MATH 409 Lecture 10

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Lecture Slides

Continuity

Given set $E\subset \mathbb{ℝ}$, a function $f:E\to \mathbb{ℝ}$, and a point $c\in E$, the function $f$ is continuous at $c$ if for any $ϵ>0$ there exists $\delta =\delta (ϵ)>0$ such that $|x-c|<\delta$ and $x\in E$ imply $|f(x)-f(c)|<ϵ$

We say that the function $f$ is continuous on a set $E_0\subset E$ if $f$ is continuous at every point $c\in E_0$. The function $f$ is continuous if it is continuous on the entire domain $E$.

Note:

• In the case $E=(a,b)$, $f$ is continuous at $c\in E$ if and only if $f(c)=\underset{x\to c}{\lim }f(x)$

Properties

Theorem. A function $f:E\to \mathbb{ℝ}$ is continuous at $c\in E$ if and only if for any sequence $\left\{x_n\right\}$ of elements of $E$, $x_n\to c$ as $n\to \mathrm{\infty }$ implies $f(x_n)\to f(c)$ as $n\to \mathrm{\infty }$.

Theorem. Suppose that functions $f,g:E\to \mathbb{ℝ}$ are both continuous at $c\in E$. Then $(f+g)$, $(f-g)$ and $fg$ are also continuous at $c$. If, additionally, $g(c)\ne 0$, then $\frac{f}{g}$ is continuous at $c$ as well.

Bounded Functions

A function $f:E\to \mathbb{ℝ}$ is bounded on a subset $E_0\subset E$ if there exists $C>0$ such that $|f(x)|\le C$ for all $x\in E_0$. In the case $E_0=E$, we say that $f$ is bounded.

The function $f$ is bounded above on $E_0$ if there exists $C\in \mathbb{ℝ}$ such that $f(x)\le C$ for all $x\in E_0$.

It is bounded below on $E_0$ if there exists $C\in \mathbb{ℝ}$ such that $f(x)\ge C$ for all $x\in E_0$.

Equivalently, $f$ is bounded on $E_0$ if the image $f(E_0)$ is a bounded subset of $\mathbb{ℝ}$. Likewise, the function $f$ is bounded above on $E_0$ if the image $f(E_0)$ is bounded above. It is bounded below on $E_0$ if $f(E_0)$ is bounded below.

Example

$h:\mathbb{ℝ}\to \mathbb{ℝ}$ $h(0)=0$, $h(x)=\frac{1}{x}$ for $x\ne 0$

The function $h$ is unbounded. At the same time, it is bounded on $[1,\mathrm{\infty })$ and on $(-\mathrm{\infty },-1]$.

Theorem. If $I=[a,b]$ is a closed, bounded interval of the real line, then any continuous function $f:I\to \mathbb{ℝ}$ is bounded.

Proof. Assume that $f:I\to \mathbb{ℝ}$ is unbounded. Then for every $n\in \mathbb{\mathbb{N}}$ there exists a point $x_n\in I$ such that $|f(x_n)|>n$. We obtain a sequence $\left\{x_n\right\}$ of elements of $I$ such that the sequence $\{f(x_n)\}$ diverges to infinity.

Since the interval $I$ is bounded, the sequence $\left\{x_n\right\}$ has a convergent subsequence $\left\{x_{n_k}\right\}$ (due to the Bolano-Weierstrass Theorem). Let $c=\underset{k\to \mathrm{\infty }}{\lim }{x}_{n_k}$. Then $c\in [a,b]$ (due to comparison theorem). Since the sequence $\{f(x_{n_k})\}$ is a subsequence of $\{f(x_n)\}$, it diverges to infinity. In particular, it does not converge to $f(c)$. It follows that $f$ is discontinuous at $c$.

Thus any continuous function on $[a,b]$ has to be bounded.

quod erat demonstrandum

Discontinuities

A function $f:E\to \mathbb{ℝ}$ is discontinuous at a point $c\in E$ if it is not continuous at $c$. There are various kinds of discontinuities including the following ones.

• $f$ has a jump discontinuity at $c$ if both one-sided limits at $c$ exist, but they are not equal: $\underset{x\to c-}{\lim }f(x)\ne \underset{x\to c+}{\lim }f(x)$.
• $f$ has a removable discontinuity at a point $c$ if the limit at $c$ exists, but it is different from the value at $c$: $\underset{x\to c}{\lim }f(x)\ne f(c)$.
• If $f$ is continuous at $c$, then it is locally bounded at $c$, which means that $f$ is bounded on the set $(c-\delta ,c+\delta )\cap E$ provided $\delta >0$ is small enough. Hence any function not locally bounded at $c$ is discontinuous at $c$.

Examples

Constant function. $f(x)=a$ for all $x\in \mathbb{ℝ}$ and some $a\in \mathbb{ℝ}$.

Since $\underset{x\to c}{\lim }f(x)=a$ for all $c\in \mathbb{ℝ}$, $f$ is continuous.

Identity function. $f(x)=x$ for all $x\in \mathbb{ℝ}$.

Since $\underset{x\to c}{\lim }f(x)=c$ for all $c\in \mathbb{ℝ}$, the function is continuous.

Step function. $f(x)=\{\begin{array}{ll}1& x>0\\ 0& x\le 0\end{array}$

Since $\underset{x\to 0-}{\lim }f(x)=0$ and $\underset{x\to 0+}{\lim }f(x)=1$, the function has a jump discontinuity at $0$. It is continuous on $\mathbb{ℝ}\setminus \left\{0\right\}$.

Harmonic function. $f(0)=0$ and $f(x)=\frac{1}{x}$ for $x\ne 0$

Since $\underset{x\to c}{\lim }f(x)=\frac{1}{c}$ for all $c\ne 0$, the function $f$ is continuous on $\mathbb{ℝ}\setminus \left\{0\right\}$. It is discontinuous at $0$ because it is not locally bounded at 0.

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

Since $\underset{x\to 0+}{\lim }f(x)$ does not exist, the function is discontinuous at 0. Notice it is neither jump nor removable discontinuous, and the function $f$ is bounded. The reason for discontinuity is oscillation.

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

Since $\underset{x\to 0}{\lim }f(x)=0$, the function is continuous at 0.

Dirichlet function. $f(x)=\{\begin{array}{ll}1& xin\mathbb{\mathbb{Q}}\\ 0& x\in \mathbb{ℝ}\setminus \mathbb{\mathbb{Q}}\end{array}$.

Since $\underset{x\to c}{\lim }f(x)$ never exists, the function has no points of continuity.

Riemann function. $f(x)=\{\begin{array}{ll}\frac{1}{q}& x=\frac{p}{q}\\ 0& x\in \mathbb{ℝ}\setminus \mathbb{\mathbb{Q}}\end{array}$

Since $\underset{x\to c}{\lim }f(x)=0$ for all $c\in \mathbb{ℝ}$, the function $f$ is continuous at irrational points and discontinuous at rational points. Moreover, all discontinuities are removable.

Extreme Value Theorem

Theorem. [Extreme Value Theorem]. if $I=[a,b]$ is a closed, bounded interval of the real line, then any continuous function $f:I\to \mathbb{ℝ}$ attains its extreme values (maximum and minimum) on $I$. To be precise, there exist points $x_{min}$ and $x_{max}$ in $I$ such that

$f(x_{min})\le f(x)\le f(x_{max})$ for all $x\in I$

Proof. Since $f$ is continuous, it is bounded on $I$. Hence $m=\underset{x\in I}{\inf }f(x)$ and $M=\underset{x\in I}{\sup }f(x)$ are well-defined numbers. In different notation, $m=\inf f(I)$ and $M=\sup f(I)$. Clearly, $m\le f(x)\le M$ for all $x\in I$.

For any $n\in \mathbb{\mathbb{N}}$, the number $M-\frac{1}{n}$ is not an upper bound of th eset $f(I)$, while $m+\frac{1}{n}$ is not a lower bound of $f(I)$. Hence we can find points $x_n,y_n\in I$ such that $f(x_n)>M-\frac{1}{n}$ and $f(y_n)<m+\frac{1}{n}$. By construction, $f(x_n)\to M$ and $f(y_n)\to m$ as $n\to \mathrm{\infty }$. The Bolzano-Weierstrass Theorem implies that the sequences $\left\{x_n\right\}$ and $\left\{y_n\right\}$ have convergent subsequences (or, in other words, they have limit points). Let $c$ be a limit point of $\left\{x_n\right\}$ and $d$ be a limit point of $\left\{y_n\right\}$. Notice that $c,d\in I$. The continuity of $f$ implies $f(c)$ is a limit point of $\{f(x_n)\}$ and $f(d)$ is a limit point of $\{f(y_n)\}$. We conclude that $f(c)=M$ and $f(d)=m$.

quod erat demonstrandum

We used the fact that the sequence is bounded when we applied the Bolzano-Weierstrass theorem.

We used the fact that the sequence is closed when we stated that the continuity of $f$ implies $f(c)$ is a limit point of $\{f(x_n)\}$ and $f(d)$ is a limit point of $\{f(y_n)\}$.

Remarks:

• The theorem may not hold if $I$ is not closed. Counterexample: $f(x)=x$ for $x\in (0,1)$. Neither maximum nor minimum is attained.
• The theorem may not hold if $I$ is not bounded. Counterexample: $f(x)=\frac{1}{x^2+1}$ for $x\in [0,\mathrm{\infty })$. The maximal value is attained at $0$, but the minimal value 0 is never attained.

Intermediate Value Theorem

Theorem. [Intermediate Value Theorem]. If a function $f:[a,b]\to \mathbb{ℝ}$ is continuous, then any number $y_0$ that lies between $f(a)$ and $f(b)$ is a value of $f$, that is, $y_0=f(x_0)$ for some $x_0\in [a,b]$.

Proof. In the case $f(a)=f(b)$, the theorem is trivial.

In the case $f(a)>f(b)$, we notice that the function $-f$ is continuous on $[a,b]$, $-f(a)<-f(b)$, and $-y_0$ lies between $-f(a)$ and $-f(b)$. Hence we acn assume without loss of generality that $f(a)<f(b)$.

Further, if $y_0$ lies between $f(a)$ and $f(b)$, then 0 lies between $f(a)-y_0$ and $f(b)-y_0$. Moreover, the function $g(x)=f(x)-y_0$ is continuous on $[a,b]$ and $g(a)<g(b)$ if and only if $f(a)<f(b)$. Hence it is no loss to assume that $y_0=0$.

Now the theorem is reduced to a simpler case:

Theorem. If a function $f:[a,b]\to \mathbb{ℝ}$ is continuous and $f(a)<0<f(b)$, then $f(x_0)=0$ for some $x_0\in (a,b)$.

Let $E=\{x\in [a,b]\mid f(x)>0\}$. The set $E$ is nonempty (as $b\in E$) and bounded (as $E\subset [a,b]$. Therefore, $x_0=\inf E$ exists. Observe that $x_0\in [a,b]$ ($x_0\le b$ as $b\in E$; $x_0\ge a$ as $a$ is a lower bound of $E$).

To complete the proof, we need the following lemma:

Lemma. If a function $f$ is continuous at a point $c$ and $f(c)\ne 0$, then $f$ maintains its sign in a sufficiently small neighborhood of $c$.

Proof. Since $f$ is continuous at $c$ and $|f(c)|>0$, there exists $\delta >0$ such that $|f(x)-f(c)|<|f(c)|$ whenever $|x-c|<\delta$. The inequality $|f(x)-f(c)|<|f(c)|$ implies $f(x)$ has the same sign as $f(c)$.

The lemma implies that $f(x_0)=0$. Indeed, if $f(x_0)\ne 0$, then for some $\delta >0$, the function $f$ maintains its sign in the interval $(x_0-\delta ,x_0+\delta )\cap [a,b]$.

• In the case $f(x_0)>0$, we obtain that $x_0>a$ and $x_0$ is not a lower bound of $E$.
• In the case $f(x_0)<0$, we obtain that $x_0<b$ and $x_0$ is not the greatest lower bound of $E$

Either way we arrive at a contradiction.

quod erat demonstrandum

Corollary. If a real-valued function $f$ is continuous on a closed bounded interval $[a,b]$, then the image $f([a,b])$ is also a closed bounded interval.

Proof. By the Extreme Value Theorem, there exist points $x_{min},x_{max}\in [a,b]$ such that $f(x_{min})\le f(x)\le f(x_{max})$ for all $x\in [a,b]$. Let $I_0$ denote the closed interval with endpoints $x_{min}$ and $x_{max}$. Let $J$ denote the closed interval with endpoints $f(x_{max})$ and $f(x_{max})$. We have that $f\left([a,b]\right)\subseteq J$. The Intermediate Value Theorem implies that $f\left(I_0\right)=J$. Since $I_0\subseteq [a,b]$, we obtain that $f\left([a,b]\right)=J$.

quod erat demonstrandum