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 $\epsilon >0$ there exists $\delta =\delta (\epsilon )>0$ such that $\left|x-c\right|<\delta$ and $x\in E$ imply $\left|f(x)-f(c)\right|<\epsilon$

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 $f(c)=\lim _{x\to c}f(x)$

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 $n\to \infty$ implies $f(x_{n})\to f(c)$ as $n\to \infty$.

Theorem. Suppose that functions $f,g:E\to \mathbb {R}$ are both continuous at $c\in E$. Then $(f+g)$, $(f-g)$ and $f\,g$ are also continuous at $c$. If, additionally, $g(c)\neq 0$, 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 $\left|f(x)\right|\leq C$ 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 $f(x)\leq C$ for all $x\in E_{0}$.

It is bounded below on $E_{0}$ if there exists $C\in \mathbb {R}$ such that $f(x)\geq C$ 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 $x\neq 0$

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

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 $n\in \mathbb {N}$ there exists a point $x_{n}\in I$ such that $\left|f(x_{n})\right|>n$. We obtain a sequence $\left\{x_{n}\right\}$ of elements of $I$ such that the sequence $\left\{f(x_{n})\right\}$ 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 $c=\lim _{k\to \infty }x_{n_{k}}$. Then $c\in \left[a,b\right]$ (due to comparison theorem). Since the sequence $\left\{f(x_{n_{k}})\right\}$ is a subsequence of $\left\{f(x_{n})\right\}$, 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 $\left[a,b\right]$ 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: $\lim _{x\to c-}f(x)\neq \lim _{x\to c+}f(x)$.
• $f$ has a removable discontinuity at a point $c$ if the limit at $c$ exists, but it is different from the value at $c$: $\lim _{x\to c}f(x)\neq f(c)$.
• If $f$ is continuous at $c$, then it is locally bounded at $c$, which means that $f$ is bounded on the set $\left(c-\delta ,c+\delta \right)\cap E$ 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 $\lim _{x\to c}f(x)=a$ for all $c\in \mathbb {R}$, $f$ is continuous.

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

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

Step function. $f(x)={\begin{cases}1&x>0\\0&x\leq 0\end{cases}}$

Since $\lim _{x\to 0-}f(x)=0$ and $\lim _{x\to 0+}f(x)=1$, 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 $x\neq 0$

Since $\lim _{x\to c}f(x)={\frac {1}{c}}$ for all $c\neq 0$, 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 $x\neq 0$.

Since $\lim _{x\to 0+}f(x)$ 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 $x\neq 0$.

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

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

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

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

Since $\lim _{x\to c}f(x)=0$ 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 $I=\left[a,b\right]$ 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

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

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

For any $n\in \mathbb {N}$, 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 $n\to \infty$. 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 $\left\{f(x_{n})\right\}$ and $f(d)$ is a limit point of $\left\{f(y_{n})\right\}$. 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 $\left\{f(x_{n})\right\}$ and $f(d)$ is a limit point of $\left\{f(y_{n})\right\}$.

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 $x\in [0,\infty )$. 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 $f:\left[a,b\right]\to \mathbb {R}$ 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 $x_{0}\in \left[a,b\right]$.

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 $\left[a,b\right]$, $-f(a)<-f(b)$, 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 $f:\left[a,b\right]\to \mathbb {R}$ is continuous and $f(a)<0<f(b)$, then $f(x_{0})=0$ for some $x_{0}\in (a,b)$.

Let $E=\left\{x\in \left[a,b\right]~\mid ~f(x)>0\right\}$. The set $E$ is nonempty (as $b\in E$) and bounded (as $E\subset \left[a,b\right]$. Therefore, $x_{0}=\inf {E}$ exists. Observe that $x_{0}\in \left[a,b\right]$ ($x_{0}\leq b$ as $b\in E$; $x_{0}\geq a$ 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 $f(c)\neq 0$, then $f$ maintains its sign in a sufficiently small neighborhood of $c$.

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

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

• 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 $\left[a,b\right]$, then the image $f(\left[a,b\right])$ is also a closed bounded interval.

Proof. By the Extreme Value Theorem, there exist points $x_{min},x_{max}\in \left[a,b\right]$ such that $f(x_{min})\leq f(x)\leq f(x_{max})$ for all $x\in \left[a,b\right]$. 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 $f\left(\left[a,b\right]\right)\subseteq J$. The Intermediate Value Theorem implies that $f\left(I_{0}\right)=J$. Since $I_{0}\subseteq \left[a,b\right]$, we obtain that $f\left(\left[a,b\right]\right)=J$.

quod erat demonstrandum