MATH 409 Lecture 11

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Lecture Slides

Continuous Functions

Review

• Definition of Continuity
• Extreme Value Theorem

Theorem. Any polynomial of odd degree has at least one real root.

Proof. Let $p(x)=a_{n}\,x^{n}+a_{n-1}\,x^{n-1}+\dots +a_{1}\,x+a_{0}$ be a polynomial of positive degree $n$. Note that $a_{n}\neq 0$. For any $x\neq 0$, we have

${\frac {p(x)}{a_{n}\,x^{n}}}=1+{\frac {a_{n-1}}{a_{n}\,x}}+\dots +{\frac {a_{1}}{a_{n}\,x^{n-1}}}+{\frac {a_{0}}{a_{n}\,x^{n}}}$

which converges to 1 as $x\to \pm \infty$. As a consequence, there exists $C>0$ such that ${\frac {p(x)}{a_{n}\,x^{n}}}\geq {\frac {1}{2}}$ if $\left|x\right|\geq C$. In particular, the numbers $p(x)$ and $a_{n}\,x^{n}$ are of the same sign if $\left|x\right|\geq C$. In the case $n$ is odd, this implies that one of the numbers $p(C)$ and $p(-C)$ is positive while the other is negative. By the Intermediate Value Theorem, we have $p(x)=0$ for some $x\in \left[-C,C\right]$.

quod erat demonstrandum

Continuity Over Domain

Theorem. Given a function $f:(a,b)\to \mathbb {R}$ and a point $c\in (a,b)$, let $f_{1}$ denote the restriction of $f$ to the interval $\left(a,c\right]$ and $f_{2}$ denote the restriction of $f$ to $\left[c,b\right)$.

The function $f$ is continuous if and only if both restrictions $f_{1}$ and $f_{2}$ are continuous.

Proof. For any $x\in (a,c)$, the continuity of $f$ at $x$ is equivalent to the continuity of $f_{1}$ at $x$. Likewise, the continuity of $f$ at a point $y\in (c,b)$ is equivalent to the continuity of $f_{2}$ at $y$. The function is continuous at $c$ if $f(x)\to f(c)$ as $x\to c$. The restriction $f_{1}$ is continuous at $c$ if $f(x)\to f(c)$ as $x\to c-$. The restriction $f_{2}$ is continuous at $c$ if $f(x)\to f(c)$ as $x\to c+$. Therefore $f$ is continuous at $c$ if and only if both $f_{1}$ and $f_{2}$ are continuous at $c$.

quod erat demonstrandum

For example, the function $f(x)=\left|x\right|$ is continuous on $\mathbb {R}$. Indeed $f$ coincides with $g(x)=x$ on $\left[0,\infty \right)$ and with the function $h(x)=-x$ on $\left(-\infty ,0\right]$.

Continuity of Compositions

Theorem. Let $f:E_{1}\to \mathbb {R}$ and $g:E_{2}\to \mathbb {R}$ be two functions. if $f(E_{1})\subseteq E_{2}$, then $(g\circ f)(x)$ is a well-defined function on $E_{1}$.

If $f$ is continuous at a point $c\in E_{1}$ and $g$ is continuous at $f(c)$, then $g\circ f$ is continuous at $c$.

Proof. Let us use the sequential characterization of continuity. Consider an arbitrary sequence $\left\{x_{n}\right\}\subset E_{1}$ converging to $c$. We have to show that

$\lim _{n\to \infty }(g\circ f)(x_{n})=(g\circ f)(c)\,\!$

Since the function $f$ is continuoous at $c$, we obtain that $f(x_{n})\to f(c)$ as $n\to \infty$. Moreover, all elements of the sequence $\left\{f(x_{n})\right\}$ belong to the set $E_{2}$. Since the function $g$ is continuous at $f(c)$, we obtain that $g(f(x_{n}))\to g(f(c))$ as $n\to \infty$.

quod erat demonstrandum

Examples

If a function $f:E\to \mathbb {R}$ is continuous at a point $c\in E$, then a function $g=\left|f(x)\right|$, $x\in E$, is also continuous at $c$.

Indeed, the function $g$ is the composition of $f$ with the continuous function $h(x)=\left|x\right|$.

If functions $f,g:E\to \mathbb {R}$ are continuous at a point $c\in E$, then functions $\max {(f,g)}$ and $\min {(f,g)}$ are also continuous at $c$.

• Indeed, $2\,\max(f(x),g(x))=f(x)+g(x)+\left|f(x)-g(x)\right|$ for all $x\in E$, and
• $2\,\min(f(x),g(x))=f(x)+g(x)-\left|f(x)-g(x)\right|$ for all $x\in E$

Trigonometric Functions

For a unit circle with angle $\theta$ measured ccw from the point $(1,0)$, we define

• $\sin {\theta }=y$
• $\cos {\theta }=x$
• $\tan {\theta }={\frac {y}{x}}$

Theorem. $0\leq \sin {\theta }\leq \theta \leq \tan {\theta }$ for all $\theta \in \left[0,{\frac {\pi }{2}}\right)$.

These are usually proved using areas, but we will use length.

Proof. The length of $\sin {\theta }$ is the length of the line from $(x,y)$ to the $x$-axis, $\theta$ is the length of the arc from $(1,0)$ to $(x,y)$, and $\tan {\theta }$ is the length of the segment from $(1,0)$ to $(1,y')$, where $\left\langle x,y\right\rangle$ is parallel to $\left\langle 1,y'\right\rangle$.

We approximate the length of an arc by summing the lengths of a bunch of lines next to the arc (similar to how $\pi$ was first estimated)

quod erat demonstrandum

Theorem. [Continuity of sine]. $\sin {x}$ is continuous for all $x\in \mathbb {R}$.

We know that $0\leq \sin {\theta }\leq \theta$ for all $\theta \in \left[0,{\frac {\pi }{2}}\right)$. Since $-\sin {\theta }=\sin(-\theta )$, we get $\left|\sin {\theta }\right|\leq \left|\theta \right|$ if $\theta <{\frac {\pi }{2}}$. In the case $\left|\theta \right|\geq {\frac {\pi }{2}}$, this estimate holds too as $\left|\sin {\theta }\right|\leq 1<{\frac {\pi }{2}}$. Now using the trigonometric formula,

$\sin {x}-\sin {c}=2\,\sin {\frac {x-c}{2}}\,\cos {\frac {x+c}{2}}$

We obtain

$\left|\sin {x}-\sin {c}\right|\leq 2\left|\sin {\frac {x-c}{2}}\right|\,\left|\cos {\frac {x+c}{2}}\right|\leq 2\left|{\frac {x-c}{2}}\right|=\left|x-c\right|$

It follows that $\sin {x}\to \sin {c}$ as $x\to c$ for every $c\in \mathbb {R}$. That is, the function $\sin {x}$ is continuous.

quod erat demonstrandum

Theorem. [Continuity of cosine.] $\cos {x}$ is continuous for all $x\in \mathbb {R}$

Proof. Since $\cos {x}=\sin \left(x+{\frac {\pi }{2}}\right)$ for all $x\in \mathbb {R}$, the function $f$ is a composition of two continuous functions $g(x)=x+{\frac {\pi }{2}}$ and $h(x)=\sin {x}$.

quod erat demonstrandum

Theorem. [Continuity of tangent]. $\tan {x}$ is continuous for all $x\in \mathbb {R}$

Proof. Since $f(x)={\frac {\sin {x}}{\cos {x}}}$, the function $f$ is continuous on its tentire domain $\mathbb {R} \setminus \left\{x\in \mathbb {R} ~\mid ~\cos {x}\right\}=\mathbb {R} \setminus \left\{{\frac {\pi }{2}}+\pi \,k~\mid ~k\in \mathbb {Z} \right\}$

quod erat demonstrandum

$f(0)=1$ and $f(x)={\frac {\sin {x}}{x}}$ for $x\neq 0$.

Proof. Since $\sin {x}$ and the identity functions are continuous, it follows that if $f$ is continuous on $\mathbb {R} \setminus \left\{0\right\}$. Further, we know that $0\leq \sin {x}\leq x\leq \tan {x}$ for $0\leq x\leq {\frac {\pi }{2}}$. Therefore $\cos {x}\leq {\frac {\sin {x}}{x}}\leq 1$ Since $\cos {0}=1$, the squeeze theorem implies that $f(x)\to 1$ as $x\to 0+$. The left-hand limit at $0$ is the same as $f(-x)=f(x)$ for all $x\in \mathbb {R}$. Thus the function $f$ is continuous at $0$ as well.

quod erat demonstrandum

Monotone Functions

Let $f:E\to \mathbb {R}$ be a function defined on a set $E\subset \mathbb {R}$.

The function $f$ is increasing if, for any $x,y\in E$, $x<y$ implies $f(x)\leq f(y)$.

It is called strictly increasing if $x<y$ implies $f(x)<f(y)$.

Similarly, $f$ is decreasing if, for any $x,y\in E$, $x>y$ implies $f(x)\geq f(y)$.

It is called strictly decreasing if $x>y$ implies $f(x)>f(y)$.

Increasing and decreasing functions are called monotone, and strictly increasing and strictly decreasing functions are called strictly monotone.

Theorem. Any monotone function defined on an open interval can have only jump discontinuities.

Theorem. A monotone function $f$ defined on an interval $I$ is continuous if and only if the image $f(I)$ is also an interval.

Theorem. A continuous function defined on a closed interval is one-to-one if and only if it is strictly monotone.

Continuity of Inverse Functions

Suppose $f:E\to \mathbb {R}$ is a strictly monotone function defined on a set $E\subset \mathbb {R}$. Then $f$ is one-to-one on $E$. Then $f$ is one-to-one on $E$ so that the inverse function $f^{-1}$ is a well-defined function on $f(E)$.

Theorem. If the domain $E$ of a strictly monotone functino $f$ is a closed interval and $f$ is continuous on $E$, then the image $f(E)$ is also a closed interval, and teh inverse function $f^{-1}$ is strictly monotone and continuous on $f(E)$.

Proof. Since $f$ is continuous on the closed interval $E$, it follows from the extreme value and intermediate value theorems that $f(E)$ is also a closed interval. The inverse function $f^{-1}$ is strictly monotone since $f$ is strictly monotone. By construction, $f^{-1}$ maps the interval $f(E)$ onto the interval $E$, which implies that $f^{-1}$ is continuous.

quod erat demonstrandum

Examples

Power function. $f(x)=x^{n}$ for $x\in \mathbb {R}$ and $n\in \mathbb {N}$.

The function $f$ is continuous on $\mathbb {R}$. It is strictly increasing on the interval $\left[0,\infty \right)$ anf $f([0,\infty ))=[0,\infty )$. In the case $n$ is odd, the function $f$ is strictly increasing on $\mathbb {R}$ and $f(\mathbb {R} )=\mathbb {R}$. We conclude that the inverse function $f^{-1}(x)=x^{\frac {1}{n}}$ is a continuous function on $[0,\infty )$ if $n$ is even and a continuous function on $\mathbb {R}$ if $n$ is odd.

$f(x)=x^{n}$ for $x\in \mathbb {R} \setminus \left\{0\right\}$ and $n\in \mathbb {Z}$.

The function $f$ is strictly decreasing on $(0,\infty )$. It is continuous on $(0,\infty )$ and maps this interval onto itself. Therefore the inverse function $f^{-1}(x)=x^{\frac {1}{n}}$ is a continuous function on $(0,\infty )$.