12 Continuity, Exponential Function and Function Sequences

Lecture from: 27.03.2024 | Video: Video ETHZ

Review: Theorem on Inverse Mappings

If $I$ is an interval and $f:I\to \mathbb{R}$ is continuous and strictly monotonic, then its inverse function $f^{-1}:f(I)\to I$ is also continuous and strictly monotonic.

Imagine two number lines, one representing the interval $I$ and the other representing the image of $I$ under $f$, which is $f(I)$. We have points $x_1,x_2,x_3,x_4,x_5$ on the interval $I$. The function $f$ maps these points to $v_1,v_2,v_3,v_4,v_5$ on $f(I)$, such that $v_n=f(x_n)$. The inverse function $f^{-1}$ goes in the opposite direction, from $f(I)$ back to $I$.

The key idea is that: $x_n$ converges to $x$ if and only if $v_n=f(x_n)$ converges to $v=f(x)$.

Example

Let $n\in {\mathbb{N}}^{\ast }$ (positive integers). Consider the function:

This function $x\mapsto x^n$ is continuous, strictly monotonically increasing, and surjective on $[0,\infty )$.

According to the theorem on inverse mappings, the inverse function also exists and is continuous, strictly monotonically increasing, and surjective. This inverse function is:

which is the $n$-th root function.

The Real Exponential Function

Recall: Definition and Properties of Exponential Function

• Definition: The exponential function $\exp (z)$ is defined by the power series: $$\exp (z)=\sum _{k=0}^{\infty }\frac{z^k}{k!}\text{for }z\in \mathbb{C}$$
• Addition Theorem: For all $z,w\in \mathbb{C}$: $$\exp (z+w)=\exp (z)\exp (w)$$
• Alternative Representation: $$\exp (z)=\underset{n\to \infty }{\lim }{\left(1+\frac{z}{n}\right)}^n$$

Theorem: Properties of Real Exponential Function

The exponential function $\exp :\mathbb{R}\to (0,\infty )$ is continuous, strictly monotonically increasing, and surjective.

Proof

For $x\ge 0$, we have:

Image of $\exp (\mathbb{R})\subseteq (0,\infty )$

Since $x\ge 0$, all terms in the series are non-negative, and the first two terms are $1+x$. Thus, for $x\ge 0$, $\exp (x)\ge 1+x\ge 1$. In particular $\exp (x)>1$ for $x>0$ and $\exp (0)=1$.

Furthermore, $1=\exp (0)=\exp (x-x)=\exp (x)\exp (-x)$. This implies $\exp (-x)=\frac{1}{\exp (x)}$. Since $\exp (x)\ge 1+x>0$ for $x\ge 0$, we have $\exp (-x)>0$ for $x\ge 0$.

Therefore, $\exp (x)>0$ for all $x\in \mathbb{R}$. Hence, the image of $\exp$ is contained in $(0,\infty )$, i.e., $\exp (\mathbb{R})\subseteq (0,\infty )$.

Strictly Monotonically Increasing

For $y\le z$, consider $\exp (z)=\exp (y+(z-y))=\exp (y)\exp (z-y)$. Since $z-y\ge 0$, we know $\exp (z-y)\ge 1$. Also, since $\exp (y)>0$, we have:

If $y<z$, then $z-y>0$, so $\exp (z-y)>1$, and therefore $\exp (z)>\exp (y)$. This shows that $\exp$ is strictly monotonically increasing.

Fact: $\exp (x)\ge 1+x$ for all $x\in \mathbb{R}$

For $n\ge 2$ and $x\ge -1$, we have the Bernoulli inequality: $(1+\frac{x}{n}{)}^n\ge 1+n\cdot (\frac{x}{n})=1+x$. Thus, for $n>|x|$, we have $(1+\frac{x}{n}{)}^n\ge 1+x$.

Taking the limit as $n\to \infty$:

Continuity at 0

Consider $x\in (-1,1)$. From the fact, we have $\exp (x)\ge 1+x$ and $\exp (-x)\ge 1-x$.

Since $\exp (-x)=\frac{1}{\exp (x)}$, we have $\frac{1}{\exp (x)}\ge 1-x$, which implies $\exp (x)\le \frac{1}{1-x}$ (since $1-x>0$ for $x<1$).

Thus, for $x\in (-1,1)$:

Consider the limits as $x\to 0$:

Since $1+x$ and $\frac{1}{1-x}$ are continuous at $0$, by the Sandwich Lemma (Squeeze Theorem) for sequential continuity, $\exp$ is continuous at $0$. (Compare with the solution to the clicker question from last time).

Continuity at $x_0$

For all $x\in \mathbb{R}$, we use the addition theorem:

Let $ϵ>0$. Due to the continuity of $\exp$ at $0$, there exists $\delta >0$ such that for all $y\in (-\delta ,\delta )$, we have $|\exp (y)-1|<\frac{ϵ}{\exp (x_0)}$.

From $(\ast )$, for all $x\in \mathbb{R}$ with $|x-x_0|<\delta$, let $y=x-x_0$, so $|y|=|x-x_0|<\delta$. Then:

Using the inequality $|\exp (y)-1|<\frac{ϵ}{\exp (x_0)}$:

Thus, for $|x-x_0|<\delta$, we have $|\exp (x)-\exp (x_0)|<ϵ$. This shows that $\exp$ is continuous at $x_0$.

Surjectivity

For $n\in \mathbb{N}$, $\exp (n)=\exp (\underset{n\text{ times}}{\underbrace{1+1+\cdots +1}})=(\exp (1){)}^n=e^n\ge 2^n$ since $e\ge 2$. So $\exp (n)\ge 2^n$.

Also, $0<\exp (-n)=\frac{1}{\exp (n)}\le \frac{1}{2^n}$.

By the Intermediate Value Theorem, for any interval $[2^{-n},2^n]$, since $\exp$ is continuous, the image of the interval $[-n,n]$ under $\exp$ must contain $[2^{-n},2^n]$. That is, $[2^{-n},2^n]\subseteq \exp ([-n,n])\subseteq \exp (\mathbb{R})$.

Since this holds for all $n\in \mathbb{N}$, we have:

But, ${\bigcup }_{n\in \mathbb{N}}[2^{-n},2^n]=(0,\infty )$.

Therefore, $(0,\infty )\subseteq \exp (\mathbb{R})$. We already knew $\exp (\mathbb{R})\subseteq (0,\infty )$. Combining both, we get $\exp (\mathbb{R})=(0,\infty )$. Thus, $\exp :\mathbb{R}\to (0,\infty )$ is surjective.

Natural Logarithm: Inverse of Exponential Function

The inverse mapping of $\exp :\mathbb{R}\to (0,\infty )$ is called the natural logarithm, denoted as $\ln$.

(also sometimes denoted as $\log :(0,\infty )\to \mathbb{R}$).

Corollary: Properties of the Natural Logarithm

The natural logarithm $\ln :(0,\infty )\to \mathbb{R}$ is strictly monotonically increasing, continuous, and bijective.y

For all $a,b\in (0,\infty )$, the following functional equation holds:

Proof

The theorem about inverse mappings gives us all properties except for the functional equation.

For $a,b>0$:

Apply $\ln$ to both sides of the equation:

Since $\ln$ is the inverse of $\exp$, we have:

General Powers

For $x>0$ and $a\in \mathbb{R}$, we define the general power:

Alternatively, we can write $x^a=(e^{\ln (x)}{)}^a=e^{a\ln (x)}$.

In particular, for $x>0$:

Corollary: Properties of General Powers

1. For $a>0$, the function:

   $$\begin{array}{rl}(0,\infty )& \to (0,\infty )\\ x& \mapsto x^a\end{array}$$

   is continuous, strictly monotonically increasing, and bijective.

2. For $a<0$, the function:

   $$\begin{array}{rl}(0,\infty )& \to (0,\infty )\\ x& \mapsto x^a\end{array}$$

   is continuous, strictly monotonically decreasing, and bijective.

3. For $x>0,a,b\in \mathbb{R}$:

   $$\ln (x^a)=a\cdot \ln (x)$$

4. For $x>0,a,b\in \mathbb{R}$:

   $$x^{a+b}=x^a\cdot x^b$$

5. For $x>0,a,b\in \mathbb{R}$:

   $$(x^a{)}^b=x^{a\cdot b}$$

Proof

Consider the composition of functions for $x\mapsto x^a$:

Since $\ln$, multiplication by $a$, and $\exp$ are continuous functions, their composition $x\mapsto x^a$ is also continuous on $(0,\infty )$.

For $a\ne 0$, all arrows in the diagram represent bijective mappings. Therefore, the composition $x\mapsto x^a$ is also bijective for $a\ne 0$.

1. For $a>0$: all arrows represent strictly monotonically increasing functions. The composition of strictly monotonically increasing functions is strictly monotonically increasing. Thus, $x\mapsto x^a$ is strictly monotonically increasing for $a>0$.

2. For $a<0$: $\ln$ and $\exp$ are strictly monotonically increasing, but multiplication by $a<0$ is strictly monotonically decreasing. So we have: (strictly increasing) $\circ$ (strictly decreasing) $\circ$ (strictly increasing) = strictly monotonically decreasing. Thus, $x\mapsto x^a$ is strictly monotonically decreasing for $a<0$.

3. Property (3): By definition, $x^a=\exp (a\ln (x))$. Taking $\ln$ of both sides:

   $$\ln (x^a)=\ln (\exp (a\ln (x)))=a\ln (x)$$

   because $\ln$ is the inverse of $\exp$.

4. Property (4):

   $$x^{a+b}=\exp ((a+b)\ln (x))=\exp (a\ln (x)+b\ln (x))$$

   Using the addition theorem for $\exp$:

   $$x^{a+b}=\exp (a\ln (x))\cdot \exp (b\ln (x))=x^a\cdot x^b$$

5. Property (5): (Exercise)

   $$(x^a{)}^b=\exp (b\ln (x^a))\stackrel{(3)}{=}\exp (b\cdot (a\ln (x)))=\exp ((a\cdot b)\ln (x))=x^{a\cdot b}$$

Convergence of Function Sequences

Let $D$ be a set. A function sequence (real-valued) is a mapping:

where $R^D$ is the set of functions from $D$ to $\mathbb{R}$. Thus, each $f_n$ is a function $f_n:D\to \mathbb{R}$.

We write a function sequence as $(f_n{)}_{n\ge 0}$ or simply $(f_n)$.

For each $x\in D$, we obtain a sequence of real numbers $(f_n(x){)}_{n\ge 0}$.

Definition: Pointwise Convergence

The function sequence $(f_n{)}_{n\ge 0}$ converges pointwise to a function $f:D\to \mathbb{R}$ if for all $x\in D$:

Example: Pointwise Convergence of $f_n(x)=x^n$

Let $D=[0,1]$, and $f_n(x)=x^n$ for $n\in \mathbb{N},x\in [0,1]$. Does $(f_n{)}_{n\ge 0}$ converge pointwise?

For $0\le x<1$: ${\lim }_{n\to \infty }{x}^n=0$. For $x=1$: ${\lim }_{n\to \infty }{1}^n=1$.

Thus, $(f_n{)}_{n\ge 0}$ converges pointwise to a function $f:[0,1]\to \mathbb{R}$ defined by:

Remark: The limit function $f$ is not continuous on $[0,1]$ (it is discontinuous at $x=1$). (Namely, discontinuous at 1).

Definition: Uniform Convergence

The sequence of functions $f_n:D\to \mathbb{R}$ converges uniformly (in $D$) to $f:D\to \mathbb{R}$ if:

$N(ϵ)$ is independent of $x\in D$!

Compare with pointwise convergence:

$N(ϵ,x)$ may depend on $x$!

Visualization of Uniform Convergence

Imagine a function $f(x)$ and an “$ϵ$-tube” around it, defined by $f(x)-ϵ$ and $f(x)+ϵ$. For uniform convergence, for any given $ϵ>0$, there exists an $N$ such that for all $n\ge N$, the graph of $f_n(x)$ lies completely within this $ϵ$-tube around $f(x)$ for all $x$ in the domain $D$.

Uniform Convergence Condition:

Example Again: Non-Uniform Convergence

Consider again $f_n(x)=x^n$ on $D=[0,1]$, with pointwise limit function $f(x)=\{\begin{array}{ll}0& 0\le x<1\\ 1& x=1\end{array}$.

Sketch the $ϵ$-tube around $f(x)$. For any $N\in \mathbb{N}$, the function $f_n(x)=x^n$ for $n\ge N$ will, near $x=1$, leave the $ϵ$-tube around $f$.

Explanation: For every $n\in \mathbb{N}$, $f_n$ leaves the $ϵ$-tube around $f$ near $x=1$.

Therefore, $f_n$ does not converge uniformly to $f$ on $[0,1]$. The convergence is only pointwise, not uniform.

Continue here: 13 Function Sequences, Continuity of Limit Function under Uniform Convergence, Cauchy Criterion for Uniform Convergence, Series of Functions, Uniform Convergence of Power Series, Trig Functions