Back to main page

On this page:

• 1. Vibe Check
• 2. Q&A
• 3. Group Discussion on Rudin Exercise 5.7
• 4. Group Discussion on Fixed Points

1. Vibe Check

[bluebox=Agenda]

• “Intro”
  • What are you hoping to get out of this?
  • What math have you done before?
  • Any topics you are interested in?
• Scheduling/logistics (we skipped this in favor of waiting till we have more people)
  • Meeting 2
  • Regular meet times?
• Rough format:
  • Mentors (chris + Mike(?)) assign readings and problems/supplementary material.
  • Question/hints: Q&A sessions or make forum post in Discord. [/bluebox]

[greenbox=Kebin] Goals: Fill in gaps for grad analysis.

Background:

• (MATH) 131, 143/144, 145, 155, quivers, dimers.
• (COMP) 047, Rest of SOECS CS

Interests:

• Functional analysis
• Fourier analysis
• Dynamical systems [/greenbox]

Since Mari/Pranav couldn’t make it, Kebin and chris vibed the following info.

[greenbox=Pranav] Goals: Todo

Background:

• (MATH) 074, 131/132, 133, 145, 155
• (COMP) Rest of SOECS CS

Interests:

• Stochastic analysis?
• Fourier analysis?
• PDEs?
• Dynamics systems? [/greenbox]

[greenbox=Mari] Goals: Todo

Background:

• (MATH) 074, 145, 131/132, 133, 155
• Grad version of 047, rest of SOECS EE?

Interests:

• Fourier analysis?
• Dynamical systems?
• PDEs? [/greenbox]

2. Q&A

We did a quick summary of the big theorems in Rudin Chapter 5.

[theorem=(Cauchy’s Mean Value Theorem – Rudin 5.9)] Let $f, g:[a, b] \to \RR$ be continuous on $[a, b]$ and differentiable on $(a, b)$. Then there exists $\xi \in (a, b)$ such that

[/theorem]

In single variable real analysis, usually this theorem is used to prove the L’Hospital rule (for instance, this is the way Rudin proves the L’Hospital rule).

Cauchy’s mean value theorem naturally arises as the version of the mean value theorem that pops up when studying tangent lines in parametric equations (for a review, see this page). See Wikipedia for a visualization.

The following result is what people usually mean by “the MVT” and can be seen as a special case of the Cauchy MVT (why?).

[theorem=(Mean Value Theorem – Rudin 5.10)] Let $f:[a, b]$ be continuous on $[a, b]$ and differentiable on $(a, b)$. Then there exists $\xi\in (a, b)$ such that

[/theorem]

The MVT is one of the most important theorems in applications of the derivative to analysis. Some of the more immediate examples are:

• Taylor’s theorem
• Integration
• Error asymptotics

The next big result in Rudin is a structural result about derivatives of functions. We rephrase it via a definition common in analysis that Rudin doesn’t use.

[definition] Let $X \subseteq \RR$. We say that a function $f:X \to \RR$ is Darboux if it satisfies the intermediate value property (IVP): For all $a, b\in X$ and for every $y$ between $f(a)$ and $f(b)$, there exists a $c \in X$ between $a$ and $b$ such that $y = f(c)$. [/definition]

The intermediate value property should look very familiar because of the following observation.

[example] All continuous functions on a closed bounded interval $[a, b]$ are Darboux by the Intermediate Value Theorem (IVT) because the IVT is precisely the statement that says that IVP holds in these cases. [/example]

As it turns out, derivatives (of differentiable functions on a closed bounded interval) are also Darboux.

[theorem=(Darboux’s Theorem – Rudin 5.12)] Let $f:[a, b] \to \RR$ be differentiable. Then $f’$ is Darboux. [/theorem]

In isolation, this probably looks like a silly “Why would anyone care?” result (which is a reasonable initial reaction). This result is important because Darboux functions only allow for certain types of discontinuities. Recall that there are three types of discontinuities in general: jump, removable, and essential.

[example=(Removable Discontinuity)] Removable discontinuities are the discontinuities that form a “hole” in the graph. An example is the function $f:([0, 2]) \to \RR$ given by $f(x) = x(x-1)(x-2)$ if $x \neq 1/2$ and $f(1/2) = 5$. The “hole” is at $x = 1/2$.

More rigorously, $x = 1/2$ is a removable discontinuity of $f(x)$ because $\lim_{x \to 1/2} f(x)$ exists but is not equal to $f(1/2)$. [/example]

[example=(Jump Discontinuity)] Jump discontinuities are the discontinuities that where the function’s graph breaks off into two or more disjoint pieces but neither piece oscillates infinitely often at the break. For example, the following code graphs the function $f:[0, 2] \to \RR$ where

More rigorously, the one-sided limits of $f$ both exist but do not agree. [/example]

[example=(Essential Discontinuity)] Essential discontinuities are the discontinuities where the function’s graph becomes infinitely oscillatory in a neighborhood of a point. An example is the function $f:[0, \infty) \to \RR$ defined by

More rigorously, one of the one-sided limits at $x = 0$ fails to exist so $f$ has an essential discontinuity at $x = 0$. [/example]

One can prove that Darboux functions are the functions which, if they have a discontinuity, must be an essential discontinuity.

[corollary] Let $f:[a, b] \to \RR$ be differentiable. Then all discontinuities (if any exist) of $f’$ are essential. [/corollary]

We leave a proof of this fact as an exercise to the reader.

[example=(Function with a derivative with an essential discontinuity)] Let $f:\RR \to \RR$ be given by

For $x\neq 0$, we can just apply chain rule and product rule. One needs to use the limit definition of the derivative to work out what happens at $x = 0$ (verify this!). It turns out that

[/example]

Then we briefly mentioned the following results.

[theorem=(L’Hospital’s Rule – Rudin 5.13)] Let $f, g$ be differentiable on $(a, b)$ (with $a = -\infty$ and $b = \infty$ being allowed possibilities). If

• $g’(x) \neq 0$ for all $x \in (a, b)$;
• $\dfrac{f(x)}{g(x)} \to L$ as $x \to a$; and
• either $f(x), g(x) \to 0$ as $x \to a$ OR $g(x) \to \infty$ as $x \to a$;

then $\dfrac{f(x)}{g(x)} \to L$ as $x \to a$. [/theorem]

[theorem=(Taylor’s Theorem – Rudin 5.15)] Let $f:[a, b] \to \RR$ be $n$-times differentiable on $(a, b)$. If $\alpha, \beta \in [a, b]$ are distinct, then there exists $\xi$ between $\alpha$ and $\beta$ so that

[/theorem]

Then we spent a little bit of time discussing the Inverse Function Theorem for $\RR$. The motivation behind this result comes from calculus: If a function $f$ is invertible and differentiable on $(a, b)$, then there is a formula for the derivative of $f^{-1}$ via the chain rule:

then divide by $f’(x)$ to get

This result can be used to compute the derivative of classic inverse functions such as the inverse trig functions (try this!). The implicit assumption that is being made here is that $f^{-1}$ is actually differentiable so, to prove the formula above rigorously, we phrase the setup carefully as follows.

[theorem=(Inverse Function Theorem for $\RR$ – Rudin Exercise 5.2)] Let $f:(a, b) \to \RR$ be differentiable with $f’(x) > 0$ for every $x \in (a, b)$. Then

1. $f$ is strictly increasing in $(a, b)$.
2. $f$ is invertible (this follows from (1) – why?).
3. $f^{-1}$ is differentiable and

[/theorem]

3. Group Discussion on Rudin Exercise 5.7

The exercise in question is as follows:

[exercise=Rudin Exercise 5.7] Let $f$ and $g$ be functions differentiable in an open interval $I$ containing the point $a$. Suppose further that $g’(a) \neq 0$ and $f(a) = g(a) = 0$. Then

[/exercise]

We mentioned a sleek way of proving this simplified version of L’Hospital’s rule based on the idea of linear approximation.

[proof] Define $\phi$ to be the function given by

Then $\phi(h) \to 0$ as $h \to 0$ (why?). We define $\eta(h)$ similarly by

Then $\eta(h) \to 0$ for the same reason. Since

and

(Make sure you understand where the first equality came from!) Taking the limit as $h \to 0$ gives precisely the desired result. [/proof]

4. Group Discussion on Fixed Points

A good reference for this material is Chapter 1 of Suli-Mayers Numerical Analysis.

One motivation for fixed points comes from solving equations: Given a function $f(x)$, we want to find its roots. That is, we are solving the equation

for $r$. We can define $g(x) \coloneqq f(x) + x$ so that

That is, $r$ is a fixed point of $g$. There is a theorem in analysis and topology which guarantees the existence of these points.

[theorem=(Brouwer Fixed Point Theorem)] Let $g:[a, b] \to [a, b]$ be continuous. Then there exists a fixed point $\xi \in [a, b]$ of $g$. [/theorem]

This particular case of the Brouwer fixed point theorem is typically proved with the Intermediate Value Theorem (try this!). From the point of view of analysis, solving for fixed points can be more ideal because they can sometimes be found through approximation methods involving contraction mappings.

[definition] The function $g:[a, b] \to \RR$ is a contraction if there exists a constant $L \in (0, 1)$ such that

for all $x, y \in [a, b]$. [/definition]

[example] Let $g(x) = x/2$ on $[-1, 1]$. There is a unique fixed point at $x = 0$ (prove this).

The map $g$ is a contraction mapping on $[-1, 1]$ because

\(|g(x) - g(y)| = \left|\frac{x}{2} - \frac{y}{2}\right| = \frac{1}{2}|x - y|.\) [/example]

Contraction mappings in general (for $\RR$) are mappings whose graphs are “flat”-ish (why?). For example, the function $f(x) = 10x$ on $[-1, 1]$ isn’t a contraction (why?).

An incredibly important result, that we closed with, is the following:

[theorem=(Banach Fixed Point Theorem)] Let $g:[a, b] \to [a, b]$ be a continuous contraction. Then $g$ has a unique fixed point $\xi$. Furthermore, we can construct a sequence $(x_n)$ that converges to $\xi$ as follows: Let $x_1$ be any element of $[a, b]$ and define $x_{n+1} = g(x_n)$ for all $n \ge 1$. Then $x_n \to \xi$. [/theorem]

[example=(Suli-Mayers Example 1.2)] The function $g(x) = \ln(2x + 1)$ on $[1, 2]$ has a unique fixed point.

[/example]

[remark] The assumption that $g$ is continuous is redundant. One can prove that contraction mappings are uniformly continuous (prove this!). [/remark]

[proof=Proof of Uniqueness (rest is an exercise)] Suppose that $\xi$ and $\eta$ are fixed points. Thus, $g(\xi) = \xi$ and $g(\eta) = \eta$. Since $g$ is also a contraction, there exists some $L$ such that

Thus,

Thus, we have that

Making the observation that

it follows that $|\xi - \eta| = 0$ (why?) so then $\xi = \eta$ (why?). Thus, $\xi$ is unique. [/proof]