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

Lecture from: 08.04.2024 | Video: Video ETHZ

Review: Pointwise and Uniform Convergence of Function Sequences

Pointwise Convergence

For a sequence of functions $f_n:D\to \mathbb{R}$ and a function $f:D\to \mathbb{R}$, we say that $(f_n)$ converges pointwise to $f$ on $D$ if for every $x\in D$, the limit of the sequence of real numbers $(f_n(x){)}_{n\in \mathbb{N}}$ is $f(x)$. In formal terms:

Uniform Convergence

A sequence of functions $f_n:D\to \mathbb{R}$ converges uniformly to $f:D\to \mathbb{R}$ on $D$ if for every $ϵ>0$, there exists an index $N\in \mathbb{N}$ such that for all $n\ge N$ and for all $x\in D$, the absolute difference between $f_n(x)$ and $f(x)$ is less than $ϵ$. Formally:

The crucial difference is that for uniform convergence, the choice of $N$ depends only on $ϵ$ and not on the point $x\in D$. This is what makes the convergence “uniform” across the entire domain $D$.

Visualization: $ϵ$-Schlauch ($ϵ$-Tube)

Imagine an “$ϵ$-tube” around the graph of the limit function $f$. This tube is defined by the region between the graphs of $y=f(x)-ϵ$ and $y=f(x)+ϵ$.

For uniform convergence, for sufficiently large $n$ (i.e., $n\ge N$), the entire graph of $f_n(x)$ for all $x\in D$ must lie within this $ϵ$-tube.

Example: $f_n(x)=x^n$ on $[0,1]$ - Pointwise but Not Uniform Convergence

Consider the sequence of functions $f_n:[0,1]\to \mathbb{R}$ defined by $f_n(x)=x^n$. We saw that this sequence converges pointwise to the function:

However, this convergence is not uniform. No matter how large we choose $n$, the function $f_n(x)=x^n$ will always “escape” the $ϵ$-tube around $f$ near $x=1$. For any $ϵ<1$, near $x=1$, $x^n$ will be close to $1$ and thus outside the $ϵ$-tube around $f(x)=0$.

Theorem: Continuity of the Limit Function under Uniform Convergence

Theorem: Let $D\subseteq \mathbb{R}$ and $f_n:D\to \mathbb{R}$ be a sequence of functions that converges uniformly on $D$ to a function $f:D\to \mathbb{R}$. If each $f_n$ is continuous at a point $x_0\in D$, then the limit function $f$ is also continuous at $x_0$.

Proof

Let $ϵ>0$. Since $f_n\to f$ uniformly on $D$, there exists an $N\in \mathbb{N}$ such that for all $n\ge N$ and for all $x\in D$, $|f_n(x)-f(x)|<ϵ$. Let’s fix such an $N$.

Since $f_N$ is continuous at $x_0$, there exists a $\delta >0$ such that for all $x\in D$ with $|x-x_0|<\delta$, we have $|f_N(x)-f_N(x_0)|<ϵ$.

Now consider any $x\in D$ with $|x-x_0|<\delta$. We want to show that $|f(x)-f(x_0)|$ is small. We can write:

Using the triangle inequality:

Now we bound each term:

1. $|f(x)-f_N(x)|<ϵ$ because of uniform convergence of $f_n$ to $f$ (for $n=N$). This holds for all $x\in D$.
2. $|f_N(x)-f_N(x_0)|<ϵ$ because $f_N$ is continuous at $x_0$ and $|x-x_0|<\delta$.
3. $|f_N(x_0)-f(x_0)|<ϵ$ because of uniform convergence of $f_n$ to $f$ (for $n=N$). This holds for all $x_0\in D$.

Therefore, for all $x\in D$ with $|x-x_0|<\delta$:

Since we can replace $ϵ$ with $ϵ/3$ in the beginning, for any $ϵ>0$, we can find a $\delta >0$ such that for $|x-x_0|<\delta$, $|f(x)-f(x_0)|<ϵ$. This shows that $f$ is continuous at $x_0$. $\square$

Corollary: If $f_n:D\to \mathbb{R}$ is a uniformly convergent sequence of continuous functions on $D$, then the limit function $f:D\to \mathbb{R}$ is also continuous on $D$.

Cauchy Criterion for Uniform Convergence

Theorem: Cauchy Criterion for Uniform Convergence

A function sequence $f_n:D\to \mathbb{R}$ is uniformly convergent on $D$ if and only if it satisfies the Cauchy criterion for uniform convergence:

Proof (Sketch)

• ($\Rightarrow$) If $f_n$ converges uniformly to $f$, then the Cauchy criterion holds. If $f_n\to f$ uniformly, then for $ϵ/2>0$, there exists $N$ such that for all $n\ge N$ and all $x\in D$, $|f_n(x)-f(x)|<ϵ/2$. Then for $m,n\ge N$:

  $$|f_n(x)-f_m(x)|=|(f_n(x)-f(x))+(f(x)-f_m(x))|\le |f_n(x)-f(x)|+|f(x)-f_m(x)|<\frac{ϵ}{2}+\frac{ϵ}{2}=ϵ$$

• ($\Leftarrow$) If the Cauchy criterion holds, then $f_n$ converges uniformly to some function $f$.

  1. For each $x\in D$, the sequence $(f_n(x){)}_{n\in \mathbb{N}}$ is a Cauchy sequence in $\mathbb{R}$ (by the Cauchy criterion). Since $\mathbb{R}$ is complete, $(f_n(x))$ converges to some limit, say $f(x)$. This defines a pointwise limit function $f:D\to \mathbb{R}$.
  2. We need to show that this pointwise convergence is uniform. From the Cauchy criterion, for $ϵ>0$, there exists $N$ such that for all $m,n\ge N$ and all $x\in D$, $|f_n(x)-f_m(x)|<ϵ$. Fix $n\ge N$ and let $m\to \infty$. Then $f_m(x)\to f(x)$.
  $$|f_n(x)-f(x)|=\underset{m\to \infty }{\lim }|f_n(x)-f_m(x)|\le ϵ$$

  This inequality holds for all $n \ge N$ and all $x \in D$. Thus, $f_n \to f$ uniformly. $\Box$
  

Series of Functions

Let $f_k:D\to \mathbb{R}$ be a sequence of functions. We can consider the series ${\sum }_{k=0}^{\infty }{f}_k$.

Definition: Pointwise and Uniform Convergence of Function Series

. The series ${\sum }_{k=0}^{\infty }{f}_k$ of functions $f_k:D\to \mathbb{R}$ converges pointwise (resp. uniformly) in $D$ if the function sequence of partial sums $S_n(x)={\sum }_{k=0}^n{f}_k(x)$ converges pointwise (resp. uniformly) in $D$.

Theorem: Comparison Test for Function Series

Let $D\subseteq \mathbb{R}$ and $f_k:D\to \mathbb{R}$ be a sequence of functions. Assume there exists a sequence of constants $(C_k{)}_{k\in {\mathbb{N}}_0}$ with $C_k\ge 0$ such that $|f_k(x)|\le C_k$ for all $x\in D$ and for all $k\in {\mathbb{N}}_0$, and that the series ${\sum }_{k=0}^{\infty }{C}_k$ converges (i.e., ${\sum }_{k=0}^{\infty }{C}_k<\infty$).

Then the series ${\sum }_{k=0}^{\infty }{f}_k$ converges uniformly (in $D$).

If all $f_k$ are continuous in $D$, then $f(x)={\sum }_{k=0}^{\infty }{f}_k(x)$ is a continuous function in $D$.

Proof

Let $ϵ>0$. Then there exists an $N\in \mathbb{N}$ (by the Cauchy criterion for series of numbers) such that for all $m\ge n\ge N$:

Then for all $m\ge n\ge N$ and all $x\in D$:

By the Cauchy criterion for uniform convergence (Satz 3.7.6), ${\sum }_{k=0}^{\infty }{f}_k$ converges uniformly. $\square$

Application: Uniform Convergence of Power Series

Recall: Power Series

A power series (in the real variable $x$) is of the form ${\sum }_{k=0}^{\infty }{c}_k{x}^k$, where $(c_k{)}_{k\in {\mathbb{N}}_0}$ is a sequence in $\mathbb{R}$.

Definition: Radius of Convergence

The power series ${\sum }_{k=0}^{\infty }{c}_k{x}^k$ has a positive radius of convergence if $R:={\mathrm{limsup}}_{k\to \infty }\sqrt[k]{|c_k|}<\infty$.

The radius of convergence $\rho$ is then defined as:

Corollary: Absolute Convergence (Weierstrass M-Test)

For all $x\in \mathbb{R}$ with $|x|<\rho$, the series ${\sum }_{k=0}^{\infty }{c}_k{x}^k$ converges absolutely.

Theorem: Uniform Convergence of Power Series within Radius of Convergence

Let ${\sum }_{k=0}^{\infty }{c}_k{x}^k$ be a power series with positive radius of convergence $\rho >0$. Then for all $0\le r<\rho$, the power series ${\sum }_{k=0}^{\infty }{c}_k{x}^k$ converges uniformly on $[-r,r]$.

In particular, $f:(-\rho ,\rho )\to \mathbb{R},x\mapsto {\sum }_{k=0}^{\infty }{c}_k{x}^k$ is continuous.

Proof

Let $f_k(x)=c_k{x}^k$ ($k\ge 0$). These functions are continuous (on $\mathbb{R}$).

For $|x|\le r<\rho$, we have $|f_k(x)|=|c_k{x}^k|=|c_k||x{|}^k\le |c_k|r^k$.

Since $r<\rho$, the series ${\sum }_{k=0}^{\infty }|c_k|r^k$ converges (absolutely convergent power series for $|x|<\rho$).

By the above corollary, ${\sum }_{k=0}^{\infty }{c}_k{x}^k$ converges uniformly in $[-r,r]$ and $f(x)={\sum }_{k=0}^{\infty }{c}_k{x}^k$ is continuous in $[-r,r]$.

This holds for all $0\le r<\rho ⟹f$ is continuous in $(-\rho ,\rho )$. $\square$

Remark

1. This provides a second proof of the continuity of the exponential function via power series.
2. In general, a power series converges uniformly on $[-r,r]$ for any $r<\rho$, but not necessarily uniformly on the entire interval of convergence $(-\rho ,\rho )$ or $[-\rho ,\rho ]$ (if the interval includes endpoints).

Example: The exponential series converges, but not uniformly on $(-\infty ,\infty )=\mathbb{R}$.

Clicker Question: $A=e^i$

Let $A=e^i=\exp (i)=1+i+\frac{i^2}{2!}+\frac{i^3}{3!}+\dots$

Direct Calculation of $|A|$:

Alternatively, using Euler’s Formula: $e^i=\cos (1)+i\sin (1)$. Thus $|e^i|=\sqrt{{\cos }^2(1)+{\sin }^2(1)}=\sqrt{1}=1$.

Trigonometric Functions: Sine and Cosine

Definition of Sine and Cosine Functions

For $z\in \mathbb{C}$, we define:

Remark: Real and Imaginary Parts of $\exp (ix)$

For $x\in \mathbb{R}$, we have the relations:

Convergence and Continuity

Using the ratio test or quotient criterion, we can show that both series for $\sin (z)$ and $\cos (z)$ converge absolutely for all $z\in \mathbb{C}$. (Compare with Example 2.7.18 for the exponential function).

The radius of convergence for both sine and cosine series is $\rho =\infty$.

Theorem: The functions $\sin :\mathbb{R}\to \mathbb{R}$ and $\cos :\mathbb{R}\to \mathbb{R}$ are continuous.

This follows from the fact that they are defined by power series with infinite radius of convergence, and power series are continuous within their radius of convergence.

Further Properties of Sine and Cosine

For all $z,w\in \mathbb{C}$:

1. Euler’s Formula:

   $$\exp (iz)=\cos (z)+i\sin (z)$$

2. Even and Odd Functions:

   $$\cos (-z)=\cos (z)\text{"cosine is an even function"}$$ $$\sin (-z)=-\sin (z)\text{"sine is an odd function"}$$

3. Exponential Representations:

   $$\cos (z)=\frac{e^{iz}+e^{-iz}}{2}$$ $$\sin (z)=\frac{e^{iz}-e^{-iz}}{2i}$$

4. Addition Theorems:

   $$\cos (z+w)=\cos (z)\cos (w)-\sin (z)\sin (w)$$ $$\sin (z+w)=\sin (z)\cos (w)+\cos (z)\sin (w)$$

5. Pythagorean Identity:

   $${\cos }^2(z)+{\sin }^2(z)=1$$

Proofs of Properties

1. Euler’s Formula:

   $$\exp (iz)=\sum _{k=0}^{\infty }\frac{(iz{)}^k}{k!}=\sum _{n=0}^{\infty }\frac{(iz{)}^{2n}}{(2n)!}+\sum _{n=0}^{\infty }\frac{(iz{)}^{2n+1}}{(2n+1)!}$$ $$=\sum _{n=0}^{\infty }(-1{)}^n\frac{z^{2n}}{(2n)!}+i\sum _{n=0}^{\infty }(-1{)}^n\frac{z^{2n+1}}{(2n+1)!}=\cos (z)+i\sin (z)$$

2. Even and Odd Properties: Follow directly from the series definitions of $\cos (z)$ and $\sin (z)$.

3. Exponential Representations:

   $$e^{iz}=\cos (z)+i\sin (z)$$ $$e^{-iz}=\cos (-z)+i\sin (-z)=\cos (z)-i\sin (z)$$

   Adding these two equations and dividing by 2 gives the formula for $\cos (z)$. Subtracting the second from the first and dividing by $2i$ gives the formula for $\sin (z)$.

4. Addition Theorems: Using Euler’s formula:

   $$e^{i(w+z)}=e^{iw}{e}^{iz}=(\cos (w)+i\sin (w))(\cos (z)+i\sin (z))$$ $$=(\cos (w)\cos (z)-\sin (w)\sin (z))+i(\sin (w)\cos (z)+\cos (w)\sin (z))$$

   Equating the real and imaginary parts of $e^{i(w+z)}=\cos (w+z)+i\sin (w+z)$ gives the addition theorems.

5. Pythagorean Identity: Setting $w=-z$ in the addition theorem for cosine:

   $$\cos (z+(-z))=\cos (z)\cos (-z)-\sin (z)\sin (-z)$$ $$\cos (0)=\cos (z)\cos (z)-\sin (z)(-\sin (z))={\cos }^2(z)+{\sin }^2(z)$$

   Since $\cos (0)=1$, we get ${\cos }^2(z)+{\sin }^2(z)=1$.

Example: $\cos (i)$

Since $e\approx 2.718$, $e^2+1\approx 7.389+1=8.389$, and $2e\approx 5.436$. Thus, $\cos (i)>1$. This shows that for complex arguments, the cosine function can take values greater than 1 in absolute value, unlike the real cosine function which is bounded by $[-1,1]$.

This concludes the lecture on uniform convergence, its applications to power series and continuity, and an introduction to trigonometric functions in the complex domain.