## Real Analysis Limits and Continuity VII

— 6.10. Global properties of continuous functions —

 Theorem 51 (Intermediate Value Theorem) Let ${I=[a,b] \in \mathbb{R}}$ and ${f: I \rightarrow \mathbb{R}}$ is a continuous function. Let ${u \in \mathbb{R}}$ such that ${\inf(I), then there exists ${c \in I}$ such that ${f(c)=u}$.Proof: Omitted. $\Box$

Intuitively speaking the previous theorem states that the graph of a continuous function doesn’t have holes in it if the domain of the function doesn’t have any holes in it too.

 Corollary 52 Let ${[a,b]}$ be an interval in ${\mathbb{R}}$, ${f:[a,b]\rightarrow\mathbb{R}}$ a continuous function. Suppose that ${f(a)f(b)<0}$. Then ${\exists c \in ]a,b[}$ such that ${f(c)=0}$.Proof: In the codomain of ${f}$ there exists values bigger than ${0}$ and values smaller than ${0}$. Hence ${\sup f(I)>0}$ and ${\inf f(I)<0}$. Therefore ${0}$ is strictly between between the infimum and supremum of the codomain of ${f}$. By hypothesis the function doesn’t vanish on the extremities of the interval, hence the ${0}$ value has to be in the interior of the interval $\Box$
 Corollary 53 Let ${I\in\mathbb{R}}$, ${f:I\rightarrow\mathbb{\mathbb R}}$ a continuous function. Then ${f(I)}$ is also an interval.Proof: Let ${\alpha=\inf(I)}$ and ${\beta=\sup(I)}$. By definition of infimum and supremum it is ${f(I)\subset [\alpha , \beta]}$. Using Theorem 51 it is ${]a,b[\subset f(I)}$.Thus we have the following four possibilities for ${f(I)}$: ${f(I)=\begin{cases}[\alpha , \beta] \\ ]\alpha , \beta] \\ [\alpha , \beta[ \\ ]\alpha , \beta[ \end{cases}}$ $\Box$

As an application let us look into ${P(x)=a_xx^n+\cdots +a_1x+a_0}$ with ${n}$ odd and ${a_n <0}$. It is ${P(x)\sim a_x^n}$ for large (positively or negatively) values of ${x}$. It is ${\displaystyle \lim_{x\rightarrow +\infty} P(x)=+\infty}$ and ${\displaystyle \lim_{x\rightarrow -\infty} P(x)=-\infty}$.

Now

• ${P(x)}$ is a continuous function.
• The domain, ${D}$ of ${P(x)}$ is ${\mathbb{R}}$ which is an interval.
• ${\sup(D)=+\infty}$ and ${\inf(D)=-\infty}$, implying ${P[\mathbb{R}]=]-\infty, +\infty[}$

By corollary 52 it is ${0\in P[\mathbb{R}]}$. Which means that every odd polynomial function has at least one ${0}$.

 Theorem 54 (Continuity of the inverse function) Let ${I}$ be an interval in ${\mathbb{R}}$ and ${f:I\rightarrow\mathbb{R}}$ a continuous function and strictly monotonous. Then ${f^{-1}}$ is continuous and strictly monotonous.Proof: Omitted. $\Box$

This theorem has important applications since it allows us to define the inverse functions of the trigonometric functions.

— 6.10.1. Arcsine function —

In ${[-\pi/2,\pi/2]}$ the function ${\sin x}$ is injective:

Sine function

Hence one can define the inverse of the sine function in this suitably restricted domain.

$\displaystyle y=\sin x\quad\mathrm{with}\quad x\in [\pi/2,\pi/2]\Leftrightarrow x=\arcsin x$

Where ${\arcsin}$ denotes the inverse of ${\sin}$.

Since ${\sin x:[-\pi/2,\pi/2]\rightarrow[-1,1]}$ it is ${\arcsin x:[-1,1]\rightarrow [-\pi/2,\pi/2]}$. Also by theorem 54 ${\arcsin}$ is continuous.

The graphical representation of ${\arcsin x}$ is

Arcsine function

and it is evident by its representation that ${\arcsin x}$ is an odd function.

— 6.10.2. Arctangent function —

In ${]-\pi/2,\pi/2[}$ the function ${\tan x}$ is injective:

Tangent function

Hence one can define the inverse of the tangent function in this suitably restricted domain.

$\displaystyle y=\tan x\quad\mathrm{with}\quad x\in ]\pi/2,\pi/2[\Leftrightarrow x=\arctan x$

Where ${\arctan}$ denotes the inverse of ${\tan}$.

Since ${\tan x:]-\pi/2,\pi/2[\rightarrow]-\infty,+\infty[}$ it is ${\arctan x:]-\infty,+\infty[\rightarrow ]-\pi/2,\pi/2[}$. Also by theorem 54 ${\arctan}$ is continuous.

The graphical representation of ${\arctan x}$ is

Arctangent function

and it is evident by its representation that ${\arctan x}$ is an odd function.

— 6.10.3. Arccosine function —

In ${[0,\pi]}$ the function ${\cos x}$ is injective:

Cosine function

Hence one can define the inverse of the cosine function in this suitably restricted domain.

$\displaystyle y=\cos x\quad\mathrm{with}\quad x\in [0,\pi]\Leftrightarrow x=\arccos x$

Where ${\arccos}$ denotes the inverse of ${\cos}$.

Since ${\cos x:[0,\pi]\rightarrow[-1,1]}$ it is ${\arccos x:[-1,1]\rightarrow [0,\pi]}$. Also by theorem 54 ${\arccos}$ is continuous.

The graphical representation of ${\arccos x}$ is

Arccosine function

Another way to define the arccosine function is to first use the relationship

$\displaystyle \cos=\sin(\pi/2-x)$

to write

$\displaystyle \arccos y=\frac{\pi}{2}-\arcsin y$

— 6.10.4. Continuous functions and intervals —

 Theorem 55 (Extreme value theorem) Let ${[a,b]\subset \mathbb{R}}$ and ${f:[a,b]\rightarrow\mathbb{R}}$. Then ${f}$ has a maximum and a minimum.Proof: Let ${E}$ be the codomain of ${f}$ and ${s=\sup E}$.By Theorem 17 in post Real Analysis – Sequences II there exists a sequence ${y_n}$ of points in ${E}$ such that ${\lim y_n=s}$. Since the terms of ${y_n}$ are points of ${f}$ for each ${n}$ there exists ${x_n\in [a,b]}$ such that ${y_=f(x_n)}$. Since ${x_n}$ is a sequence of points in the compact interval (see definition 22 in post Real Analysis – Sequences IV) ${[a,b]}$, by Corollary 27 (also in post Real Analysis – Sequences IV) there exists a subsequence ${x_{\alpha n}}$ of ${x_n}$ that converges to a point in ${[a,b]}$. Let ${c\in [a,b]}$ such that ${x_n\rightarrow c}$. Since ${f}$ is continuous in ${c}$ it is, by definition of continuity, (see definition 34) ${\lim f(x_{\alpha n})=f(c)}$. But ${f(x_{\alpha n})=y_{\alpha n}}$, which is a subsequence of ${y_n}$. Since ${y_n\rightarrow s}$ it also is ${y_{\alpha n}\rightarrow s}$. But ${y_{\alpha n}=f(x_{\alpha n})\rightarrow f(c)}$. In conclusion it is ${s=f(c)}$, hence ${s\in E}$. That is ${s=\max E}$. For the minimum one can construct a similar proof. This proof is left as an exercise for the reader. $\Box$

One easy way to remember the previous theorem is:

Continuous functions have a maximum and a minimum in compact intervals.

 Theorem 56 Let ${I}$ be a compact interval of ${\mathbb{R}}$ and ${f:I\rightarrow\mathbb{R}}$ continuous. Then ${f(I)}$ is a compact interval.Proof: By corollary 53 ${f(I)}$ is an interval.By theorem 55 ${f(I)}$ has a maximum and a minimum. Hence ${f(I)}$ is of the form ${[\alpha , \beta]}$. Thus ${f(I)}$ is a limited and closed interval, which is the definition of a compact interval. $\Box$

One easy way to remember the previous corollary is:

Compactness is preserved under a continuous map.