Mathematics for Applied Sciences (Osnabrück 2023-2024)/Part I/Lecture 10/refcontrol

Continuous functions

We denote the distance between two real numbers ${}x$ and ${}x'$ by ${}d{\left(x,x'\right)}:=\vert {x-x'}\vert$. For a given function

$f\colon \mathbb {R} \longrightarrow \mathbb {R} ,$

one may ask whether it is possible to control the distance in the image (the target) by controlling the distance in the domain. Let ${}x\in \mathbb {R}$ and let ${}y=f(x)$ be its image. One would like that for points ${}x'$ which are close to ${}x$, also their image points ${}f(x')$ are close to ${}f(x)$. Already linear functions with different slopes show that the measure for being close in the image and in the domain can not be the same. A reasonable question is whether for any desired exactness in the image there exists at all an exactness in the domain which ensures that the values of the function are lying inside the desired exactness. To make this intuitive idea more precise, let ${}\epsilon >0$ be given. This ${}\epsilon$ represents an accuracy for the target. The question is whether one can find a ${}\delta >0$ (representing an accuracy for the domain) such that for all ${}x'$ fulfilling ${}d{\left(x,x'\right)}\leq \delta$, also the relation ${}d{\left(f(x),f(x')\right)}\leq \epsilon$ holds. This idea leads to the concept of a continuous function.

Definition

Let ${}D\subseteq \mathbb {R}$ be a subset,

$f\colon D\longrightarrow \mathbb {R}$

a function,MDLD/function and ${}x\in D$ a point. We say that ${}f$ is continuous in the point ${}x$, if for every ${}\epsilon >0$, there exists a ${}\delta >0$, such that for all ${}x'\in D$ fulfilling ${}\vert {x-x'}\vert \leq \delta$, the estimate ${}\vert {f(x)-f(x')}\vert \leq \epsilon$ holds. We say that ${}f$ continuous, if it is continuous in every point

${}x\in D$

${}D$ stands for the set of definition of the function. Typically, ${}D$ is just ${}\mathbb {R}$, or an interval, or ${}\mathbb {R}$ with finitely many points removed. Instead of working with the real numbers ${}\epsilon$ and ${}\delta$, one might also work just with the unit fractions ${}{\frac {1}{n}}$ and ${}{\frac {1}{m}}$.

Example Create referencenumber

A constant function

$\mathbb {R} \longrightarrow \mathbb {R} ,x\longmapsto c,$

is continuous.MDLD/continuous (R) For every given ${}\epsilon$, one can choose an arbitrary ${}\delta$, since

${}d(f(x),f(x'))=d(c,c)=0\leq \epsilon \,$

holds anyway.

The identity

$\mathbb {R} \longrightarrow \mathbb {R} ,x\longmapsto x,$

is also continuous.MDLD/continuous (R) For every given ${}\epsilon$, one can take ${}\delta =\epsilon$, yielding the tautology: If

${}d(x,x')\leq \delta =\epsilon \,,$

then

${}d(f(x),f(x'))=d(x,x')\leq \epsilon \,$

holds.

Example Create referencenumber

We consider the function

$f\colon \mathbb {R} \longrightarrow \mathbb {R} ,$

given by

${}f(x)={\begin{cases}0,{\text{ if }}x<0\,,\\1,{\text{ if }}x\geq 0\,.\end{cases}}\,$

This function is not continuous in ${}0$. For ${}\epsilon ={\frac {1}{2}}$ and every positive ${}\delta$, there exists a negative number ${}x'$ such that ${}d(0,x')=\vert {x'}\vert \leq \delta$. But for such a number we have ${}d(f(0),f(x'))=d(1,0)=1\not \leq {\frac {1}{2}}$.

It is not possible to draw any continuous function, even if you allow arbitrary enlargements. The image shows an approximation of a so-called Weierstraß-function, which is continuous, but nowhere differentiable. For a continuous function one can control the extent of the fluctuation in the image by reducing the seize of the definition interval, but the number of the fluctuations (how often the direction of the graph is changing) can not be controlled.

The following statement relates continuity to convergent sequences.

LemmaLemma 10.4 change

Let ${}D\subseteq \mathbb {R}$ be a subset,

$f\colon D\longrightarrow \mathbb {R}$

a functionMDLD/function and ${}x\in D$

a point. Then the following statements are equivalent.

1. ${}f$ is continuousMDLD/continuous (R) in the point ${}x$.
2. For every convergent sequenceMDLD/convergent sequence (R) ${}{\left(x_{n}\right)}_{n\in \mathbb {N} }$ in ${}D$ with ${}\lim _{n\rightarrow \infty }x_{n}=x$, also the image sequenceMDLD/image sequence ${}{\left(f(x_{n})\right)}_{n\in \mathbb {N} }$ is convergent with limit ${}f(x)$.

Proof

Suppose that (1) is fulfilled and let ${}{\left(x_{n}\right)}_{n\in \mathbb {N} }$ be a sequence in ${}D$ converging to ${}x$. We have to show that

${}\lim _{n\rightarrow \infty }f(x_{n})=f(x)\,$

holds. To show this, let ${}\epsilon >0$ be given. Due to (1), there exists a ${}\delta >0$ fulfilling the estimation property (from the definition of continuity) and because of the convergence of ${}{\left(x_{n}\right)}_{n\in \mathbb {N} }$ to ${}x$ there exists a natural number ${}n_{0}$ such that for all ${}n\geq n_{0}$ the estimate

${}d(x_{n},x)\leq \delta \,$

holds. By the choice of ${}\delta$ we have

$d(f(x_{n}),f(x))\leq \epsilon {\text{ for all }}n\geq n_{0},$

so that the image sequence converges to ${}f(x)$.

Suppose now that (2) is fulfilled. We assume that ${}f$ is not continuous. Then there exists an ${}\epsilon >0$ such that for all ${}\delta >0$ there exist elements ${}z\in D$ such that their distance to ${}x$ is at most ${}\delta$, but such that the distance of their value ${}f(z)$to ${}f(x)$ is larger than ${}\epsilon$. This holds in particular for every ${}\delta =1/n$, ${}n\in \mathbb {N} _{+}$. This means that for every natural number ${}n\in \mathbb {N} _{+}$, there exists a ${}x_{n}\in D$ with

$d(x_{n},x)\leq {\frac {1}{n}}{\text{ and with }}d(f(x_{n}),f(x))>\epsilon .$

This sequence ${}{\left(x_{n}\right)}_{n\in \mathbb {N} }$ converges to ${}x$, but the image sequence does not converge to ${}f(x)$, since the distance of its members to ${}f(x)$ is always at least ${}\epsilon$. This contradicts condition (2).

$\Box$

Rules for continuous functions

LemmaCreate referencenumber

Let ${}D\subseteq \mathbb {R}$ and ${}E\subseteq \mathbb {R}$ be subsets and let

$f\colon D\longrightarrow \mathbb {R}$

and

$g\colon E\longrightarrow \mathbb {R}$

denote functions with

${}f(D)\subseteq E$. Then the following statements hold.

1. If ${}f$ is continuousMDLD/continuous (R) in ${}x\in D$ and ${}g$ is continuous in ${}f(x)$, then also the compositionMDLD/composition ${}g\circ f$ is continuous in ${}x$.
2. If ${}f$ and ${}g$ are continuous, so is ${}g\circ f$.

Proof

The first statement follows from Lemma 10.4 . This implies also the second statement.

$\Box$

LemmaLemma 10.6 change

Let ${}D\subseteq \mathbb {R}$ be a subset and let

$f,g\colon D\longrightarrow \mathbb {R}$

be continuous functions.MDLD/continuous functions (R) Then also the functions

$f+g\colon D\longrightarrow \mathbb {R} ,x\longmapsto f(x)+g(x),$

$f-g\colon D\longrightarrow \mathbb {R} ,x\longmapsto f(x)-g(x),$

$f\cdot g\colon D\longrightarrow \mathbb {R} ,x\longmapsto f(x)\cdot g(x),$

are continuous. For a subset ${}U\subseteq D$ such that ${}g$ has no zero in ${}U$, also

$f/g\colon U\longrightarrow \mathbb {R} ,x\longmapsto f(x)/g(x),$

is continuous.

Proof

This follows from Lemma 10.4 and Lemma 8.1 .

$\Box$

CorollaryCorollary 10.7 change

Polynomial functionsMDLD/Polynomial functions (K)

$P\colon \mathbb {R} \longrightarrow \mathbb {R} ,x\longmapsto P(x),$

are continuous.MDLD/continuous (R)

Proof

Due to Example 10. and Lemma 10.6 , the powers

$\mathbb {R} \longrightarrow \mathbb {R} ,x\longmapsto x^{n},$

are continuous for every ${}n\in \mathbb {N}$. Hence, also the functions

$\mathbb {R} \longrightarrow \mathbb {R} ,x\longmapsto ax^{n},$

are continuous for every ${}a\in \mathbb {R}$ and therefore, again due to Lemma 10.6 , the functions

$\mathbb {R} \longrightarrow \mathbb {R} ,x\longmapsto a_{n}x^{n}+a_{n-1}x^{n-1}+\cdots +a_{1}x+a_{0},$

are continuous.

$\Box$

Rational functions are continuous on their domain.

CorollaryCreate referencenumber

Let ${}P,Q\in \mathbb {R} [X]$ be polynomialsMDLD/polynomials (K) and let ${}U:={\left\{x\in \mathbb {R} \mid Q(x)\neq 0\right\}}$. Then the rational functionMDLD/rational function (R)

$U\longrightarrow \mathbb {R} ,x\longmapsto {\frac {P(x)}{Q(x)}},$

is

continuous.MDLD/continuous (R)

Proof

This follows from Corollary 10.7 and Lemma 10.6 .

$\Box$

Limit of a function

Quite often, functions are not defined in certain points, e.g. if the used functional term is not defined there. However, it makes a huge difference whether only the functional term is not defined in a point, but has a useful (continuous) extension, or whether the function does not have a useful extension in this point, e.g. because it has a pole or an even more chaotic behavior. The following concept is in particular relevant for the definition of differentiability (if the difference quotient has a useful limit, then it is called differential quotient).

Definition

Let ${}T\subseteq \mathbb {R}$ denote a subset and ${}a\in \mathbb {R}$ a point. Let

$f\colon T\longrightarrow \mathbb {R}$

be a function.MDLD/function Then ${}b\in \mathbb {R}$ is called limit of ${}f$ in ${}a$, if for every ${}\epsilon >0$ there exists some ${}\delta >0$ such that for all ${}x\in T$ fulfilling

${}\vert {x-a}\vert \leq \delta \,,$

the estimate

${}\vert {f(x)-b}\vert \leq \epsilon \,$

holds. In this case, we write

${}\operatorname {lim} _{x\rightarrow a}\,f(x)=b\,.$

This concept is basically only useful if there exists at least some sequence within ${}T$ converging to ${}a$. A typical situation is the following: Let ${}I$ denote a real interval, ${}a\in I$ a point and let ${}T=I\setminus \{a\}$. The function is defined on ${}T$ but not in ${}a$, and we are dealing with the question whether ${}f$ can be extended to a function ${}{\tilde {f}}$ defined on the whole ${}I$. Here, ${}{\tilde {f}}(a)$ should be determined by ${}f$.

LemmaLemma 10.10 change

Let ${}T\subseteq \mathbb {R}$ denote a subset and ${}a\in \mathbb {R}$ a point. Let

$f\colon T\longrightarrow \mathbb {R}$

be a functionMDLD/function and ${}b\in \mathbb {R}$

a point. Then the following statements are equivalent.

1. We have

   ${}\operatorname {lim} _{x\rightarrow a}\,f(x)=b\,.$

2. For every sequence ${}{\left(x_{n}\right)}_{n\in \mathbb {N} }$ in ${}T$ which convergesMDLD/converges (R) to ${}a$, also the image sequence ${}{\left(f(x_{n})\right)}_{n\in \mathbb {N} }$ converges to ${}b$.

Proof

See Exercise 10.24 .

$\Box$

This implies for a continuous function ${}f\colon T\rightarrow \mathbb {R}$ that it can be extended to a continuous function ${}{\tilde {f}}\colon T\cup \{a\}\rightarrow \mathbb {R}$ (by ${}{\tilde {f}}(a)=b$) if and only if the limit of ${}f$ in ${}a$ equals ${}b$.

LemmaLemma 10.11 change

Let ${}T\subseteq \mathbb {R}$ denote a subset and ${}a\in \mathbb {R}$ a point. Let ${}f\colon T\rightarrow \mathbb {R}$ and ${}g\colon T\rightarrow \mathbb {R}$ denote functions,MDLD/functions such that the limitsMDLD/limits (function R) ${}\operatorname {lim} _{x\rightarrow a}\,f(x)$ and ${}\operatorname {lim} _{x\rightarrow a}\,g(x)$

exist. Then the following statements hold.

1. The sum ${}f+g$ has in ${}a$ the limit

   ${}\operatorname {lim} _{x\rightarrow a}\,(f(x)+g(x))=\operatorname {lim} _{x\rightarrow a}\,f(x)+\operatorname {lim} _{x\rightarrow a}\,g(x)\,.$

2. The product ${}f\cdot g$ has in ${}a$ the limit

   ${}\operatorname {lim} _{x\rightarrow a}\,(f(x)\cdot g(x))=\operatorname {lim} _{x\rightarrow a}\,f(x)\cdot \operatorname {lim} _{x\rightarrow a}\,g(x)\,.$

3. Suppose that ${}g(x)\neq 0$ for all ${}x\in T$ and ${}\operatorname {lim} _{x\rightarrow a}\,g(x)\neq 0$. Then the quotient ${}f/g$ has in ${}a$ the limit

   ${}\operatorname {lim} _{x\rightarrow a}\,{\frac {f(x)}{g(x)}}={\frac {\operatorname {lim} _{x\rightarrow a}\,f(x)}{\operatorname {lim} _{x\rightarrow a}\,g(x)}}\,.$

Proof

This follows from Lemma 10.10 and from Lemma 8.1 .

$\Box$

Example Create referencenumber

We consider the limit

$\operatorname {lim} _{x\rightarrow 0}\,{\frac {{\sqrt {x+4}}-2}{x}},$

where ${}x\in \mathbb {R} \setminus \{0\},\,x\geq -4$. For ${}x=0$, this term is not defined, and from this term one can not read of directly whether the limit exists. It is however possible to multiply the numerator and the denominator by ${}{\sqrt {x+4}}+2$, then we get

${}{\begin{aligned}{\frac {{\sqrt {x+4}}-2}{x}}&={\frac {{\left({\sqrt {x+4}}-2\right)}{\left({\sqrt {x+4}}+2\right)}}{x{\left({\sqrt {x+4}}+2\right)}}}\\&={\frac {x+4-4}{x{\left({\sqrt {x+4}}+2\right)}}}\\&={\frac {x}{x{\left({\sqrt {x+4}}+2\right)}}}\\&={\frac {1}{{\sqrt {x+4}}+2}}.\end{aligned}}$

Due to the rules for limits, we can determine the limit in the numerator and in the denominator separately, where for the denominator we use the continuity of the square root according to Exercise 10.4 . Hence, the limit is ${}1/4$.

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