Uniform Continuity: An Introduction

1. Formula
   Observation

   Written as f in the definition and example.

f

A function mapping a set E to the real numbers.

Functions from E to R

1. Formula
   Observation

   Written as E in the definition.

E

The domain set on which the function f is defined and evaluated for uniform continuity.

Subsets of R

1. Formula
   Observation

   Written as R in the definition and example.

R

The set of real numbers.

Real numbers

1. Formula
   Observation

   Written as epsilon in the definition and proof.

$\varepsilon$

An arbitrary positive real number representing the tolerance for the difference in function values.

(0, $\infty$)

1. Formula
   Observation

   Written as delta in the definition and proof.

$\delta$

A positive real number representing the tolerance for the difference in input values.

(0, $\infty$)

1. Formula
   Observation

   Written as x and y in the definition and proof.

x, y

Arbitrary elements of the domain set E (or R in the example).

Elements of E or R

1. Formula
   Observation

   Written as m and b in the example $f(x)$ = mx + b.

m, b

Constants defining a linear function, where m is the slope and b is the y-intercept.

Real numbers

1. Formula
   Observation

   Pre-written board text defines f:E→R and uses $ε>0$, $δ>0$, |x-y|<δ ⇒ |$f(x)-f(y)$|<ε for all x,$y\in E$.

2. Audio
   Observation

   The lecturer reads the negation aloud as 'there is epsilon zero greater than zero such that for all delta greater than zero ... for some x and y in the set'.

f:E→R

Function from a set E to the real numbers.

E is the domain subset under discussion; codomain is R.

1. Formula
   Observation

   Yellow line writes ∃ ε₀>0.

2. Audio
   Observation

   'epsilon zero greater than zero'.

ε₀

A fixed positive tolerance witnessing failure of uniform continuity.

ε₀∈(0,$\infty$).

1. Formula
   Observation

   Yellow line writes 'for all $δ>0$'.

2. Audio
   Observation

   'for all delta greater than zero'.

δ

Arbitrary positive distance threshold in the negated statement.

$δ\in (0,\infty )$.

1. Formula
   Observation

   Yellow line writes sequences {$x_n$} and {$y_n$} with |$x_n-y_n$|<$1/n$ and |$f(x_n)-f(y_n)$|≥ε₀.

2. Audio
   Observation

   'a pair of sequences $x_n$ and $y_n$ ... with their difference less than one over n, and $f(x_n)-f(y_n)$ bigger than or equal to epsilon zero'.

{$x_n$}, {$y_n$}

Sequences used to convert the quantified negation into a sequential test.

Each $x_n$,$y_n\in E$ (the lecturer verbally notes the general negation first says 'some x and y in R', then specializes to sequences in E).

1. Formula
   Observation

   Pink line writes Ex: $f(x)=x$² is not uniformly continuous on ℝ.

2. Audio
   Observation

   'Example f of x which is x squared is not uniformly continuous on R'.

$f(x)=x$²

Concrete example function for which uniform continuity on ℝ will be disproved.

f:ℝ→ℝ.

1. Formula
   Observation

   Def: We say that f: E -> R is uniformly continuous on E if for every epsilon > 0 there is delta > 0 s.t. |x - y| < delta implies |$f(x) - f(y)$| < epsilon for all x, y in E.

A function f mapping a set E to the real numbers is uniformly continuous on E if, for every positive epsilon, there exists a positive delta such that the absolute difference between any two points x and y in E being less than delta guarantees that the absolute difference between their function values $f(x)$ and $f(y)$ is less than epsilon.

1. f: $E \to \mathbb{R}$

2. $\varepsilon > 0$

3. $\delta > 0$

4. x, $y \in E$

1. Audio
   Observation

   Speaker states: 'Now to find an example of a function which is not uniformly continuous we need to do the following. So let me look at the logical negation of this statement over here.'

2. Formula
   Observation

   Written as: f fails to be uniformly continuous

1. The full logical negation formula is not completed on the board within the provided clip.

To prove that a function is not uniformly continuous, one must analyze and utilize the logical negation of the formal definition of uniform continuity.

1. Definition of Uniform Continuity

1. Formula
   Observation

   Yellow text: 'f fails to be uniformly continuous on E if there is ∃ ε₀>0 s.t. ∀ $δ>0$, |x-y|<δ ⇒ |$f(x)-f(y)$|≥ε₀, ∃ x,$y\in E$'.

2. Audio
   Observation

   Lecturer states the same negation in words.

The video explicitly writes the logical negation of the uniform-continuity definition. Instead of requiring every $ε>0$ to admit some $δ>0$ that works uniformly for all pairs in E, failure means there exists one positive ε₀ such that no matter how small $δ>0$ is chosen, one can still find a pair x,$y\in E$ with |x-y|<δ but |$f(x)-f(y)$|≥ε₀.

1. f:E→R

2. ε₀>0

3. $δ>0$

4. x,$y\in E$

1. Formula
   Observation

   Yellow text: 'Taking $δ=1/n$, we can find a pair of sequences {$x_n$} and {$y_n$} with |$x_n-y_n$|<$1/n$, |$f(x_n)-f(y_n)$|≥ε₀'.

2. Audio
   Observation

   'if I take a sequence of delta getting small ... taking delta equals one over n ... we can find ... a pair of sequences $x_n$ and $y_n$ ... with their difference less than one over n, and $f(x_n)-f(y_n)$ bigger than or equal to epsilon zero'.

From the quantified negation, the lecturer derives a practical test: choose $δ=1/n$ for each n, then the negation guarantees a corresponding pair $x_n$,$y_n\in E$ with |$x_n-y_n$|<$1/n$ while |$f(x_n)-f(y_n)$|≥ε₀. Thus non-uniform continuity can be witnessed by two sequences whose inputs become arbitrarily close but whose images stay separated by at least ε₀.

1. Use the previously stated negation of uniform continuity

2. Take $δ=1/n$ for each n∈ℕ

1. Negation of uniform continuity on E

1. Formula
   Observation

   Pink text: 'Ex: $f(x)=x$² is not uniformly continuous on ℝ.'

2. Audio
   Observation

   'So here's the example of a function which is not uniformly continuous. Example f of x which is x squared is not uniformly continuous on R.'

1. The full verification is not completed within this clip; only the setup and sequence choice are shown.

The lecturer introduces $f(x)=x$² as the concrete example to which the sequential criterion will be applied. He begins constructing the witnessing sequences by choosing $y_n=n$ and first $x_n=n+1/n$, then revises to $x_n=n+2/n$ so that |$x_n-y_n$|=$2/n$ becomes smaller while still tending to 0.

1. Domain is ℝ

2. Goal is to show failure of uniform continuity using the sequential criterion

1. Sequential reformulation of failure of uniform continuity

1. Formula
   Observation

   Top green line on the board defines uniform continuity: 'We say that f:E→R is uniformly continuous on E if for every $ε>0$ there is $δ>0 s$.t. |x-y|<δ ⇒ |$f(x)-f(y)$|<ε for all x,$y\in E$.'

2. Audio
   Observation

   The spoken explanation in this clip focuses on the counterexample, while the definition remains visible on the board throughout.

The board gives the standard epsilon-delta definition: a function f from E to R is uniformly continuous on E when one can choose a single δ depending only on ε (not on the particular points) so that any two inputs within δ of each other have outputs within ε of each other.

1. f:E→R

2. $ε>0$

3. $δ>0$

4. x,$y\in E$

1. Formula
   Observation

   Yellow text states: 'f fails to be uniformly continuous on E if there is ε₀>0 s.t. for all $δ>0$, |x-y|<δ ⇒ |$f(x)-f(y)$|≥ε₀ ∀x,$y\in R$.'

2. Audio
   Observation

   At 00:56-01:09 the lecturer summarizes the same idea verbally: he found sequences whose difference is less than $1/n$ but whose squared values differ by at least a fixed ε₀, proving non-uniform continuity.

1. The displayed negation line ends with '∀x,$y\in R$' even though the sentence begins 'on E'; the clip does not explain this mismatch.

Failure of uniform continuity is characterized by the existence of a fixed positive output gap ε₀ such that no matter how small a δ is chosen, one can still find two points closer than δ whose images differ by at least ε₀.

1. ε₀>0

2. $δ>0$

3. the witness points may depend on δ

1. Definition of uniform continuity on a set E

1. Formula
   Observation

   Yellow line below the negation says: 'Taking $δ=1/n$, we can find a sequences {$x_n$} and {$y_n$} with |$x_n-y_n$|<$1/n$, |$f(x_n)-f(y_n)$|≥ε₀.'

2. Audio
   Observation

   The final spoken summary explicitly uses this sequence formulation to conclude that $x^2$ is not uniformly continuous on R.

1. The board writes 'a sequences', apparently a minor wording slip; mathematically it means two sequences {$x_n$} and {$y_n$}.

By specializing δ to $1/n$, the negated epsilon-delta condition can be rewritten as the existence of two sequences whose input distances tend to 0 while their output distances stay bounded below by some fixed ε₀.

1. Use the negation of uniform continuity

2. Choose $δ=1/n$

3. The sequences must lie in the domain

1. Negation of uniform continuity

1. Formula
   Observation

   Pink line states 'Ex $f(x)$=mx+b is uniformly continuous on R' and gives the proof sketch choosing $δ=ε$/|m|.

2. Formula
   Observation

   The displayed computation is '|mx+b-(my+b)|=|m||x-y|<|m|$δ=ε$'.

1. The board does not separately discuss the case $m=0$, although the formula $δ=ε$/|m| implicitly requires $m\ne 0$.

For $f(x)$=mx+b, the output difference is controlled directly by the input difference through the factor |m|, so choosing $δ=ε$/|m| makes |$f(x)-f(y)$|<ε whenever |x-y|<δ.

1. $f(x)$=mx+b

2. domain is R

3. in the displayed proof, $m\ne 0$ because $δ=ε$/|m| is used

1. Definition of uniform continuity on a set E

1. Formula
   Observation

   Board directly contrasts the pre-written definition with the yellow negation line.

2. Audio
   Observation

   Lecturer reads the negation as the condition for 'fails to be uniformly continuous on E'.

A function f:E→R fails to be uniformly continuous on E iff there exists ε₀>0 such that for every $δ>0$ there are x,$y\in E$ with |x-y|<δ and |$f(x)-f(y)$|≥ε₀.

1. f:E→R

∃ε₀>0 ∀$δ>0$ ∃x,$y\in E$

1. Formula
   Observation

   Yellow line states 'Taking $δ=1/n$, we can find a pair of sequences {$x_n$} and {$y_n$} with |$x_n-y_n$|<$1/n$, |$f(x_n)-f(y_n)$|≥ε₀'.

2. Audio
   Observation

   Lecturer explains that showing this condition gives an example of a function that fails to be uniformly continuous on E.

If there exist ε₀>0 and sequences {$x_n$},{$y_n$}⊂E such that |$x_n-y_n$|<$1/n$ and |$f(x_n)-f(y_n)$|≥ε₀ for all n, then f is not uniformly continuous on E.

1. f:E→R

2. ε₀>0

3. $x_n$,$y_n\in E$ for all n

∃ε₀>0 ∃{$x_n$},{$y_n$} ∀n∈ℕ

1. Formula
   Observation

   Purple heading states 'Ex $f(x)=x^2$ is not uniformly continuous on R.'

2. Audio
   Observation

   At 01:04-01:09 the lecturer says this proves that $x^2$ is not uniformly continuous on the real line.

There exist sequences $x_n=n+1/(2n)$ and $y_n=n$ such that |$x_n-y_n$|<$1/n$ while |$x_n^2-y_n^2$|>1, so $f(x)=x^2$ fails the uniform continuity condition on R.

1. Consider $f(x)=x^2$ on R

2. Use the sequence criterion for failure of uniform continuity

∃ sequences {$x_n$},{$y_n$}⊂R such that ∀n, |$x_n-y_n$|<$1/n$ and |$x_n^2-y_n^2$|≥ε₀ for a fixed ε₀ (here one may take ε₀=1).

1. Formula
   Observation

   Pink line asserts '$f(x)$=mx+b is uniformly continuous on R' and displays a direct epsilon-delta proof.

1. The displayed proof chooses $δ=ε$/|m| without explicitly treating $m=0$.

For $f(x)$=mx+b on R, given $ε>0$ one can choose $δ=ε$/|m| so that |x-y|<δ implies |$f(x)-f(y)$|<ε.

1. $f(x)$=mx+b

2. domain is R

3. in the written proof, $m\ne 0$ because division by |m| appears

∀$ε>0$ ∃$δ>0$ ∀x,$y\in R$, |x-y|<δ ⇒ |$f(x)-f(y)$|<ε.

1. Formula
   Observation

   Ex: $f(x)$ = mx + b is uniformly continuous on R. Pr: Let epsilon > 0 and choose delta = epsilon / m. Then if |x - y| < delta, then |mx + b - (my + b)| = |m||x - y| < |m|delta = epsilon.

1. The speaker verbally corrects the choice of delta to include absolute value around m, but initially writes delta = epsilon / m.

1. Expression

   $$\text{Let } \varepsilon > 0 \text{ and choose } \delta = \frac{\varepsilon}{|m|}.$$
   Explanation

   Set up the proof by choosing a specific delta based on the given epsilon and the slope of the linear function.

   Justification

   Definition of uniform continuity requires finding a delta for any given epsilon.

   Shown in the video
2. Expression

   $$\text{If } |x - y| < \delta, \text{ then } |(mx + b) - (my + b)|$$
   Explanation

   Evaluate the absolute difference of the function values for arbitrary x and y in R.

   Justification

   Substitution of the function definition $f(x)$ = mx + b.

   Shown in the video
3. Expression

   $$= |m(x - y)| = |m||x - y|$$
   Explanation

   Simplify the expression by factoring out m and applying the property of absolute values.

   Justification

   Algebraic manipulation and properties of absolute value.

   Shown in the video
4. Expression

   $$< |m|\delta = |m|\left(\frac{\varepsilon}{|m|}\right) = \varepsilon$$
   Explanation

   Apply the assumption |x - y| < delta and substitute the chosen value of delta to show the result is strictly less than epsilon.

   Justification

   Inequality properties and substitution.

   Shown in the video

The linear function $f(x)$ = mx + b satisfies the definition of uniform continuity on the set of all real numbers R.

1. Audio
   Observation

   'And so what this implies, if I take a sequence of delta getting small ... taking delta equals one over n ... we can find ... a pair of sequences $x_n$ and $y_n$ ...'

2. Formula
   Observation

   Yellow board line records the resulting sequence condition.

1. Expression

   $$$$
   Explanation

   Start from the negation of uniform continuity: there is ε₀>0 such that for every $δ>0$ one can find x,$y\in E$ with |x-y|<δ and |$f(x)-f(y)$|≥ε₀.

   Justification

   This is the definition written immediately above on the board.

   Shown in the video
2. Expression

   $$\delta=\frac1n$$
   Explanation

   Choose a specific decreasing family of tolerances $δ=1/n$, one for each natural number n.

   Justification

   Since the negated statement holds for all $δ>0$, it holds in particular for $δ=1/n$.

   Shown in the video
3. Expression

   $$\exists x_n,y_n\in E:\ |x_n-y_n|<\frac1n,\ |f(x_n)-f(y_n)|\ge \varepsilon_0$$
   Explanation

   For each n, apply the negated statement with $δ=1/n$ to obtain a pair $x_n$,$y_n\in E$ satisfying the two inequalities.

   Justification

   Direct instantiation of ∀$δ>0$ ∃x,$y\in E$ from the negation.

   Shown in the video
4. Expression

   $$$$
   Explanation

   Collecting these pairs over all n produces two sequences {$x_n$} and {$y_n$} whose input distances tend to 0 while output distances remain bounded below by ε₀.

   Justification

   Because $1/n\to 0$, the condition |$x_n-y_n$|<$1/n$ forces the sequences to get arbitrarily close; the lower bound |$f(x_n)-f(y_n)$|≥ε₀ prevents uniform control.

   Derived from the video

Failure of uniform continuity can be tested by exhibiting sequences {$x_n$},{$y_n$}⊂E with |$x_n-y_n$|→0 but |$f(x_n)-f(y_n)$| not tending to 0; in the video's form, |$x_n-y_n$|<$1/n$ and |$f(x_n)-f(y_n)$|≥ε₀.

1. Formula
   Observation

   Pink lines write Ex: $f(x)=x$² is not uniformly continuous on ℝ, then Choose $x_n=n+1/n$, $y_n=n$, later revised to $x_n=n+2/n$.

2. Audio
   Observation

   Lecturer says he needs to find the sequence and chooses $x_n$ and $y_n$ accordingly.

1. The algebraic verification that |$f(x_n)-f(y_n)$|≥ε₀ is not completed in this clip.

1. Expression

   $$f(x)=x^2,\quad E=\mathbb{R}$$
   Explanation

   Introduce the example function and its domain.

   Justification

   Stated explicitly on the board and in speech.

   Shown in the video
2. Expression

   $$y_n=n$$
   Explanation

   Choose the second sequence to be $y_n=n$.

   Justification

   This is the lecturer's explicit choice for the counterexample.

   Shown in the video
3. Expression

   $$x_n=n+\frac1n$$
   Explanation

   Initially choose $x_n=n+1/n$.

   Justification

   Written on the board before revision.

   Shown in the video
4. Expression

   $$x_n=n+\frac2n$$
   Explanation

   Revise the first sequence to $x_n=n+2/n$.

   Justification

   The lecturer says putting $2/n$ 'won't change anything' and rewrites the choice on the board.

   Shown in the video
5. Expression

   $$|x_n-y_n|=\frac2n$$
   Explanation

   With the revised choice, the input separation is exactly $2/n$, which tends to 0.

   Justification

   Direct subtraction: ($n+2/n$)-n=$2/n$.

   Derived from the video

The clip sets up the standard sequence-based disproof for $f(x)=x$² on ℝ by choosing $y_n=n$ and $x_n=n+2/n$ so that |$x_n-y_n$|→0; the remaining step of checking |$x_n$²-y_n²|≥ε₀ is not shown within this excerpt.

1. Formula
   Observation

   Purple work shows the chosen sequences and the full algebraic computation leading to |$x_n^2-y_n^2$|=$1+1/(4n^2)>1$.

2. Audio
   Observation

   From 00:00 to 01:09 the lecturer verbally walks through the same computation and concludes non-uniform continuity.

1. Expression

   $$x_n=n+\frac{1}{2n},\qquad y_n=n$$
   Explanation

   Choose two sequences of real numbers depending on n.

   Justification

   These are the sequences explicitly written on the board under 'Choose'.

   Shown in the video
2. Expression

   $$|x_n-y_n|=\left|\left(n+\frac{1}{2n}\right)-n\right|=\frac{1}{2n}<\frac{1}{n}$$
   Explanation

   Their mutual distance is smaller than $1/n$.

   Justification

   Direct subtraction and simplification shown on the board and spoken aloud at 00:00-00:07.

   Shown in the video
3. Expression

   $$|x_n^2-y_n^2|=\left|\left(n+\frac{1}{2n}\right)^2-n^2\right|$$
   Explanation

   Now compute the difference of the squared values, matching $f(x)=x^2$.

   Justification

   This is the next line written on the board at about 00:12-00:18.

   Shown in the video
4. Expression

   $$\left(n+\frac{1}{2n}\right)^2=n^2+2\cdot n\cdot \frac{1}{2n}+\frac{1}{4n^2}=n^2+1+\frac{1}{4n^2}$$
   Explanation

   Expand the square using ($a+b$)^$2=a^2+2$ab+$b^2$.

   Justification

   The lecturer verbally explains the middle term at 00:23-00:35 and the board displays the expanded expression.

   Shown in the video
5. Expression

   $$|x_n^2-y_n^2|=\left|n^2+1+\frac{1}{4n^2}-n^2\right|=1+\frac{1}{4n^2}$$
   Explanation

   Subtract $n^2$ and simplify; the absolute value is unnecessary because the result is positive.

   Justification

   Shown on the board around 00:44-00:54 and stated aloud as the $n^2$ terms cancel.

   Shown in the video
6. Expression

   $$1+\frac{1}{4n^2}>1$$
   Explanation

   For every n, the quantity is strictly larger than 1.

   Justification

   Immediate from positivity of $1/(4n^2)$; written on the board at the end of the computation.

   Shown in the video
7. Expression

   $$\exists \varepsilon_0=1\ \text{such that}\ |x_n-y_n|<\frac1n\ \text{but}\ |x_n^2-y_n^2|\ge \varepsilon_0$$
   Explanation

   Thus the sequence criterion for failure of uniform continuity is satisfied with ε₀=1.

   Justification

   This is the conclusion drawn by the lecturer at 00:56-01:09 from the preceding inequalities.

   Derived from the video