Definition 1.4.1. A function $f$ is continuous at $x=a$ if
\[\lim_{x\to a}f(x)=f(a).\]And, $f$ is discontinuous at $x=a$ if $f$ is not continuous at $x=a$.
Example 1.4.2. Where is the following function discontinuous?
\[f(x) = \left \lfloor x \right \rfloor.\]Solution. Let $a\in \mathbb R$ be given and $n=\left \lfloor a \right \rfloor$. The necessary and sufficient condition for $(\forall \varepsilon>0\ \vert f(x)-f(a)\vert<\varepsilon)$ is $(n\le x<n+1)$. The necessary and sufficient condition for $(0<\vert x-a\vert <\delta \Rightarrow n\le x<n+1)$ is $(n\le a-\delta \wedge a+\delta \le n+1)$, which is equivalent to $\delta\le \min\lbrace a-n,n+1-a\rbrace$. And $a-n>0$ is the necessary and sufficient condition for this. Thus, $n<a<n+1$ is the necessary and sufficient condition for ‘$f$ is continuous at $x=a$’. Therefore, $a\not\in \mathbb Z$ is the necessary and sufficient condition for continuity at $x=a$, and, in other words, $f$ is discontinuous at integers.
Definition 1.4.3. A function $f$ is continuous from the right at $x=a$ if
\[\lim_{x\to a^+}f(x)=f(a),\]and $f$ is continuous from the left at $x=a$ if
\[\lim_{x\to a^-}f(x)=f(a).\]Theorem 1.4.4. If $f$ and $g$ are continuous at $x=a$ and $c$ is a constant, then the functions $f+g, f-g, c\cdot f, f\cdot g,$ and (if $g(a)\neq 0$) $f/g$ are also continuous at $x=a$.
Definition 1.4.5. A function $f$ is continuous on an interval $I$ if it is continuous at every $x\in I$. (If $f$ is defined only on one side of an endpoint of $I$, we understand continuous at the endpoint to mean continuous from the right or continuous from the left.)
Theorem 1.4.6. If $f$ and $g$ are continuous on an interval $I$, then so are the functions $f+g, f-g, c\cdot f, f\cdot g,$ and (if $g$ is never $0$) $f/g$.
Definition 1.4.7. A function $f$ is a continuous function if $f$ is continuous on its domain.
Theorem 1.4.8. The following types of functions are continuous functions.
- polynomial functions
- rational functions
- root functions
- trigonometric functions
Theorem 1.4.9. If functions $f$ and $g$ are continuous function, then the following functions are also continuous functions.
- $f+g$
- $f-g$
- $c\cdot f$ (where $c$ is a constant)
- $f\cdot g$
- $f/g$
- $f\circ g$
Example 1.4.10. Let $F : D \to \mathbb R$ be the function such that
\[F(x)=\frac{1}{\sqrt{x^2+7}-4}\]for every input $x\in D$, where $D=\lbrace x\in \mathbb R : x\neq \pm 3\rbrace$. Where is $F$ continuous?
Solution. For every $x\in D$
where $f(x)=1/x, g(x)=x-4, h(x)=\sqrt x,$ and $k(x)=x^2+7$. That is, $F=f\circ g\circ h\circ k$. Because $f, g, h, k$ are continuous functions, $f\circ g\circ h\circ k$ is also a continuous function by Theorem 1.4.9. Thus $F$ is a continuous function.
Theorem 1.4.11 (The Intermediate Value Theorem). Suppose that a function $f$ is continuous on an closed interval $[a,b]$ with $f(a)\neq f(b)$. Then, for any number $k$ between $f(a)$ and $f(b)$, there exists a number $x$ in $(a,b)$ such that
\[f(x)=k.\]Remark 1.4.12. function concept을 도입해서 input-function-output 구조를 생각하지 않고, 간단히 independent-dependent만으로 볼 때 ‘for any $k$, there exists $a$ such that if $x=a$ then $f(a)=k$’로 의미를 이해할 수 있다.
Example 1.4.13. Show that there is a solution of the equation
\[4x^3-6x^2+3x-2=0\]between $1$ and $2$.
Solution. Let $f(x)=4x^3-6x^2+3x-2$. Then $f$ is a continuous function. So, it is enough to show that $0$ is between $f(1)$ and $f(2)$, and because $f(1)=-1$ and $f(2)=12$ it’s simply true. Thus, by Theorem 4.11, there exists such a number $x$ in $(1,2)$.
Exercises
Exercise 1.4.1. Prove that the function $f(x)=1-\sqrt{1-x^2}$ is continuous on the interval $[-1,1]$.
Proof. We first prove two elementary estimates.
Lemma 1. Let $x\in[-1,1]$ and $a\in[0,1)$. Then
Proof of Lemma 1. Since $a\in[0,1)$, we have $\sqrt{1-a^2}>0$, so the denominator is positive. Also, since $x\in[-1,1]$ and $a\geq 0$, we have $\vert x+a\vert \leq 1+a$. Therefore,
\[\frac{|x+a|}{\sqrt{1-x^2}+\sqrt{1-a^2}} \leq \frac{1+a}{\sqrt{1-a^2}}=\sqrt{\frac{1+a}{1-a}}. \tag*{\(\Diamond\)}\]Lemma 2. Let $x\in[-1,1]$ and $a\in(-1,0)$. Then
\[\frac{|x+a|}{\sqrt{1-x^2}+\sqrt{1-a^2}} \leq \sqrt{\frac{1-a}{1+a}}.\]Proof of Lemma 2. Since $a\in(-1,0)$, we have $\sqrt{1-a^2}>0$, so the denominator is positive. Also, since $x\in[-1,1]$ and $a<0$, we have $\vert x+a\vert \leq 1-a$. Therefore,
\[\frac{|x+a|}{\sqrt{1-x^2}+\sqrt{1-a^2}} \leq \frac{1-a}{\sqrt{1-a^2}}=\sqrt{\frac{1-a}{1+a}}. \tag*{\(\Diamond\)}\]We now prove that $f$ is continuous on $[-1,1]$. We prove continuity relative to the domain $[-1,1]$. Let $a\in[-1,1]$ and let $\varepsilon>0$ be given. First suppose that $a\in(-1,1)$. If $a\in[0,1)$, choose
\[\delta=\varepsilon\sqrt{\frac{1-a}{1+a}}.\]Suppose that $x\in[-1,1]$ and $\vert x-a\vert <\delta$. Then, by Lemma 1, we have
\[\begin{aligned} |f(x)-f(a)| &= \left| \left(1-\sqrt{1-x^2}\right) - \left(1-\sqrt{1-a^2}\right) \right| \\ &= \frac{|a^2-x^2|} {\sqrt{1-x^2}+\sqrt{1-a^2}} \\ &< \delta\sqrt{\frac{1+a}{1-a}} \\ &= \varepsilon. \end{aligned}\]Therefore $f$ is continuous at $a$. If $a\in(-1,0)$, choose
\[\delta=\varepsilon\sqrt{\frac{1+a}{1-a}}.\]Suppose that $x\in[-1,1]$ and $\vert x-a\vert <\delta$. Then, by Lemma 2, we have
\[\begin{aligned} |f(x)-f(a)| &= \left| \left(1-\sqrt{1-x^2}\right) - \left(1-\sqrt{1-a^2}\right) \right| \\ &= \frac{|a^2-x^2|} {\sqrt{1-x^2}+\sqrt{1-a^2}} \\ &< \delta\sqrt{\frac{1-a}{1+a}} \\ &= \varepsilon. \end{aligned}\]Therefore $f$ is continuous at $a$. Thus $f$ is continuous at every point $a\in(-1,1)$.
It remains to check the endpoints. At $a=-1$, we have $f(-1)=1$. Let $\varepsilon>0$ be given. Choose $\delta={\varepsilon^2}/{2}$. Suppose that $x\in[-1,1]$ and $\vert x-(-1)\vert =\vert x+1\vert <\delta.$ Since $x\in[-1,1]$, we have $x+1\geq 0$ and $1-x\leq 2$. Hence
\[\begin{aligned} |f(x)-f(-1)| &= \left|1-\sqrt{1-x^2}-1\right| \\ &= \sqrt{(1-x)(1+x)} \\ &\leq \sqrt{2(1+x)} \\ &= \sqrt{2|x+1|} \\ &< \sqrt{2\delta} \\ &= \varepsilon. \end{aligned}\]Therefore $f$ is continuous from the right at $-1$. At $a=1$, we have $f(1)=1$. Let $\varepsilon>0$ be given. Choose $\delta={\varepsilon^2}/{2}$. Suppose that $x\in[-1,1]$ and $\vert x-1\vert <\delta$. Since $x\in[-1,1]$, we have $1-x\geq 0$ and $1+x\leq 2$. Hence
\[\begin{aligned} |f(x)-f(1)| &= \left|1-\sqrt{1-x^2}-1\right| \\ &= \sqrt{(1-x)(1+x)} \\ &\leq \sqrt{2(1-x)} \\ &= \sqrt{2|x-1|} \\ &< \sqrt{2\delta} \\ &= \varepsilon. \end{aligned}\]Therefore $f$ is continuous from the left at $1$. Hence, $f$ is continuous on $[-1,1]$.$\square$
Exercises 1.4.2. Prove that if functions $f$ and $g$ are continuous at $x=a$, then $f+g$ is also continuous at $x=a$.
Solution. 먼저 continuous는 ‘when $x$ approaches to $a$, $f(x)$ approaches to $f(a)$’를 의미한다. 그리고 $f+g$는 input $x$에 대하여 $f(x)+g(x)$를 출력하는 function이다. 그렇다면 ‘when $x$ approaches to $a$, $f(x)+g(x)$ approaches to $f(a)+g(a)$’를 보이면 되는 것이다. 그런데 이는 자명하다: Because $f(x)$ approaches to $f(a)$ and $g(x)$ approaches to $g(a)$, $f(x)+g(x)$ approaches to $f(a)+g(a)$.
Exercise 1.4.3. Prove that any polynomial is continuous everywhere.
Solution. $f$가 polynomial function이라는 것의 의미는 ‘there exist numbers $n\in \mathbb N$ and $c_n,\dots, c_0\in \mathbb R$ such that for all input $x$, $f(x)=c_nx^n+c_{n-1}x^{n-1}+\cdots+c_1x+c_0$’이다. 그렇다면, $f$가 polynomial function이라면, 모든 $a\in \mathbb R$에 대하여 ‘$f$ is continuous at $x=a$’가 성립할까? 성립하며 이유는 다음과 같다: Theorem 1.3.6과 Theorem 1.3.7에 의하여
Exercise 1.4.4. Let $f:\mathbb R \to \mathbb R$ be the function such that
\[f(x)=\frac{\sin x}{2 + \cos x}\]for every input $x\in \mathbb R$. Evaluate $\lim_{x\to \pi}f(x)$.
Solution. $g(x)=\sin x, h(x)=2+\cos x$로 둘 때, $f=g/h$이다. 그리고 Theorem 1.4.8과 Theorem 1.4.9로부터 $g$와 $h$가 continuous function이며 따라서 $f$도 continuous function임을 알 수 있다. $\pi \in \operatorname{dom}(f)$이므로 $\lim_{x\to \pi}f(x)=f(\pi)=0$.