# MATH 409 Lecture 10

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Lecture Slides

## Continuity

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

We say that the function ${\displaystyle f}$ is continuous on a set ${\displaystyle E_{0}\subset E}$ if ${\displaystyle f}$ is continuous at every point ${\displaystyle c\in E_{0}}$. The function ${\displaystyle f}$ is continuous if it is continuous on the entire domain ${\displaystyle E}$.

Note:

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

### Properties

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

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

### Bounded Functions

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

The function ${\displaystyle f}$ is bounded above on ${\displaystyle E_{0}}$ if there exists ${\displaystyle C\in \mathbb {R} }$ such that ${\displaystyle f(x)\leq C}$ for all ${\displaystyle x\in E_{0}}$.

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

Equivalently, ${\displaystyle f}$ is bounded on ${\displaystyle E_{0}}$ if the image ${\displaystyle f(E_{0})}$ is a bounded subset of ${\displaystyle \mathbb {R} }$. Likewise, the function ${\displaystyle f}$ is bounded above on ${\displaystyle E_{0}}$ if the image ${\displaystyle f(E_{0})}$ is bounded above. It is bounded below on ${\displaystyle E_{0}}$ if ${\displaystyle f(E_{0})}$ is bounded below.

#### Example

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

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

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

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

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

Thus any continuous function on ${\displaystyle \left[a,b\right]}$ has to be bounded.

quod erat demonstrandum

## Discontinuities

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

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

## Examples

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

## Extreme Value Theorem

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

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

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

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

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 ${\displaystyle f}$ implies ${\displaystyle f(c)}$ is a limit point of ${\displaystyle \left\{f(x_{n})\right\}}$ and ${\displaystyle f(d)}$ is a limit point of ${\displaystyle \left\{f(y_{n})\right\}}$.

Remarks:

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

## Intermediate Value Theorem

Theorem. [Intermediate Value Theorem]. If a function ${\displaystyle f:\left[a,b\right]\to \mathbb {R} }$ is continuous, then any number ${\displaystyle y_{0}}$ that lies between ${\displaystyle f(a)}$ and ${\displaystyle f(b)}$ is a value of ${\displaystyle f}$, that is, ${\displaystyle y_{0}=f(x_{0})}$ for some ${\displaystyle x_{0}\in \left[a,b\right]}$.

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

In the case ${\displaystyle f(a)>f(b)}$, we notice that the function ${\displaystyle -f}$ is continuous on ${\displaystyle \left[a,b\right]}$, ${\displaystyle -f(a)<-f(b)}$, and ${\displaystyle -y_{0}}$ lies between ${\displaystyle -f(a)}$ and ${\displaystyle -f(b)}$. Hence we acn assume without loss of generality that ${\displaystyle f(a).

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

Now the theorem is reduced to a simpler case:

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

Let ${\displaystyle E=\left\{x\in \left[a,b\right]~\mid ~f(x)>0\right\}}$. The set ${\displaystyle E}$ is nonempty (as ${\displaystyle b\in E}$) and bounded (as ${\displaystyle E\subset \left[a,b\right]}$. Therefore, ${\displaystyle x_{0}=\inf {E}}$ exists. Observe that ${\displaystyle x_{0}\in \left[a,b\right]}$ (${\displaystyle x_{0}\leq b}$ as ${\displaystyle b\in E}$; ${\displaystyle x_{0}\geq a}$ as ${\displaystyle a}$ is a lower bound of ${\displaystyle E}$).

To complete the proof, we need the following lemma:

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

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

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

• In the case ${\displaystyle f(x_{0})>0}$, we obtain that ${\displaystyle x_{0}>a}$ and ${\displaystyle x_{0}}$ is not a lower bound of ${\displaystyle E}$.
• In the case ${\displaystyle f(x_{0})<0}$, we obtain that ${\displaystyle x_{0} and ${\displaystyle x_{0}}$ is not the greatest lower bound of ${\displaystyle E}$

Either way we arrive at a contradiction.

quod erat demonstrandum

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

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

quod erat demonstrandum