For a subset of the real numbers $D$, we denote a function $f$

$$f:D\to \mathbb{R}$$

As an operation assigning a value to all $x\in D$, denoted $f(x)$.

Two concepts essential to an analysis of functions are **continuity and differentiability. We discuss continuity here.

Continuity §

Defining Continuity §

Let $f$ be a function. We say $f$ is continuous at a point $x_0\in D$, if whenever any sequence $\{x_n\}\in D$ converges to $x_0$, so too does the sequence $\{f(x_n)\}$.

$$\{x_n\}\to x_0⟹\{f(x_n)\}\to f(x_0)$$

In other words, if a point $x\in D$ is sufficiently close to $x_0$, then the distance $f(x)-f(x_0)$ should become arbitrarily small.

We can similarly define this in terms of limits,

$$\underset{n\to \infty }{\lim }f(x_n)=f\left(\underset{n\to \infty }{\lim }{x}_n\right)=f(x_0)$$

Example: Continuity Proof

Prove $f:\mathbb{R}\to \mathbb{R},f(x)=x^3$ is continuous.

Take any real number $x_0$, and take any convergent sequence $\{x_n\}\to x_0$. Then, applying the product property of the limits,

$$\underset{n\to \infty }{\lim }(x_n{)}^3=(\underset{n\to \infty }{\lim }{x}_n{)}^3=(x_0{)}^3$$

Example: Continuity Proof (2)

Let $f:\mathbb{R}\to \mathbb{R}$ such that

$$f(x)=\{\begin{array}{ll}2x& x\le 5\\ 0& x>5\end{array}$$

Show that $f(x)$ is continuous at $x=3$.

We want to show that for any $\{x_n\}\to x_0$, $\{f(x_n)\}\to f(x_0)$.

Let $\{x_n\}\to 3$. By definition, $\forall ϵ>0$, $\exists N\in \mathbb{N}$ such that $\forall n\ge N$,

$$|x_n-3|<ϵ$$

Let $ϵ=1$, and choose the $N$ associated with this epsilon such that $\forall n\ge N$,

$$|x_n-3|<1$$

Now take the expression $|f(x_n)-f(3)|$. Because $\forall n\ge N$, $2<x_n<4$, we always apply the $2x$ part of the function.

$$\begin{array}{rl}|f(x_n)-f(3)|& =|2x_n-6|\\ & =2|x_n-3|\end{array}$$

Because $\{x_n\}\to 3$, and $|f(x_n)-f(3)|\le 2|x_n-3|$, by the squeeze theorem, $\{f(x_n)\}\to f(3)$.

Example: Continuity Disproof

Prove that $f:\mathbb{R}\to \mathbb{R}$

$$f=\{\begin{array}{ll}1& x<0\\ 2& x\ge 0\end{array}$$

is not continuous.

Let us have sequence $\{-1/n\}$, which converges to 0. However, $f(-1/n)$ will converge to 1, which is not equal to $f(0)=2$!

Theorem: Properties of Continuity

Let $f$ and $g$ be continuous functions. Then $fg$, $\frac{f}{g}$, $f+g$, $f-g$, $f\circ g$ are also continuous.

These properties trivially follow from the limit properties.

Example: Continuity and Sequential Compactness

Let $f:D\to \mathbb{R}$ be continuous where $D$ is sequentially compact. Prove that $f(D)$ is sequentially compact.

To show that $f(D)$, is sequentially compact, we want to show that for any $\{y_n\}\in f(D)$, there exists a subsequence $\{y_{n_k}\}\to y_0$ such that $y_0\in f(D)$.

As all terms of $\{y_n\}$ are in $f(D)$, for all $y_n$, there exists an $x_n$ such that $f(x_n)=y_n$. By sequential compactness on $D$, there exists a subsequence $\{x_{n_k}\}\to x_0$ such that $x_0\in D$.

By continuity on $f$, as $\{x_{n_k}\}\to x_0$, $\{f(x_{n_k})\}\to f(x_0)$. Letting this $f(x_0)=y_0$, we have a subsequence $\{x_{n_k}\}=y_{n_k}$ converging to a $y_0\in f(D)$.

Geometric Properties of Continuity §

Some important implications of continuity are given below: the extreme value theorem, and the intermediate value theorem.

Let $f$ be a function. For $f$, we define a set

$$f(D)\equiv \{y|y=f(x),x\in D\}$$

As the image (output) of $f$.

Now, we say $f$ has a maximum given that it’s image has a minimum - that is, there exists a point $x_0\in D$ such that

$$\forall x\in D,f(x)\le f(x_0)$$

Such a point $x_0$ is called a maximizer.

Similarly, we say $f$ has a minimum given that it’s image has a minimum - that is, there exists a point $x_0\in D$ such that

$$\forall x\in D,f(x)\ge f(x_0)$$

Such a point $x_0$ is called a minimizer.

Continuity guarantees the existence of a maximum and/or minimum along bounded domains of a continuous function. This is known as the Extreme Value Theorem.

Theorem: Extreme Value Theorem

Let $f:[a,b]\to \mathbb{R}$ be continuous, where $[a,b]$ is a closed bounded interval. Then, the maximum and minimum of $f$ exist.

In this theorem, the two key assumptions we need are compactness of $[a,b]$, and the continuity on $f$.

Proof

We prove a maximum (proof for minimum follows similarly). Let $D=[a,b]$. First, we prove that our image is bounded above, then we prove that the supremum is a functional value.

Upper Bound Proof §

If there exists an upper bound for the image $f(D)$, then

$$\forall x\in [a,b],f(x)\le M$$

By way of contradiction, assume there is no such $M$. Now let $n$ be any natural number. By assumption,

$$\forall x\in [a,b],f(x)\le n$$

Thus, there must exist a point $x\in [a,b]$ such that $f(x)>n$. We call this point $x_n$.

Continuing this argument for many $n$, we define a sequence $\{x_n\}$ with the property that $f(x_n)>n$ for all $n$.

1. By the Sequential Compactness Theorem, we can choose a subsequence $\{x_{n_k}\}$ converging to a point $x_0\in [a,b]$.
2. By definition of continuity, the image of the sequence, $\{f(x_{n_k})\}$, converges to $f(x_0)$.

However, because $\{f(x_{n_k})\}$ converges, it must be bounded! This contradicts the property that

$$\forall k\in \mathbb{N},f(x_{n_k})>n_k\ge k$$

Thus, there must exist an upper bound for the image $f(D)$.

Supremum Value Proof §

Let $S\equiv f([a,b])$. From above, we know that $S$ is bounded above, and because of this, must have a supremum.

Let $c$ be this supremum. We now show a point $x_0\in [a,b]$ where $c=f(x_0)$.

Let $n\in \mathbb{N}$. By assumption,

$$c-\frac{1}{n}$$

is smaller than $c$ and is therefore not an upper bound for $S$.

By the epsilon definition of bounds, this means that there is a point $x\in [a,b]$ such that $f(x)>c-\frac{1}{n}$! Let this point be $x_n$. So,

$$c-\frac{1}{n}<f(x_n)\le c$$

Continuing this for many $n$, we obtain a sequence $\{x_n\}$ such that $\{f(x_n)\}\to c$!

1. By the Sequential Compactness Theorem, we can find a subsequence $\{x_{n_k}\}$ converging to a point $x_0\in [a,b]$.
2. By continuity, if $\{x_{n_k}\}\to x_0$, then $\{f(x_{n_k})\}\to f(x_0)$.
3. But because $\{f(x_n)\}\to c$, this means that $f(x_0)=c$!

Thus, $x_0$ is a value within the function, and must exist.

Continuity also guarantees the existence of values within a range of a continuous function.

Theorem: Intermediate Value Theorem

Let $f:[a,b]\to \mathbb{R}$ be a continuous function.

Pick any $c$ in the interval $f(x_1)<c<f(x_2)$, where $x_1$ and $x_2$ are in the interior of $[a,b]$. Then, there exists an $x$ strictly between $x_1$ and $x_2$ such that $f(x)=c$.

The proof is ommitted, as it is long and technical. It uses the Nested Interval Theorem and Bisection Method.

Finally, continuity is maintained along intervals. We define an interval $I$ as the set with bounds $u,v$, such that all real numbers $x$ between the bounds exist within the set.

$$\begin{array}{rlr}& [u,v]& u\le x\le v\\ & [u,v)& u\le x<v\\ & (u,v)& u<x<v\\ & (u,v]& u<x\le v\end{array}$$

Theorem: Continuity along Intervals

If $f:I\to \mathbb{R}$ is a continuous function where $I$ is an interval, then the output $f(I)$ is also an interval.

Proof

Let $y_1,y_2\in f(I)$, and choose any value $C\in (y_1,y_2)$. Without loss of generality, suppose $y_1<y_2$. We wish to show that $C$ exists within $(y_1,y_2)$, which by definition shows the existence of the interval between $y_1,y_2$.

We know that

$$f(x_1)<c<f(x_2)$$

And thus, by the intermediate value theorem, there exists some $x$ between $x_1$ and $x_2$ such that $f(x)=c$,

Note that for this proof to work, we assumed that $[x_1,x_2]$ is an interval.

Uniform Continuity §

This topic is essential for defining integrals!

We say $f:D\to \mathbb{R}$ is uniformly continuous on $D$ if for any two sequences $\{x_n\},\{y_n\}\in D$ such that if their difference converges to 0,

$$\{x_n-y_n\}\to 0$$

Then $\{f(x_n)-f(y_n)\}\to 0$. In other words, for any two sequences that become arbitrarily close, so too should our function along these points $\{f(x_n)\}$ and $\{f(y_n)\}$.

Our sequences don’t even have to converge!

The following theorem may help inform us when to expect that a function is uniformly continuous. It will be given later formally, as we haven’t yet defined a derivative; but is listed here for our understanding.

Theorem: Uniform Continuity

If $f:D\to \mathbb{R}$ has a bounded derivative on $D$, then $f$ is uniformly continuous.

Uniform continuity is a fairly strong result! While continuity does not necessarily imply uniform continuity; uniform continuity can imply continuity!

Theorem: Continuity and Uniform Continuity

If $f:[a,b]\to \mathbb{R}$ is continuous, where $[a,b]$ is a closed bounded interval, then it is uniformly continuous.

This only works because of the compact set on $f$’s domain!

Proof

By way of contradiction, let $\{x_n-y_n\}\to 0$, and suppose $\{f(x_n)-f(y_n)\nrightarrow 0$.

So, there exists an $ϵ>0$ such that for all $N\in \mathbb{N}$, there exists an $n\ge N$ such that

$$|f(x_n)-f(y_n)|\ge ϵ$$

For $N=1$, choose $n_1\ge 1$ such that

$$|f(x_{n_1})-f(y_{n_1})|\ge ϵ$$

Now, choose $n_2>n_1$ such that

$$|f(x_{n_2})-f(y_{n_2})|\ge ϵ$$

Continuing this, we’ve constructed a sequence on $n$. Letting $\{x_{n_k}\}$, $\{y_{n_k}\}$ be subsequences using this sequence on $n$.

$$|f(x_{n_k})-f(y_{n_k})|\ge ϵ$$

But we have $\{x_{n_k}-y_{n_k}\}\to 0$!

So, by the Sequential Compactness Theorem, we can show that

$$\underset{k\to \infty }{\lim }{x}_{n_k}\to u_0$$

And because the difference of $x_n$ and $y_n$ converge, this forces

$$\underset{k\to \infty }{\lim }{y}_{n_k}\to u_0$$

By continuity, this means that $f(x_{n_k})\to f(u_0)$ and $f(y_{n_k})\to f(u_0)$! So,

$$|f(x_{n_k})-f(y_{n_k})|\to 0$$

But $ϵ>0$, which is a contradiction! Thus, $f$ is uniformly continuous on $[a,b]$.

However, if $f:D\to \mathbb{R}$ is uniformly continuous on $D$, then $f$ is continuous.

Proof

Fix $x_0\in D$. Define $\{x_n\}$ as a sequence converging to $x_0$, and let $\{y_n\}$ be a sequence that is only $x_0$. Then $f(x_n)\to f(x_0)$ by definition of uniform continuity.

Example: Uniform Continuity

Let $f:\mathbb{R}\to \mathbb{R}$, $f(x)=2x$.

We show that $f$ is uniformly continuous on $\mathbb{R}$. Let $x_n$ and $\{y_n\}$ be sequences such that $\{x_n-y_n\}\to 0$. Then,

$$|f(x_n)-f(y_n)|=|2x_n-2y_n|=2|x_n-y_n|\to 0$$

Example: Uniform Continuity (2)

Let $f:\mathbb{R}\to \mathbb{R}$, $f(x)=|x|$.

We show that $f$ is uniformly continuous on $\mathbb{R}$. Let $x_n$ and $\{y_n\}$ be sequences such that $\{x_n-y_n\}\to 0$. Then, applying the reverse triangle inequality,

$$|f(x_n)-f(y_n)|=||x_n|-|y_n||\le |x_n-y_n|\to 0$$

Example: Uniform Continuity (3)

Let $f:[0,\infty )\to \mathbb{R}$, $f(x)=\sqrt{x}$.

We should expect this to be uniform continuous, as our derivative is bounded between $[c,\infty )$, and our function is continuous in the compact set $[0,c]$!

Using the definition, we show this as follows.

$$\begin{array}{rl}|\sqrt{x_n}-\sqrt{y_n}|& \le |\sqrt{x_n}-\sqrt{y_n}{|}^2=|\sqrt{x_n}-\sqrt{y_n}|\cdot |\sqrt{x_n}-\sqrt{y_n}|\\ & \le |\sqrt{x_n}-\sqrt{y_n}|\cdot |\sqrt{x_n}+\sqrt{y_n}|\\ & \le |x_n-y_n|\to 0\end{array}$$

Example: Uniform Continuity Disproof

Let $f:\mathbb{R}\to \mathbb{R}$, $f(x)=x^2$.

While we have uniform continuity for any compact subset $[a,b]$, our theorem won’t apply for infinity! In fact, the end points are where we fail to satisfy the uniform continuity definition.

Choose $\{x_n\}=n$, $\{y_n\}=n+1/n$. Then, while they become arbitrarily close,

$$\left\{n^2-(n+1/n{)}^2\right\}=\left\{-2-\frac{1}{n^2}\right\}\to -2$$

Example: Uniform Continuity Disproof (2)

Let $f:(0,\infty )\to \mathbb{R}$, $f(x)=\frac{1}{x}$.

We should expect this to not be uniformly continuous, as our slope blows up as we tend towards 0!

Choose $\{x_n\}=\{1/n\}$, $\{y_n\}=\{1/n^2\}$. Then, these two sequences both converge to 0, yet

$$\left\{1/\frac{1}{n^2}-1/\frac{1}{n}\right\}=\{n^2-n\}\to \infty$$

$ϵ-\delta$ Criterion for Continuity §

Recall how we defined continuity. We propose below an alternative, yet equivalent, definition that may be useful.

Let $f:D\to \mathbb{R}$ be a function. Then, we say $f$ satisfies the $ϵ$-$\delta$ criterion at $x_0\in D$, if for all $ϵ>0$, there exists a $\delta >0$ such that for $x\in D$,

$$|x-x_0|<\delta \to |f(x)-f(x_0)|<ϵ$$

In other words, this is saying that for any interval in $f$’s range, we can choose an interval in $f$’s domain that contains all of the values of this range!

The following theorem establishes a connection between the $ϵ$-$\delta$ criterion and continuity.

Theorem: Continuity and $ϵ$-$\delta$

Let $f:D\to \mathbb{R}$ be a function, and let $x_0\in D$. Then, the following two assertions are equivalent.

1. The function $f:D\to \mathbb{R}$ is continuous at $x_0$.
2. The $ϵ$-$\delta$ criterion at $x_0$ holds.

Example: Disproof

Define $f(x)$ as the function where

$$f(x)=\{\begin{array}{ll}0& x<0\\ 1& x\ge 0\end{array}$$

We want to show by the $ϵ$-$\delta$ proof that the function is not continuous at $x=0$.

Choose $ϵ=\frac{1}{2}$. Then, there is no $\delta$ such that $|x|<\delta \to |f(x)-f(0)|$.

Example: Epsilon-Delta Continuity Proof

Let $f:\mathbb{R}\to \mathbb{R}$ be defined as $f(x)=x^2-4$. Verify using the $ϵ-\delta$ definition that $f$ is continuous at point $x_0=0$.

Scratch Work §

Pick any arbitrary $ϵ>0$. For this $ϵ$, we want to find a $\delta >0$ satisfying the relation

$$|x-3|<\delta \to |f(x)-f(3)|<ϵ$$

Let’s find this $\delta$.

$$\begin{array}{rl}|f(x)-f(3)|& =|x^2-4-(3^2-4)|=|x^2-9|\\ & =|x-3|\cdot |x+3|\end{array}$$

We know that $|x-3|<\delta$ by assumption, but $|x+3|$ is not bounded! Thus, we need to establish a bound for it.

Suppose $\delta =1$. Then, for this $\delta$ bound, we know that

$$|x-3|<1\to 2<x<4\to 5<x+3<7$$

Thus, for all $\delta \le 1$, $|x+3|<7$! Assume our $\delta \le 1$. Then,

$$=|x-3|\cdot |x+3|<\delta \cdot 7$$

Let $\delta =\min (\frac{ϵ}{7},1)$. Then, for this $\delta$, we successfully have bounded $|f(x)-f(3)|<ϵ$.

Proof §

Let $ϵ>0$, and let $\delta =\min (\frac{ϵ}{7},1)$. Then, for all $x$ such that $|x-3|<\delta$,

$$|f(x)-f(3)|=|x^2-9|=|x-3|\cdot |x+3|<\delta \cdot 7<ϵ$$

Example: Proof

Let $f(x)=x^2$. Prove it is continuous at $x_0\in \mathbb{R}$ by $ϵ$-$\delta$.

Scratch Work §

Fix any $ϵ>0$. We start from our $y$ constraint and work backwards to choose a $\delta$.

$$\begin{array}{rl}|f(x)-f(x_0)|& =|x^2-x_0^2|=|x-x_0|\cdot |x+x_0|\\ & <\delta |x+x_0|\end{array}$$

If $\delta =1$, then

$$|x-x_0|<1⟹|x|-|x_0|<1⟹|x|<1+|x_0|$$

$$\begin{array}{rl}|f(x)-f(x_0)|& <\delta |x+x_0|\\ & \le \delta (|x|+|x_0|)\\ & \le \delta (2|x_0|+1)<ϵ\end{array}$$

Provided that $\delta <\frac{ϵ}{2|x_0|+1}$.

Proof §

Let $ϵ>0$. Let $\delta =\min \left(1,\frac{ϵ}{2|x_0|+1}\right)$.

Then, for all $x\in \mathbb{R}$, if $|x-x_0|<\delta$, we get $|x^2-x_0^2|<ϵ$ (show from above inequalities).

Example: Proof

Suppose $f(x_0)>0$ and $f:\mathbb{R}\to \mathbb{R}$ is continuous. Prove $f(x)>0$ for all $x$ on some interval $(x_0-\delta ,x_0+\delta )$.

$$\begin{array}{r}|f(x)-f(x_0)|<ϵ\\ f(x_0)-ϵ<f(x)<f(x_0)+ϵ\end{array}$$

Choose $ϵ=f(x_0)/2$.

$$\frac{f(x_0)}{2}<f(x)$$

For all $x\in (x_0-\delta ,x_0+\delta )$.

Similarly, we can establish a connection between the $ϵ$-$\delta$ criterion and uniform continuity.

We say $f$ satisfies the $ϵ$-$\delta$ criterion on the domain $D$, if for all $ϵ>0$, there is a $\delta >0$ such that for all $u,v\in D$,

$$|u-v|<\delta |f(u)-f(v)|<ϵ$$

Theorem: Uniform Continuity and $ϵ$-$\delta$.

Let $f:D\to \mathbb{R}$ be a function, and let $x_0\in D$. Then, the following two assertions are equivalent.

1. The function $f:D\to \mathbb{R}$ is uniformly continuous at $x_0$.
2. The $ϵ$-$\delta$ criterion on the domain $D$ holds.

Example: Proof

Prove $f(x)=\sqrt{x}$ is uniformly continuous on $[0,2]$ using $ϵ$-$\delta$.

$$\begin{array}{rl}|\sqrt{x}-\sqrt{y}|& =|\sqrt{x}-\sqrt{y}|\frac{|\sqrt{x}+\sqrt{y}|}{|\sqrt{x}+\sqrt{y}|}\\ & =\frac{|x-y|}{|\sqrt{x}+\sqrt{y}|}\end{array}$$

But this doesn’t work, as we can’t control the denominator! So, we approach this differently, and begin with a squared term.

$$\begin{array}{rl}|\sqrt{x}-\sqrt{y}{|}^2& =|\sqrt{x}-\sqrt{y}|\cdot |\sqrt{x}-\sqrt{y}|\\ & \le |\sqrt{x}-\sqrt{y}|\cdot |\sqrt{x}+\sqrt{y}|\\ & \le |x-y|\end{array}$$

Given this, we take the square root of all terms to find

$$|\sqrt{x}-\sqrt{y}|<|x-y{|}^{\frac{1}{2}}<{\delta }^{\frac{1}{2}}<ϵ$$

So we choose $\delta <{ϵ}^2$ to finish our proof.

Inverses, Continuity, and Monotonicity §

Monotonicity §

Recall in the intermediate value theorem section, the theorem on the image of continuous functions.

Theorem: Intervals of Images

Let $f:I\to \mathbb{R}$ be continuous, where $I$ is an interval. Then, $f(I)$ is an interval.

Note that the converse of this is not true in normal cases, but it is when $f$ is monotone! This is expressed below.

Theorem: Continuity of Monotone Functions

If $f:D\to \mathbb{R}$ is monotone, then if $f(D)$ is an interval, $f$ is continuous!

Note that $D$ does not have to be an interval. We skip the proof as it is long and technical.

Function Inverses §

We say a function $f:D\to \mathbb{R}$ is injective (one-to-one) if any $y\in f(D)$ has exactly one $x$ value. Formally, if we have any two $x$-values $x_1,x_2$, then

$$x_1\ne x_2\to f(x_1)\ne f(x_2)$$

In other words, the function passes the horizontal line test - if we draw any horizontal line, it only crosses the function once.

Theorem: Monotonicity and Injections

If $f$ is strictly monotone, then $f$ is injective.

Proof

Choose any two $x$-values $x_1,x_2$. If $f$ is strictly monotone, then $f(x_1)<f(x_2)$, or $f(x_1)>f(x_2)$. In either case, $f(x_1)\ne f(x_2)$.

Note that the converse is not true, as we can choose a piecewise function to violate monotonicity. However, this changes if $f$ is continuous!

Theorem: Continuity, Injections, and Monotonicity

Let $f:I\to f(I)$ be continuous. Then, $f$ is injective if and only if $f$ is strictly monotone.

The continuity and interval clauses are extremely important, and without them, this theorem won’t hold.

We say that function $f:D\to f(D)$ is invertible provided that $f$ is injective. In the event this is true, we define the inverse function as

$$f^{-1}:f(D)\to D$$

Such that $x=f^{-1}(y)$ if and only if $f(x)=y$.

Theorem: Monotonicity of Inverses

If $f:D\to f(D)$ is strictly monotone, then $f^{-1}$ is also strictly monotone.

Proof

Without loss of generality, suppose $f$ is strictly increasing. Then, for two domain values $u,v\in D$ such that $u<v$,

$$u<v⟺f(u)<f(v)$$

Let $u=f^{-1}(x)$, $v=f^{-1}(y)$. Then,

$$f^{-1}(x)<f^{-1}(y)⟺f^{-1}(f(u))<f^{-1}(f(v))⟹u<v$$

So, $f^{-1}$ is strictly increasing.

Theorem: Continuity of Inverses

Let $f:I\to f(I)$ be strictly monotone. Then, $f^{-1}$ is continuous on $f(I)$.

Proof

Since $f$ is a strictly monotone, so is $f^{-1}$. Because $f^{-1}$ maps to an interval, and is strictly monotone, it is continuous by an earlier theorem.