11 Continuity, Extreme Value Theorem, Composition of Continuous Functions, Continuity of Inverse Functions

Lecture from: 25.03.2024 | Video: Video ETHZ

This lecture focuses on the concept of continuity for functions, exploring its definition, key properties, and fundamental theorems such as the Intermediate Value Theorem and the Min-Max Theorem. We will also examine the behavior of continuous functions under composition and inversion.

Clicker Question: Properties of a Function on [0, 1]

Let $f:[0,1]\to \mathbb{R}$ be a function with $f(0)=1$ and $f(1)=0$. Which of the following statements are true? (Multiple answers may be correct).

• Statement 1: If $f$ is continuous, then there exists an $x_0\in [0,1]$ with $f(x_0)=\frac{1}{2}$.
• Statement 2: If $(1-x{)}^2\le f(x)\le 1-x$ for all $x\in [0,1]$, then $f$ is continuous at points 0 and 1.
• Statement 3: If $f$ is continuous at all points $x_0\in [0,1)$, then $f$ is bounded.
• Statement 4: If $f$ is unbounded, then $f$ is discontinuous at least at one point $x_0\in [0,1]$.

Let’s analyze each statement:

Analysis of Statement 1: Intermediate Value Theorem

Statement 1: If $f$ is continuous, then there exists an $x_0\in [0,1]$ with $f(x_0)=\frac{1}{2}$.

Answer: True!

Explanation: This statement is a direct application of the Intermediate Value Theorem (Zwischenwertsatz). Since $f$ is continuous on the closed interval $[0,1]$, and $f(0)=1>\frac{1}{2}$ and $f(1)=0<\frac{1}{2}$, the Intermediate Value Theorem guarantees that there must be at least one point $x_0\in [0,1]$ such that $f(x_0)=\frac{1}{2}$. The value $\frac{1}{2}$ lies between $f(0)$ and $f(1)$.

Analysis of Statement 2: Continuity via Squeeze Theorem (Sandwichlemma)

Statement 2: If $(1-x{)}^2\le f(x)\le 1-x$ for all $x\in [0,1]$, then $f$ is continuous at points 0 and 1.

Answer: True!

Explanation: This statement uses the Squeeze Theorem (Sandwichlemma) in the context of sequential continuity (Folgenstetigkeit). Let’s check continuity at $x=1$.

Consider a sequence $(x_n{)}_{n\ge 1}$ in $[0,1]$ such that ${\lim }_{n\to \infty }{x}_n=1$. We need to show that ${\lim }_{n\to \infty }f(x_n)=f(1)=0$.

We are given the inequalities: $(1-x{)}^2\le f(x)\le 1-x$. Therefore, for each $x_n$:

Now consider the limits of the bounding functions as $n\to \infty$ (and $x_n\to 1$):

• ${\lim }_{n\to \infty }(1-x_n{)}^2=(1-{\lim }_{n\to \infty }{x}_n{)}^2=(1-1{)}^2=0$
• ${\lim }_{n\to \infty }(1-x_n)=(1-{\lim }_{n\to \infty }{x}_n)=(1-1)=0$

By the Squeeze Theorem (Sandwichlemma) for sequences, since $(1-x_n{)}^2\le f(x_n)\le 1-x_n$ and both bounding sequences converge to 0, we must have:

This shows that $f$ is sequentially continuous at $x=1$, and thus continuous at $x=1$.

Similarly, we can check continuity at $x=0$. Consider a sequence $(x_n{)}_{n\ge 1}$ in $[0,1]$ with ${\lim }_{n\to \infty }{x}_n=0$.

• ${\lim }_{n\to \infty }(1-x_n{)}^2=(1-0{)}^2=1$
• ${\lim }_{n\to \infty }(1-x_n)=(1-0)=1$

Again by the Squeeze Theorem, ${\lim }_{n\to \infty }f(x_n)=1=f(0)$. Thus, $f$ is continuous at $x=0$.

Analysis of Statement 3: Boundedness and Continuity on $[0,1)$

Statement 3: If $f$ is continuous at all points $x_0\in [0,1)$, then $f$ is bounded.

Answer: False!

Explanation: Continuity on a half-open interval $[0,1)$ does not guarantee boundedness on the entire interval $[0,1]$.

Counterexample: Consider the function $f(x)=\frac{1}{1-x}$ for $x\in [0,1)$ and we can define $f(1)=0$ to complete the definition on $[0,1]$.

• For $x\in [0,1)$, $f(x)=\frac{1}{1-x}$ is continuous on $[0,1)$. As $x$ approaches 1 from the left ($x\to 1^{-}$), $f(x)=\frac{1}{1-x}\to \infty$.
• Thus, $f$ is unbounded on $[0,1)$. Although the statement is about boundedness on $[0,1]$, the unboundedness on $[0,1)$ itself shows that continuity on $[0,1)$ does not imply boundedness on $[0,1]$ (or even $[0,1)$ itself). Defining $f(1)=0$ does not change the unboundedness on $[0,1)$ and the continuity on $[0,1)$.

Analysis of Statement 4: Unboundedness and Discontinuity

Statement 4: If $f$ is unbounded, then $f$ is discontinuous at least at one point $x_0\in [0,1]$.

Answer: True!

Explanation: This statement is related to the Min-Max Theorem (Min-Max-Satz), which we will discuss later. The contrapositive of this statement is: “If $f$ is continuous on $[0,1]$, then $f$ is bounded.” This contrapositive is a consequence of the Min-Max Theorem (and more generally, the Extreme Value Theorem).

If $f$ were continuous on $[0,1]$, then by the Min-Max Theorem, $f$ would have to be bounded on $[0,1]$. Therefore, if $f$ is unbounded, it cannot be continuous on the entire interval $[0,1]$. This implies there must be at least one point in $[0,1]$ where $f$ is discontinuous. In fact, the point of discontinuity must be related to the source of unboundedness. In our counterexample $f(x)=\frac{1}{1-x}$ for $x\in [0,1)$, the function becomes unbounded as $x\to 1^{-}$, and we can see the discontinuity is “at” or “near” $x=1$ (though technically, in the way the example is set up it is discontinuous at $x=1$ because the limit from the left is infinity, not $f(1)=0$). We will refine this idea later.

Summary of Clicker Question Answers

• Statement 1: True (Intermediate Value Theorem)
• Statement 2: True (Squeeze Theorem and Sequential Continuity)
• Statement 3: False (Counterexample: $f(x)=\frac{1}{1-x}$ on $[0,1)$)
• Statement 4: True (Related to Min-Max Theorem - contrapositive is true)

Remark: Reformulation of Intermediate Value Theorem

The Intermediate Value Theorem can be reformulated in terms of image of intervals:

Remark (Intermediate Value Theorem - Image of Interval): The image of an interval under a continuous function is an interval.

More precisely, if $f:I\to \mathbb{R}$ is continuous on an interval $I$, and if for some $a,b\in I$ and $c$ between $f(a)$ and $f(b)$, then there exists a $z\in I$ such that $f(z)=c$. This means that the set $f(I)$ must be an interval. [If $f(a)\le c\le f(b)$, then $\exists z$ such that $f(z)=c$].

Corollary of Intermediate Value Theorem: Existence of Nullstelle for Polynomials

Corollary (Existence of Nullstelle for Polynomials): Let $P(x)=a_n{x}^n+a_{n-1}{x}^{n-1}+\cdots +a_{1}x+a_0$ be a polynomial with $a_n\ne 0$ and $n$ being an odd integer. Then $P$ has at least one real root (Nullstelle) in $\mathbb{R}$.

Explanation:

For large positive $x$, the term $a_n{x}^n$ dominates the polynomial’s behavior. If $a_n>0$ and $n$ is odd, then as $x\to \infty$, $P(x)\to \infty$. As $x\to -\infty$, $P(x)\to -\infty$. If $a_n<0$ and $n$ is odd, then as $x\to \infty$, $P(x)\to -\infty$ and as $x\to -\infty$, $P(x)\to \infty$. In either case, the polynomial takes on both very large positive and very large negative values.

Proof: Assume, without loss of generality, that $a_n=1$ (we can divide the entire polynomial by $a_n$ without changing the roots). So $P(x)=x^n+a_{n-1}{x}^{n-1}+\cdots +a_0$.

We can rewrite $P(x)$ as:

Let $Q(x)=1+\frac{a_{n-1}}{x}+\frac{a_{n-2}}{x^2}+\cdots +\frac{a_0}{x^n}$. Then ${\lim }_{x\to \pm \infty }Q(x)=1$. This means for sufficiently large $|x|$, $Q(x)$ is close to 1 (and thus positive).

• As $x\to \infty$, $Q(x)\to 1>0$, and since $n$ is odd, $x^n\to \infty$. Thus, ${\lim }_{x\to \infty }P(x)={\lim }_{x\to \infty }{x}^{n}Q(x)=\infty$. So for some large $N\in {\mathbb{N}}^{\ast }$, $P(N)>0$.

• As $x\to -\infty$, $Q(x)\to 1>0$, and since $n$ is odd, $x^n\to -\infty$. Thus, ${\lim }_{x\to -\infty }P(x)={\lim }_{x\to -\infty }{x}^{n}Q(x)=-\infty$. So for some large $N\in {\mathbb{N}}^{\ast }$, $P(-N)<0$.

Since $P(N)>0$ and $P(-N)<0$, and $P$ is a polynomial (hence continuous on $\mathbb{R}$), by the Intermediate Value Theorem applied to the interval $[-N,N]$, there exists some $x_0\in [-N,N]\subseteq \mathbb{R}$ such that $P(x_0)=0$. Therefore, $P$ has a real root.

Remark: For $c>0$, the quadratic polynomial $Q(x)=x^2+c$ has no real roots because $x^2+c\ge c>0$ for all $x\in \mathbb{R}$. This shows that the odd degree condition is necessary for the corollary to hold for real roots.

The Min-Max Theorem (Min-Max-Satz) - also known as Extreme Value Theorem

Statement of the Min-Max Theorem

Theorem (Min-Max Theorem / Extreme Value Theorem): Let $I\subseteq \mathbb{R}$ be a compact interval (i.e., a closed and bounded interval) and let $f:I\to \mathbb{R}$ be a continuous function on $I$. Then $f$ is bounded on $I$, and $f$ attains its minimum and maximum values on $I$.

This means there exist points $u,v\in I$ such that for all $x\in I$:

• $f(u)$ is the minimum value of $f$ on $I$.
• $f(v)$ is the maximum value of $f$ on $I$.

Examples Illustrating the Necessity of Compactness

The conditions of compactness (closed and bounded interval) and continuity are crucial for the Min-Max Theorem to hold. Let’s see examples where the theorem fails if these conditions are relaxed.

1. Unbounded Interval: Consider $f:[0,\infty )\to \mathbb{R},x\mapsto x^2$.

   • $[0,\infty )$ is an interval but not bounded (hence not compact).
   • $f(x)=x^2$ is continuous on $[0,\infty )$.
   • $f$ is unbounded on $[0,\infty )$ (as $x\to \infty ,f(x)\to \infty$). It attains a minimum at $x=0$, $f(0)=0$, but no maximum.

2. Half-Open Interval: Consider $f:(0,1]\to \mathbb{R},x\mapsto \frac{1}{x}$.

   • $(0,1]$ is an interval but not closed (hence not compact).
   • $f(x)=\frac{1}{x}$ is continuous on $(0,1]$.
   • $f$ is unbounded on $(0,1]$ as $x\to 0^{+}$, $f(x)\to \infty$. It attains a minimum at $x=1$, $f(1)=1$, but no maximum.

3. Half-Open Interval - Bounded but No Maximum: Consider $f:[0,1)\to \mathbb{R},x\mapsto x$.

   • $[0,1)$ is an interval but not closed (hence not compact).
   • $f(x)=x$ is continuous on $[0,1)$.
   • $f$ is bounded on $[0,1)$ (e.g., by 1). It attains a minimum at $x=0$, $f(0)=0$, but does not attain a maximum on $[0,1)$. While the supremum of $f(x)$ on $[0,1)$ is 1, there is no $v\in [0,1)$ such that $f(v)=1$.

Definition: Compact Interval

Definition: Compact Interval in $\mathbb{R}$ An interval $I$ in $\mathbb{R}$ is compact if it is of the form $I=[a,b]$ for some $a,b\in \mathbb{R}$ (i.e., closed and bounded interval).

Lemma: Characterization of Compact Intervals via Sequences

Lemma (Characterization of Compact Intervals): Let $I\subseteq \mathbb{R}$ be a non-empty interval. Then $I$ is compact if and only if every sequence $(x_n{)}_{n\ge 1}$ in $I$ has a convergent subsequence whose limit is also in $I$.

This lemma provides a sequential characterization of compactness, linking it to the behavior of sequences and subsequences within the interval.

Proof of the Min-Max Theorem

Theorem (Min-Max Theorem / Extreme Value Theorem): Let $I\subseteq \mathbb{R}$ be a compact interval (i.e., a closed and bounded interval) and let $f:I\to \mathbb{R}$ be a continuous function on $I$. Then $f$ is bounded on $I$, and $f$ attains its minimum and maximum values on $I$.

We will break the proof into three parts:

1. Boundedness Above: We will show that $f$ is bounded above on $I$.
2. Existence of Maximum: We will show that $f$ attains its maximum value on $I$.
3. Existence of Minimum and Boundedness Below: We will show that $f$ is bounded below and attains its minimum value on $I$. (The proof for boundedness below and minimum will be analogous to boundedness above and maximum, but we will explicitly show it for completeness.)

Really good explanation of the proof below…

(1) Proof of Boundedness Above: (By contradiction)

Assume, for contradiction, that $f$ is not bounded above on $I$. This means for every $n\in \mathbb{N}$, there exists a point $x_n\in I$ such that $f(x_n)>n$. This constructs a sequence $(x_n{)}_{n\ge 1}$ in $I$ with the property that $f(x_n)\to \infty$ as $n\to \infty$.

Since $I$ is a compact interval, by the sequential characterization of compactness, the sequence $(x_n{)}_{n\ge 1}$ has a convergent subsequence $(x_{e(n)}{)}_{n\ge 1}$ (where $e:{\mathbb{N}}^{\ast }\to {\mathbb{N}}^{\ast }$) whose limit $r={\lim }_{n\to \infty }{x}_{e(n)}$ is also in $I$.

Since $f$ is continuous on $I$, it is continuous at the point $r\in I$. By sequential continuity, we must have:

Since $f(r)$ is a real number, ${\lim }_{n\to \infty }f(x_{e(n)})$ must be finite. However, from the construction of $(x_n)$, we have $f(x_{e(n)})>e(n)\ge n$. As $n\to \infty$, $f(x_{e(n)})\to \infty$.

This leads to a contradiction: we have ${\lim }_{n\to \infty }f(x_{e(n)})=f(r)$ (a finite value) and ${\lim }_{n\to \infty }f(x_{e(n)})=\infty$.

This contradiction invalidates our assumption that $f$ is not bounded above. Therefore, $f$ must be bounded above on $I$.

(2) Proof of Existence of Maximum:

Since $f$ is bounded above on $I$ (from Part 1), the supremum of the image set $f(I)$ exists as a finite real number. Let $M=\sup \{f(x)\mid x\in I\}$. We want to show that there exists a point $v\in I$ such that $f(v)=M$ (i.e., the supremum is attained as a maximum value).

Since $M=\sup \{f(x)\mid x\in I\}$, for every $n\in {\mathbb{N}}^{\ast }$, $M-\frac{1}{n}$ is not an upper bound for $f(I)$. Therefore, there exists a point $v_n\in I$ such that $f(v_n)>M-\frac{1}{n}$. This creates a sequence $(v_n{)}_{n\ge 1}$ in $I$.

Since $I$ is compact, the sequence $(v_n{)}_{n\ge 1}$ has a convergent subsequence $(v_{e(n)}{)}_{n\ge 1}$ whose limit $v={\lim }_{n\to \infty }{v}_{e(n)}$ is also in $I$.

Since $f$ is continuous on $I$, it is continuous at $v\in I$. By sequential continuity:

From the construction, we have $f(v_{e(n)})>M-\frac{1}{e(n)}$. Taking the limit as $n\to \infty$:

So, $f(v)={\lim }_{n\to \infty }f(v_{e(n)})\ge M$. However, since $M=\sup \{f(x)\mid x\in I\}$ is the supremum of all values of $f$ on $I$, we must also have $f(v)\le M$ (as $v\in I$).

Combining $f(v)\ge M$ and $f(v)\le M$, we conclude $f(v)=M$. Thus, $f$ attains its maximum value $M$ at the point $v\in I$.

(3) Proof of Existence of Minimum and Boundedness Below:

To show that $f$ is bounded below and attains its minimum value, we can consider the function $g(x)=-f(x)$. If $f$ is continuous, then $g$ is also continuous on $I$.

From Part 1, since $g$ is continuous on the compact interval $I$, $g$ is bounded above on $I$. This means there exists some $C\in \mathbb{R}$ such that $g(x)\le C$ for all $x\in I$. Substituting $g(x)=-f(x)$, we get $-f(x)\le C$, which means $f(x)\ge -C$ for all $x\in I$. Thus, $f$ is bounded below on $I$.

From Part 2, since $g$ is continuous on the compact interval $I$, $g$ attains its maximum value on $I$. Let $u\in I$ be a point where $g$ attains its maximum, so $g(u)=\max \{g(x)\mid x\in I\}$. This means $g(u)\ge g(x)$ for all $x\in I$. Substituting $g(x)=-f(x)$, we have $-f(u)\ge -f(x)$, which means $f(u)\le f(x)$ for all $x\in I$.

Therefore, $f$ attains its minimum value $f(u)$ at the point $u\in I$.

Combining Parts 1, 2, and 3, we have shown that a continuous function $f$ on a compact interval $I$ is bounded, attains its maximum value, and attains its minimum value. This completes the proof of the Min-Max Theorem. $\square$

Composition of Continuous Functions

Theorem: Continuity of Composition

Theorem (Continuity of Composition): Let $D_1,D_2\subseteq \mathbb{R}$. Let $f:D_1\to D_2$ be continuous at $x_0\in D_1$, and let $g:D_2\to \mathbb{R}$ be continuous at $f(x_0)\in D_2$. Then the composition $g\circ f:D_1\to \mathbb{R}$ is continuous at $x_0$.

Proof (using Sequential Continuity):

Let $(x_n{)}_{n\ge 1}$ be a sequence in $D_1$ such that ${\lim }_{n\to \infty }{x}_n=x_0$. We want to show that ${\lim }_{n\to \infty }(g\circ f)(x_n)=(g\circ f)(x_0)$.

1. Continuity of $f$ at $x_0$: Since $f$ is continuous at $x_0$ and $(x_n)\to x_0$ in $D_1$, we have ${\lim }_{n\to \infty }f(x_n)=f(x_0)$. Let $y_n=f(x_n)$ and $y_0=f(x_0)$. Then ${\lim }_{n\to \infty }{y}_n=y_0$. Note that $y_n=f(x_n)\in D_2$ for all $n$, and $y_0=f(x_0)\in D_2$.

2. Continuity of $g$ at $f(x_0)=y_0$: Since $g$ is continuous at $y_0=f(x_0)$ and $(y_n)\to y_0$ in $D_2$, we have ${\lim }_{n\to \infty }g(y_n)=g(y_0)$.

Combining these, we get:

Thus, $g\circ f$ is sequentially continuous at $x_0$, and therefore continuous at $x_0$.

Corollary: Global Continuity of Composition

Corollary (Global Continuity of Composition): If $f:D_1\to D_2$ is continuous on $D_1$ and $g:D_2\to \mathbb{R}$ is continuous on $D_2$, then the composition $g\circ f:D_1\to \mathbb{R}$ is continuous on $D_1$.

Continuity of Inverse Functions

Theorem: Continuity of Inverse of Strictly Monotonic Continuous Function

Theorem (Continuity of Inverse Function): Let $I\subseteq \mathbb{R}$ be an interval, and let $f:I\to \mathbb{R}$ be a continuous and strictly monotonic function. Then:

1. Image is Interval: The image $J=f(I)$ is also an interval.
2. Bijectivity: $f:I\to J=f(I)$ is bijective.
3. Strict Monotonicity of Inverse: The inverse function $f^{-1}:J\to I$ is also strictly monotonic (same type of monotonicity as $f$).
4. Continuity of Inverse: The inverse function $f^{-1}:J\to I$ is also continuous.

Proof

1. (Image is Interval): If $f$ is continuous on an interval $I$, by the Intermediate Value Theorem (reformulated), the image $f(I)$ must also be an interval.

2. (Bijectivity): Strict monotonicity (either strictly increasing or strictly decreasing) implies injectivity. Since we consider $f$ mapping to its image $J=f(I)$, it is surjective by definition. Thus, $f:I\to J$ is bijective.

3. (Strict Monotonicity of Inverse): Assume $f$ is strictly monotonically increasing (the decreasing case is similar). Let $y_1,y_2\in J$ with $y_1<y_2$. Let $x_1=f^{-1}(y_1)$ and $x_2=f^{-1}(y_2)$. If $x_1\ge x_2$, then since $f$ is strictly increasing, $f(x_1)\ge f(x_2)$, i.e., $y_1\ge y_2$, contradicting $y_1<y_2$. Thus, we must have $x_1<x_2$, i.e., $f^{-1}(y_1)<f^{-1}(y_2)$. So $f^{-1}$ is also strictly monotonically increasing.

4. (Continuity of Inverse): We prove continuity of $f^{-1}$ using sequential continuity. Let $(v_n{)}_{n\ge 1}$ be a sequence in $J$ converging to $v\in J$. We want to show that ${\lim }_{n\to \infty }{f}^{-1}(v_n)=f^{-1}(v)$. Let $x_n=f^{-1}(v_n)$ and $x=f^{-1}(v)$. We need to show ${\lim }_{n\to \infty }{x}_n=x$.

   Assume, for contradiction, that $(x_n)$ does not converge to $x$. Then there exists an $ϵ>0$ such that for infinitely many $n$, $x_n\notin (x-ϵ,x+ϵ)$. This means for infinitely many $n$, either $x_n\le x-ϵ$ or $x_n\ge x+ϵ$.

   Assume, without loss of generality, that for infinitely many $n$, $x_n\ge x+ϵ$. Since $f$ is strictly monotonically increasing, $v_n=f(x_n)\ge f(x+ϵ)>f(x)=v$. Thus, for infinitely many $n$, $v_n\ge f(x+ϵ)$. This contradicts the assumption that $v_n\to v=f(x)<f(x+ϵ)$. A similar contradiction arises if we assume $x_n\le x-ϵ$ for infinitely many $n$.

   Therefore, our assumption that $(x_n)$ does not converge to $x$ must be false. Hence, ${\lim }_{n\to \infty }{x}_n=x$, i.e., ${\lim }_{n\to \infty }{f}^{-1}(v_n)=f^{-1}(v)$. Thus, $f^{-1}$ is sequentially continuous at $v$, and hence continuous at $v$.

This lecture has laid the groundwork for a deeper understanding of continuous functions, exploring their fundamental properties, boundedness on compact intervals, behavior under composition, and the continuity of their inverses when strictly monotonic.

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