NoteNextra-origin/content/Math4111/Math4111_L25.md

Lecture 25

Review

Are the following statements true or false? You do not need to give a rigorous proof.

1. \exists N\in \mathbb{N}, \forall n\geq N, \left(\frac{1}{2}\right)^n < 0.001
   • True, let N = 10
2. \forall \epsilon > 0, \exists N\in \mathbb{N}, \forall n\geq N, \left(\frac{1}{2}\right)^n < \epsilon
   • True, let N = \lceil \log_{\frac{1}{2}} \epsilon \rceil
3. \forall x\in [0, 1), \forall \epsilon > 0, \exists N\in \mathbb{N}, \forall n\geq N, x^n < \epsilon
   • True, let N = \lceil \log_x \epsilon \rceil
4. \forall \epsilon > 0, \exists N\in \mathbb{N}, \forall n\geq N, \forall x\in [0, 1), x^n < \epsilon
   • False, x can be arbitrarily close to 1
5. \forall \epsilon > 0, \forall x\in \left[0, \frac{1}{2}\right], \exists N\in \mathbb{N}, \forall n\geq N, x^n < \epsilon
   • True, let N = \lceil \log_{\frac{1}{2}} \epsilon \rceil

New Materials

Sequences and series of functions

Definition 7.1

Let (f_n) be a sequence of functions E\to \mathbb{R}. Let f:E\to \mathbb{R} be a function. We say (f_n) converges pointwise to f if \forall x\in E, \lim_{n\to \infty} f_n(x) = f(x).

i.e. \forall x\in E, \forall \epsilon > 0, \exists N\in \mathbb{N}, \forall n\geq N, |f_n(x) - f(x)| < \epsilon.

Example:

The sequence from the warm up exercise converges pointwise to f(x) = \begin{cases} 0 & x\in [0, 1) \\ 1 \text{ otherwise} \end{cases}.

To check if a series of functions converges pointwise, we can take the limit of the series as n\to \infty.

Example:

Let g_n(x):\mathbb{R}\to \mathbb{R} be defined by g_n(x) = \sum_{k=0}^{n} \frac{x^2}{(1+x^2)^n}.

g_n(x) = \sum_{k=0}^{n} \left(\frac{x^2}{(1+x^2)^n}\right)

And this is a geometric series with first term \frac{x^2}{1+x^2} and common ratio \frac{x^2}{1+x^2}.

Then \left(g_n\right) converges pointwise to g(x) =\begin{cases} 0 & x = 0 \\ 1+x^2 & \text{otherwise} \end{cases}.

This example shows that pointwise convergence does not preserve continuity. But if (f_n) converges uniformly to f, then f is continuous.

Definition 7.7

Let (f_n) be a sequence of functions E\to \mathbb{R}. We say (f_n) converges uniformly to f if \forall \epsilon > 0, \exists N\in \mathbb{N}, \forall n\geq N, \forall x\in E, |f_n(x) - f(x)| < \epsilon.

i.e. \forall \epsilon > 0, \exists N\in \mathbb{N}, such that \forall n\geq N, \forall x\in E, |f_n(x) - f(x)| < \epsilon.

f_n must always be within [f-\epsilon, f+\epsilon] for all n\geq N.

Example:

Let f_n(x) = \frac{1}{n}\sin(n^2 x).

Ideas of proof:

Given \epsilon > 0, let N = \lceil \frac{1}{\epsilon} \rceil. (This choice of N does not depend on x.)

Then \forall n\geq N, \forall x\in \mathbb{R}, |\frac{1}{n}\sin(n^2 x)| < \epsilon.

Example:

Let f_n:[0, 1]\to \mathbb{R}, f_n(x) = x^n.

Then (f_n) does not converge uniformly to f(x) = \begin{cases} 0 & x<1 \\ 1 & x = 0 \end{cases}.

f_n does not lie in the region [0, 1] for all n.

Theorem 7.12 (Corollary of Theorem 7.11) (Uniform limit theorem)

Let (f_n) be a sequence of continuous functions E\to \mathbb{R}. If (f_n) converges uniformly to f on E, then f is continuous.

Proof:

Suppose f_n\to f uniformly and \forall n, f_n is continuous.

Let p\in E and \epsilon > 0.

Since f_n\to f uniformly, \exists N\in \mathbb{N}, \forall n\geq N, \forall x\in E, |f_n(x) - f(x)| < \epsilon.

Since f_N is continuous at p, \exists \delta > 0, such that \forall x\in E, if |f_N(x) - f_N(p)| < \epsilon.

Suppose x\in B_\delta(p). Then

\begin{aligned}
|f(x) - f(p)| &= |f(x) - f_N(x) + f_N(x) - f_N(p) + f_N(p) - f(p)| \\
&\leq |f(x) - f_N(x)| + |f_N(x) - f_N(p)| + |f_N(p) - f(p)| \\
&< 3\epsilon
\end{aligned}

The \epsilon bound would not hold if we only had pointwise convergence.

|f_N(x) - f_N(p)| < 3\epsilon.

QED

Recall: If (s_n) is a sequence in \mathbb{R}, then (s_n) converges to s if and only if it is Cauchy.
i.e. \forall \epsilon > 0, \exists N\in \mathbb{N}, \forall n, m\geq N, |s_n - s_m| < \epsilon.

Theorem 7.9 (Cauchy criterion for uniform convergence)

Let (f_n) be a sequence of functions E\to \mathbb{R}. (f_n) converges uniformly to f on E if and only if (f_n) is uniformly Cauchy.

i.e. \forall \epsilon > 0, \exists N\in \mathbb{N}, \forall n, m\geq N, \forall x\in E, |f_n(x) - f_m(x)| < \epsilon.

Proof:

Exercise.

Theorem 7.10 (Weierstrass M-test)

Let (f_n) be a sequence of functions E\to \mathbb{R} (or \mathbb{C}). Suppose

• \forall n, \exists M_n\in \mathbb{R}_{\geq 0}, such that \forall x\in E, |f_n(x)| \leq M_n.
• \sum_{n=1}^{\infty} M_n converges.

Then \sum_{n=1}^{\infty} f_n(x) converges uniformly on E.

i.e. The sequence of partial sums s_n(x) = \sum_{k=1}^{n} f_k(x) converges uniformly on E.

Remark:

The proof is nearly identical to the proof of the comparison test in Chapter 3.

Proof:

By Theorem 7.8, it's enough to show that (s_n) is uniformly Cauchy.

i.e. \forall \epsilon > 0, \exists N\in \mathbb{N}, \forall m\geq n\geq N, \forall x\in E, |\sum_{k=n}^{m} f_k(x)| < \epsilon.

Let \epsilon > 0. Since \sum_{n=1}^{\infty} M_n converges, \exists N\in \mathbb{N}, \forall m\geq n\geq N, \sum_{k=n}^{m} M_k < \epsilon.

Suppose m\geq n\geq N and x\in E.

\begin{aligned}
|\sum_{k=n}^{m} f_k(x)| &\leq \sum_{k=n}^{m} |f_k(x)| \\
&\leq \sum_{k=n}^{m} M_k \\
&< \epsilon
\end{aligned}

QED

Example:

Let f_n(x) = \sum_{n=0}^{\infty} \frac{1}{2^n} \cos(3^n x).

\left|\frac{1}{2^n} \cos(3^n x)\right| \leq \frac{1}{2^n}=M_n

By Weierstrass M-test, f_n(x) converges uniformly on \mathbb{R}.

By Theorem 7.12, f(x) = \sum_{n=0}^{\infty} \frac{1}{2^n} \cos(3^n x) is continuous on \mathbb{R}.

Fun fact: f(x) is not differentiable at any x\in \mathbb{R}.