to\infty} s_n = 0 \quad\text{but}\quad \limsup_{n\to\infty} \,\abs{s_n} = 2.

I also saw something like:

lim sup⁡n→∞ ∣sn∣=−lim inf⁡n→∞ ∣−sn∣\displaystyle\limsup_{n\to\infty} \,\abs{s_n} = -\liminf_{n\to\infty} \,\abs{-s_n}

The following statement is true:

lim sup⁡n→∞tn=−lim inf⁡n→∞ (−tn).\limsup_{n\to\infty} t_n = -\liminf_{n\to\infty} \,\p{-t_n}.

In this problem, we have tn=∣sn∣t_n = \abs{s_n}, so substituting in,

lim sup⁡n→∞ ∣sn∣=−lim inf⁡n→∞ (−∣sn∣),\limsup_{n\to\infty} \,\abs{s_n} = -\liminf_{n\to\infty} \,\p{-\abs{s_n}},

i.e., the negative sign goes outside of the absolute values. As a counterexample, the sequence sn=1s_n = 1 works:

lim sup⁡n→∞ ∣sn∣=1but−lim inf⁡n→∞ ∣−sn∣=−1.\limsup_{n\to\infty} \,\abs{s_n} = 1 \quad\text{but}\quad -\liminf_{n\to\infty} \,\abs{-s_n} = -1.

Continuous Functions

Since things don't seem to be too hard this week, I decided to do a fun example.

Example 1.

Let f ⁣:R→R\func{f}{\R}{\R} be a continuous function such that f(1)=1f\p{1} = 1 such that f(1)=1f\p{1} = 1 and f(nx)=nf(x)f\p{nx} = nf\p{x} for any n∈Zn \in \Z and x∈Rx \in \R. Then f(x)=xf\p{x} = x for all x∈Rx \in \R.
Solution.

The idea here is that it's hard to show directly for any x∈Rx \in \R that this equality is true, since it's hard to connect irrational numbers to the conditions given to you. Instead, however, we can show that it's true for x∈Qx \in \Q, and by a proposition from last week, it follows that f(x)=xf\p{x} = x everywhere.

First, if we set x=1x = 1, then for n∈Zn \in \Z, we get

f(n)=f(n⋅1)=nf(1)=n.f\p{n} = f\p{n \cdot 1} = nf\p{1} = n.

Next, we can let x=1nx = \frac{1}{n} to get

f(1)=f(n⋅1n)=nf(1n)  ⟹  f(1n)=1n.f\p{1} = f\p{n \cdot \frac{1}{n}} = nf\p{\frac{1}{n}} \implies f\p{\frac{1}{n}} = \frac{1}{n}.

Finally, if m,n∈Nm, n \in \N, then

f(mn)=mf(1n)=mn.f\p{\frac{m}{n}} = mf\p{\frac{1}{n}} = \frac{m}{n}.

Thus, f(q)=qf\p{q} = q for all q∈Qq \in \Q, so by density, we have f(x)=xf\p{x} = x for all x∈Rx \in \R.

Properties

Definition

Let f ⁣:D→R\func{f}{D}{\R} be a function. Then the image of a set A⊆DA \subseteq D under ff is f(A)={f(x)∣x∈A}f\p{A} = \set{f\p{x} \mid x \in A}.

Basically, the image tells you all the points that ff attains when you restrict the domain to AA.

Example 2.

Let f(x)=x2f\p{x} = x^2. Then

  • f(R)=[0,∞)f\p{\R} = \pco{0, \infty}
  • f({−1,0,1,2,3})={0,1,4,9}f\p{\set{-1, 0, 1, 2, 3}} = \set{0, 1, 4, 9}

Next are two very important theorems about continuous functions:

Theorem (extreme value theorem)

If f ⁣:[a,b]→R\func{f}{\br{a,b}}{\R} is continuous, then ff is bounded and there exist s0,s1∈[a,b]s_0, s_1 \in \br{a, b} such that f(s0)≤f(x)≤f(s1)f\p{s_0} \leq f\p{x} \leq f\p{s_1} for all x∈[a,b]x \in \br{a, b}.

In short, this tells you that on a closed and bounded interval, ff is bounded and attains both its minimum and maximum. (In general, we say that the image of a compact set under a continuous function is compact, but this is beyond the scope of this class.)

Theorem (intermediate value theorem)

Let I⊆RI \subseteq \R be an interval. If f ⁣:I→R\func{f}{I}{\R} is a continuous function, then f(I)f\p{I} is an interval.

This basically tells you that continuous functions can't "jump" on an interval. (In general, the continuous image of a connected set is connected, but like compactness, this is beyond our class.)

Example 3.
(18.5)

Let ff and gg be continuous functions on [a,b]\br{a, b} such that f(a)≥g(a)f\p{a} \geq g\p{a} and f(b)≤g(b)f\p{b} \leq g\p{b}. Prove f(x0)=g(x0)f\p{x_0} = g\p{x_0} for at least one x0∈[a,b]x_0 \in \br{a, b}.
Solution.

For problems like these, you usually want to use the intermediate value theorem. That only involves a single function, though, so you need to turn ff and gg into a single function. Notice that

f(x)=g(x)  ⟺  f(x)−g(x)=0,f\p{x} = g\p{x} \iff f\p{x} - g\p{x} = 0,

so we'll set h(x)=f(x)−g(x)h\p{x} = f\p{x} - g\p{x}. Then

h(a)=f(a)−g(a)≥0andh(b)=f(b)−g(b)≤0.h\p{a} = f\p{a} - g\p{a} \geq 0 \quad\text{and}\quad h\p{b} = f\p{b} - g\p{b} \leq 0.

Thus, by the intermediate value theorem, hh attains every value in the interval [h(b),h(a)]\br{h\p{b}, h\p{a}}, which contains 00. In other words, there exists x0∈[a,b]x_0 \in \br{a, b} such that h(x0)=0h\p{x_0} = 0, which means f(x0)=g(x0)f\p{x_0} = g\p{x_0}.
Example 4.
(18.8)

Suppose ff is a real-valued continuous function on R\R and f(a)f(b)<0f\p{a}f\p{b} < 0 for some a,b∈Ra, b \in \R. Prove there exists xx between aa and bb such that f(x)=0f\p{x} = 0.
Solution.
(Hint)

f(a)f(b)<0f\p{a}f\p{b} < 0 tells you that a≠ba \neq b and f(a),f(b)f\p{a}, f\p{b} have different signs (you should prove this). From there, you should break up the problem into different cases based on the signs of f(a),f(b)f\p{a}, f\p{b} and apply the intermediate value theorem.

Uniform Continuity

Definition

Let f ⁣:D→R\func{f}{D}{\R}. Then ff is uniformly continuous on DD if for all ε>0\epsilon > 0, there exists δ>0\delta > 0 such that ∣f(x)−f(y)∣<ε\abs{f\p{x} - f\p{y}} < \epsilon whenever x,y∈Dx, y \in D satisfy ∣x−y∣<δ\abs{x - y} < \delta.
Remark.

Compare this with our usual definition of continuity:

For all x∈Dx \in D and ε>0\epsilon > 0, there exists δ>0\delta > 0 such that ∣f(x)−f(y)∣<ε\abs{f\p{x} - f\p{y}} < \epsilon when y∈Dy \in D satisfies ∣x−y∣<δ\abs{x - y} < \delta.

Notice that δ\delta may depend on xx, i.e., given a different x∈Dx \in D, you may have to pick a different δ\delta for regular continuity. However, for uniform continuity, you can find a δ\delta that works for all values of x∈Dx \in D.

This concept is useful when we need to bound a lot of values of ff at the same time, e.g., when finding inequalities for integrals or sums relating to continuous functions.
Example 5.

Show that x3x^3 is uniformly continuous on [0,1]\br{0, 1} with the ε\epsilon-δ\delta definition.
Solution.

Let ε>0\epsilon > 0 and x,y∈[0,1]x, y \in \br{0, 1}. Then

∣x3−y3∣=∣x−y∣∣x2+xy+y2∣(difference of cubes)≤∣x−y∣(∣x∣2+∣x∣∣y∣+∣y∣2)(triangle inequality)≤3∣x−y∣,\begin{aligned} \abs{x^3 - y^3} &= \abs{x - y}\abs{x^2 + xy + y^2} && \p{\text{difference of cubes}} \\ &\leq \abs{x - y}\p{\abs{x}^2 + \abs{x}\abs{y} + \abs{y}^2} && \p{\text{triangle inequality}} \\ &\leq 3\abs{x - y}, \end{aligned}

since ∣x∣,∣y∣≤1\abs{x}, \abs{y} \leq 1. Thus, let δ=ε3\delta = \frac{\epsilon}{3} (which only depends on ε\epsilon and not the actual values of xx and yy) so that if ∣x−y∣<δ\abs{x - y} < \delta, then

∣x3−y3∣≤3∣x−y∣<3δ=ε,\abs{x^3 - y^3} \leq 3\abs{x - y} < 3\delta = \epsilon,

so x3x^3 is uniformly continuous on [0,1]\br{0, 1}.

True or False

Example 6.

True or false: If ff and gg are uniformly continuous on R\R, then so is fgfg.
Solution.

False. If f(x)=g(x)=xf\p{x} = g\p{x} = x, then f(x)g(x)=x2f\p{x}g\p{x} = x^2, which is not uniformly continuous on R\R.
Example 7.

True or false: If ff and gg are uniformly continuous on R\R, then so is f+gf + g.
Solution.

True. Let ε>0\epsilon > 0 and notice that by the triangle inequality,

∣(f+g)(x)−(f+g)(y)∣=∣(f(x)−f(y))+(g(x)−g(y))∣≤∣f(x)−f(y)∣+∣g(x)−g(y)∣.\begin{aligned} \abs{\p{f + g}\p{x} - \p{f + g}\p{y}} &= \abs{\p{f\p{x} - f\p{y}} + \p{g\p{x} - g\p{y}}} \\ &\leq \abs{f\p{x} - f\p{y}} + \abs{g\p{x} - g\p{y}}. \end{aligned}

Since ff is uniformly continuous on R\R, there exists δ1>0\delta_1 > 0 such that if ∣x−y∣<δ1\abs{x - y} < \delta_1, then ∣f(x)−f(y)∣<ε2\abs{f\p{x} - f\p{y}} < \frac{\epsilon}{2}. Similarly, because gg is uniformly continuous on R\R, there exists δ2>0\delta_2 > 0 so that if ∣x−y∣<δ2\abs{x - y} < \delta_2, then ∣g(x)−g(y)∣<ε2\abs{g\p{x} - g\p{y}} < \frac{\epsilon}{2}.

Let δ=min⁡ {δ1,δ2}\delta = \min\,\set{\delta_1, \delta_2}, so if ∣x−y∣<δ\abs{x - y} < \delta, then ∣x−y∣<δ1\abs{x - y} < \delta_1 and ∣x−y∣<δ2\abs{x - y} < \delta_2, and so

∣(f+g)(x)−(f+g)(y)∣≤∣f(x)−f(y)∣+∣g(x)−g(y)∣<ε2+ε2=ε,\abs{\p{f + g}\p{x} - \p{f + g}\p{y}} \leq \abs{f\p{x} - f\p{y}} + \abs{g\p{x} - g\p{y}} < \frac{\epsilon}{2} + \frac{\epsilon}{2} = \epsilon,

so f+gf + g is uniformly continuous on R\R.