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)<u<\sup(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

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

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

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

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

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

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.

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2 Responses to “Real Analysis Limits and Continuity VII”

  1. […] Proof: Since is continuous in the compact interval it has a maximum and a minimum in (see Extreme Value Theorem which is theorem 55 in Real Analysis – Limits and Continuity VII). […]

  2. […] contínua no intervalo compacto sabemos que tem um máximo e um mínimo em (teorema 55 no artigo Análise Matemática – Limites e Continuidade VII […]

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