10 Continuity, Intermediate Value Theorem

Lecture from: 20.03.2024 | Video: Video ETHZ

Real-Valued Functions

Clicker Question: Exploring Function Behavior Near Zero

Let’s begin our exploration of continuous functions with a clicker question to illustrate some key ideas. Consider two functions:

1. $f(x)=\frac{1}{x^2+1}$
2. $g(x)=\{\begin{array}{ll}1,& x>0\\ 0,& x\le 0\end{array}$

And a sequence $x_n=-\frac{1}{n}$.

Let’s analyze their behavior:

• Function $f(x)=\frac{1}{x^2+1}$:

  • Value at $x=0$: $f(0)=\frac{1}{0^2+1}=1$.
  • Limit as $x_n\to 0$: ${\lim }_{n\to \infty }f(x_n)={\lim }_{n\to \infty }\frac{1}{(-\frac{1}{n}{)}^2+1}={\lim }_{n\to \infty }\frac{1}{\frac{1}{n^2}+1}=\frac{1}{0+1}=1$.

• Function $g(x)=\{\begin{array}{ll}1,& x>0\\ 0,& x\le 0\end{array}$:

  • Value at $x=0$: $g(0)=0$.
  • Limit as $x_n\to 0$: Since $x_n=-\frac{1}{n}\le 0$ for all $n$, $g(x_n)=0$ for all $n$. Therefore, ${\lim }_{n\to \infty }g(x_n)=0$.

Graphical Interpretation:

• Function $f(x)$: The graph of $f(x)=\frac{1}{x^2+1}$ is a smooth curve. As $x$ approaches 0, $f(x)$ approaches $f(0)=1$. We can draw the graph “without lifting the pen” near $x=0$.

• Function $g(x)$: The graph of $g(x)$ has a “jump” or “step” discontinuity at $x=0$. Even though the limit of $g(x_n)$ as $x_n\to 0$ (from the negative side) equals $g(0)=0$ in this specific case, if we approach 0 from the positive side, the limit would be 1, not equal to $g(0)$. This function “jumps” at $x=0$.

These examples intuitively illustrate the concept of continuity: a function is continuous at a point if its values approach the function value at that point as the input approaches that point, regardless of the direction of approach.

Continuity (Stetigkeit)

Formal $ϵ-\delta$ Definition of Continuity

To make the intuitive idea of “no jumps” mathematically rigorous, we use the $ϵ-\delta$ definition of continuity.

Definition: Continuity at a Point

Let $D\subseteq \mathbb{R}$, $x_0\in D$. A function $f:D\to \mathbb{R}$ is continuous at $x_0$ if for every $ϵ>0$, there exists a $\delta >0$ such that for all $x\in D$, if $|x-x_0|<\delta$, then $|f(x)-f(x_0)|<ϵ$.

Explanation of the $ϵ-\delta$ Definition:

• $ϵ$ (Tolerance): The value $ϵ>0$ represents an arbitrarily small tolerance or allowed error in the function value $f(x)$ around $f(x_0)$. We want $f(x)$ to be within $ϵ$ of $f(x_0)$.
• $\delta$ (Neighborhood): The value $\delta >0$ represents a neighborhood around $x_0$. We need to find a $\delta$-neighborhood around $x_0$ such that for any $x$ within this neighborhood (and in the domain $D$), the function value $f(x)$ is within the desired $ϵ$-tolerance of $f(x_0)$.
• For Every $ϵ$: The condition must hold for every possible tolerance $ϵ>0$, no matter how small. This ensures that we can make $f(x)$ arbitrarily close to $f(x_0)$ by choosing $x$ sufficiently close to $x_0$.
• Existence of $\delta$: For each $ϵ$, we must be able to find a corresponding $\delta$. The value of $\delta$ typically depends on $ϵ$ and the function $f$, and the point $x_0$.

Geometric Interpretation:

Imagine a “target band” of width $2ϵ$ around $f(x_0)$ (i.e., the interval $(f(x_0)-ϵ,f(x_0)+ϵ)$). Continuity at $x_0$ means that we can find a “source interval” of width $2\delta$ around $x_0$ (i.e., $(x_0-\delta ,x_0+\delta )$) such that if we take any $x$ from the intersection of this source interval with the domain $D$, then the function value $f(x)$ will fall within the target band.

Definition: Continuity on a Set

A function $f:D\to \mathbb{R}$ is continuous on $D$ (or simply continuous) if it is continuous at every point $x_0\in D$.

Examples: Proving Continuity Using $ϵ-\delta$ Definition

Example (1): Continuity of $f(x)=x^n$ and $n=1$

Let $f:\mathbb{R}\to \mathbb{R},f(x)=x$. We want to show that $f$ is continuous at any $x_0\in \mathbb{R}$.

Given $ϵ>0$. We need to find $\delta >0$ such that if $|x-x_0|<\delta$, then $|f(x)-f(x_0)|<ϵ$.

Consider $|f(x)-f(x_0)|=|x-x_0|$. We want $|x-x_0|<ϵ$. So, we can choose $\delta =ϵ$.

Proof: Let $ϵ>0$ be given. Choose $\delta =ϵ$. Then for any $x\in \mathbb{R}$, if $|x-x_0|<\delta =ϵ$, then $|f(x)-f(x_0)|=|x-x_0|<ϵ$. Thus, $f(x)=x$ is continuous at $x_0$. Since $x_0$ was arbitrary, $f(x)=x$ is continuous on $\mathbb{R}$.

Example (2): Continuity of $f(x)=|x|$

Let $f:\mathbb{R}\to \mathbb{R},f(x)=|x|$. We want to show that $f$ is continuous at any $x_0\in \mathbb{R}$.

Given $ϵ>0$. We need to find $\delta >0$ such that if $|x-x_0|<\delta$, then $|f(x)-f(x_0)|<ϵ$.

Consider $|f(x)-f(x_0)|=||x|-|x_0||$. By the reverse triangle inequality, $||x|-|x_0||\le |x-x_0|$. We want $||x|-|x_0||<ϵ$. Since $||x|-|x_0||\le |x-x_0|$, if we choose $\delta =ϵ$, then whenever $|x-x_0|<\delta =ϵ$, we will have $||x|-|x_0||\le |x-x_0|<ϵ$.

Proof: Let $ϵ>0$ be given. Choose $\delta =ϵ$. Then for any $x\in \mathbb{R}$, if $|x-x_0|<\delta =ϵ$, then $|f(x)-f(x_0)|=||x|-|x_0||\le |x-x_0|<ϵ$. Thus, $f(x)=|x|$ is continuous at $x_0$. Since $x_0$ was arbitrary, $f(x)=|x|$ is continuous on $\mathbb{R}$.

Example (3): Discontinuity of the Rounding Function (Abrundungsfunktion) $f(x)=\lfloor x\rfloor$

Let $f:\mathbb{R}\to \mathbb{R},f(x)=\lfloor x\rfloor =\max \{m\in \mathbb{Z}\mid m\le x\}$. We want to show that $f$ is discontinuous at $x_0\in \mathbb{Z}$.

Consider $x_0=0$. Let $ϵ=\frac{1}{2}$. We want to show that there is no $\delta >0$ such that for all $x\in \mathbb{R}$, if $|x-0|<\delta$, then $|\lfloor x\rfloor -\lfloor 0\rfloor |<\frac{1}{2}$.

Suppose such a $\delta >0$ exists. Choose $x=-\frac{\delta }{2}$. Then $|x-0|=|-\frac{\delta }{2}|=\frac{\delta }{2}<\delta$. However, $\lfloor x\rfloor =\lfloor -\frac{\delta }{2}\rfloor =-1$ (since $-\frac{\delta }{2}<0$ and close to 0). And $\lfloor x_0\rfloor =\lfloor 0\rfloor =0$. Thus, $|f(x)-f(x_0)|=|\lfloor -\frac{\delta }{2}\rfloor -\lfloor 0\rfloor |=|-1-0|=1$.

But we require $|f(x)-f(x_0)|<ϵ=\frac{1}{2}$. We have $1\nless \frac{1}{2}$. This is a contradiction. Therefore, $f(x)=\lfloor x\rfloor$ is not continuous at $x_0=0$. Similarly, it is discontinuous at any integer $x_0\in \mathbb{Z}$.

However, $f(x)=\lfloor x\rfloor$ is continuous on $\mathbb{R}\setminus \mathbb{Z}$.

Example (4): Dirichlet-like Function

Consider $f:\mathbb{R}\to \mathbb{R}$ defined by:

This function is continuous only at $x_0=\frac{1}{2}$. For any other point, it is discontinuous. (Proof omitted here, but can be shown using the density of rationals and irrationals).

Sequential Criterion for Continuity (Folgenstetigkeit)

An alternative and often more convenient way to characterize continuity is using sequences.

Theorem: Sequential Criterion for Continuity

Let $D\subseteq \mathbb{R}$, $x_0\in D$, and $f:D\to \mathbb{R}$. The function $f$ is continuous at $x_0$ if and only if for every sequence $(a_n{)}_{n\ge 1}$ in $D$ such that ${\lim }_{n\to \infty }{a}_n=x_0$, we have ${\lim }_{n\to \infty }f(a_n)=f(x_0)$.

This theorem states that continuity at $x_0$ is equivalent to preserving limits of sequences converging to $x_0$.

Proof

• ($\Rightarrow$) Assume $f$ is continuous at $x_0$. Let $(a_n)$ be a sequence in $D$ with ${\lim }_{n\to \infty }{a}_n=x_0$. We want to show ${\lim }_{n\to \infty }f(a_n)=f(x_0)$.

  Given $ϵ>0$. By continuity of $f$ at $x_0$, there exists $\delta >0$ such that for all $x\in D$, if $|x-x_0|<\delta$, then $|f(x)-f(x_0)|<ϵ$.

  Since ${\lim }_{n\to \infty }{a}_n=x_0$, there exists $N\in \mathbb{N}$ such that for all $n>N$, $|a_n-x_0|<\delta$.

  Combining these, for all $n>N$, since $|a_n-x_0|<\delta$, we have $|f(a_n)-f(x_0)|<ϵ$. This shows that ${\lim }_{n\to \infty }f(a_n)=f(x_0)$.

• ($\Leftarrow$) Assume sequential continuity holds. Suppose $f$ is not continuous at $x_0$. Then the $ϵ-\delta$ definition fails. This means there exists some ${ϵ}_0>0$ such that for every $\delta >0$, there exists some $x\in D$ with $|x-x_0|<\delta$ but $|f(x)-f(x_0)|\ge {ϵ}_0$.

  For each $n\in {\mathbb{N}}^{\ast }$, let ${\delta }_n=\frac{1}{n}$. Then there exists $a_n\in D$ such that $|a_n-x_0|<\frac{1}{n}$ but $|f(a_n)-f(x_0)|\ge {ϵ}_0$.

  The sequence $(a_n)$ satisfies $|a_n-x_0|<\frac{1}{n}\to 0$ as $n\to \infty$, so ${\lim }_{n\to \infty }{a}_n=x_0$. However, $|f(a_n)-f(x_0)|\ge {ϵ}_0$ for all $n$, so $(f(a_n))$ does not converge to $f(x_0)$. This contradicts the assumption of sequential continuity. Therefore, $f$ must be continuous at $x_0$.

Corollary: Algebraic Operations and Continuity

Continuity is preserved under algebraic operations.

Corollary: Algebraic Properties of Continuous Functions

Let $D\subseteq \mathbb{R}$, $x_0\in D$, $\lambda \in \mathbb{R}$, and let $f,g:D\to \mathbb{R}$ be functions continuous at $x_0$. Then the following functions are also continuous at $x_0$:

1. Sum: $f+g$

2. Scalar Product: $\lambda \cdot f$

3. Product: $f\cdot g$

4. Quotient: If $g(x_0)\ne 0$, then the quotient function $\frac{f}{g}:\{x\in D\mid g(x)\ne 0\}\to \mathbb{R}$, defined by $\frac{f}{g}(x)=\frac{f(x)}{g(x)}$, is continuous at $x_0$.

Proof (using Sequential Criterion):

We use the Sequential Criterion for continuity. Let $(a_n)$ be a sequence in $D$ with ${\lim }_{n\to \infty }{a}_n=x_0$.

1. Sum: ${\lim }_{n\to \infty }(f+g)(a_n)={\lim }_{n\to \infty }(f(a_n)+g(a_n))={\lim }_{n\to \infty }f(a_n)+{\lim }_{n\to \infty }g(a_n)=f(x_0)+g(x_0)=(f+g)(x_0)$. Thus, $f+g$ is continuous at $x_0$.

2. Scalar Product: ${\lim }_{n\to \infty }(\lambda f)(a_n)={\lim }_{n\to \infty }(\lambda f(a_n))=\lambda {\lim }_{n\to \infty }f(a_n)=\lambda f(x_0)=(\lambda f)(x_0)$. Thus, $\lambda f$ is continuous at $x_0$.

3. Product: ${\lim }_{n\to \infty }(f\cdot g)(a_n)={\lim }_{n\to \infty }(f(a_n)\cdot g(a_n))=({\lim }_{n\to \infty }f(a_n))\cdot ({\lim }_{n\to \infty }g(a_n))=f(x_0)\cdot g(x_0)=(f\cdot g)(x_0)$. Thus, $f\cdot g$ is continuous at $x_0$.

4. Quotient: If $g(x_0)\ne 0$, then since $g$ is continuous at $x_0$, ${\lim }_{n\to \infty }g(a_n)=g(x_0)\ne 0$. For $n$ large enough, $g(a_n)\ne 0$. Then ${\lim }_{n\to \infty }\frac{f}{g}(a_n)={\lim }_{n\to \infty }\frac{f(a_n)}{g(a_n)}=\frac{{\lim }_{n\to \infty }f(a_n)}{{\lim }_{n\to \infty }g(a_n)}=\frac{f(x_0)}{g(x_0)}=\frac{f}{g}(x_0)$. Thus, $\frac{f}{g}$ is continuous at $x_0$.

Definition: Polynomial Functions

A polynomial function $P:\mathbb{R}\to \mathbb{R}$ is a function of the form:

where $a_0,a_1,\dots ,a_n\in \mathbb{R}$ are coefficients, and $n\in \mathbb{N}\cup \{0\}$ is the degree (if $a_n\ne 0$).

Corollary: Continuity of Polynomial Functions

Polynomial functions are continuous on $\mathbb{R}$.

Proof: The constant function $f(x)=c$ and the identity function $g(x)=x$ are continuous on $\mathbb{R}$. Using Corollary (Algebraic Properties of Continuous Functions) repeatedly for sums and products, we can show that any polynomial function, built from constants and $x$ using addition and multiplication, is continuous on $\mathbb{R}$.

Corollary: Continuity of Rational Functions

Let $P,Q$ be polynomial functions on $\mathbb{R}$, and assume $Q$ is not the zero polynomial. Let $x_1,\dots ,x_m$ be the zeros (roots) of $Q$. Then the rational function $\frac{P}{Q}:\mathbb{R}\setminus \{x_1,\dots ,x_m\}\to \mathbb{R}$, defined by $\frac{P}{Q}(x)=\frac{P(x)}{Q(x)}$, is continuous on its domain $\mathbb{R}\setminus \{x_1,\dots ,x_m\}$.

Proof: Polynomial functions $P$ and $Q$ are continuous on $\mathbb{R}$. By Corollary (Algebraic Properties of Continuous Functions) (quotient property), $\frac{P}{Q}$ is continuous wherever $Q(x)\ne 0$, i.e., on $\mathbb{R}\setminus \{x_1,\dots ,x_m\}$.

Example: $f(x)=\frac{1}{1-x}$ is a rational function, continuous on $\mathbb{R}\setminus \{1\}$.

Remark: If we try to “extend” a discontinuous rational function by assigning an arbitrary value at a point of discontinuity, the extended function typically remains discontinuous. For example, if we define $f(x)=\frac{1}{1-x}$ for $x\ne 1$ and try to set $f(1)=0$ (or any other value), the resulting function is still discontinuous at $x=1$.

The Intermediate Value Theorem (Zwischenwertsatz)

Definition: “Between” for Real Numbers

Definition: For $x_1,x_2\in \mathbb{R}$, we say a number $c$ is between $x_1$ and $x_2$ if $c\in [\min \{x_1,x_2\},\max \{x_1,x_2\}]$. In other words, $c$ lies in the closed interval with endpoints $x_1$ and $x_2$.

Theorem: Intermediate Value Theorem (IVT)

Theorem (Intermediate Value Theorem): Let $I\subseteq \mathbb{R}$ be an interval, and let $f:I\to \mathbb{R}$ be a continuous function on $I$. Let $a,b\in I$, and let $c$ be any value between $f(a)$ and $f(b)$. Then there exists at least one value $z$ between $a$ and $b$ (i.e., $z\in [\min \{a,b\},\max \{a,b\}]\subseteq I$) such that $f(z)=c$.

Intuitive Interpretation:

If a continuous function takes on two values $f(a)$ and $f(b)$, it must take on every value in between as well, at some point in between $a$ and $b$. Graphically, if you draw a continuous curve from $(a,f(a))$ to $(b,f(b))$, you cannot avoid hitting any horizontal line between $y=f(a)$ and $y=f(b)$. There are no “jumps” or “gaps” in the range of a continuous function on an interval.

Proof:

Without loss of generality, assume $a<b$. (If $a>b$, we can swap them; if $a=b$, the theorem is trivially true). Also, without loss of generality, assume $f(a)\le f(b)$. (If $f(a)>f(b)$, we can consider $-f$ instead of $f$, or reverse the roles of $a$ and $b$). We want to find $z\in [a,b]$ such that $f(z)=c$, where $f(a)\le c\le f(b)$.

We use interval bisection to construct a nested sequence of intervals.

1. Initialization: Let $I_1=[a_1,b_1]=[a,b]$. We have $f(a_1)=f(a)\le c$ and $f(b_1)=f(b)\ge c$.

2. Iteration: Given an interval $I_n=[a_n,b_n]$ such that $f(a_n)\le c$ and $f(b_n)\ge c$. Consider the midpoint $m_n=\frac{a_n+b_n}{2}$.

   • Case 1: $f(m_n)\le c$. Then set $a_{n+1}=m_n$ and $b_{n+1}=b_n$. Now $f(a_{n+1})\le c$ and $f(b_{n+1})\ge c$.
   • Case 2: $f(m_n)>c$. Then set $a_{n+1}=a_n$ and $b_{n+1}=m_n$. Now $f(a_{n+1})\le c$ and $f(b_{n+1})\ge c$.
   • Case 3: $f(m_n)=c$. We found our value! Set $z=m_n$, and we are done.

3. Nested Intervals: We obtain a sequence of nested closed intervals $I_1\supseteq I_2\supseteq I_3\supseteq \dots$ with $I_n=[a_n,b_n]$ such that for all $n$:

   • Length of intervals: $L(I_n)=b_n-a_n=\frac{1}{2^{n-1}}L(I_1)=\frac{b-a}{2^{n-1}}$. So $L(I_n)\to 0$ as $n\to \infty$.
   • Function values at endpoints: $f(a_n)\le c$ and $f(b_n)\ge c$.

4. Nested Interval Theorem: By the Nested Interval Theorem, there exists a unique point $z\in {\bigcap }_{n=1}^{\infty }{I}_n$. Also, ${\lim }_{n\to \infty }{a}_n={\lim }_{n\to \infty }{b}_n=z$.

5. Continuity and Limit: Since $f$ is continuous at $z$, by sequential continuity:

   • ${\lim }_{n\to \infty }f(a_n)=f(z)$. Since $f(a_n)\le c$ for all $n$, ${\lim }_{n\to \infty }f(a_n)\le c$, so $f(z)\le c$.
   • ${\lim }_{n\to \infty }f(b_n)=f(z)$. Since $f(b_n)\ge c$ for all $n$, ${\lim }_{n\to \infty }f(b_n)\ge c$, so $f(z)\ge c$.

Combining $f(z)\le c$ and $f(z)\ge c$, we get $f(z)=c$. And $z\in {\bigcap }_{n=1}^{\infty }{I}_n\subseteq I_1=[a,b]$, so $z$ is between $a$ and $b$.

Conclusion: We have shown that there exists a value $z$ between $a$ and $b$ such that $f(z)=c$. $\square$

This concludes our lecture on continuity, providing a rigorous definition, examples, and the fundamental Intermediate Value Theorem.

Continue here: 11 Continuity, Extreme Value Theorem, Composition of Continuous Functions, Continuity of Inverse Functions