Monday, March 18, 2013

Cauchy Criterion for function sequences, introduction to the norm

1Criteria for a sequence of functions not to uniformly converge¶

Lemma: A sequence of functions $(f_n)$ on $A \subseteq \mathbb{R}$ to $\mathbb{R}$ does not converge uniformly on $A_0 \subseteq A$ to $f : A_0 \to \mathbb{R}$ $\iff$ for some $\epsilon > 0$, there exists a subsequence $(f_{n_k})$ of $(f_n)$ and a sequence $(x_k) \in A_0$ such that $|f_{n_k}(x_k) - f(x_k)| \geq \epsilon$ for all $k \in \mathbb{N}$.

Definition: Let $A \subseteq \mathbb{R}$, $\varphi : A \to \mathbb R$ a function. If $\varphi$ is bounded on $A$, we define the uniform norm of $\varphi$ on $A$, denoted

$$\|\varphi\|_A = \sup \{|\varphi(x)| : x \in A \}$$

note that $\| \varphi \|_A \leq \epsilon \iff |\varphi(x)| \leq \epsilon$ for all $x \in A$.

Lemma: A sequence of bounded functions $(f_n)$ on $A \subseteq \mathbb R$ converges uniformly to $f$ on $A$ $\iff$ $\| f_n - f \|_A \to 0$.

Proof: ($\Rightarrow$) then, for a given $\epsilon > 0$ there exists some $N \in \mathbb N$ such that if $n \geq N$ then $|f_n(x) - f(x)| < \epsilon / 2$ for all $x \in A$. Then

$$\sup \{ | f_n(x) - f(x)| : x \in A \} = \|f_n - f\|_A \leq \frac{\epsilon}{2} < \epsilon$$

i.e. $|\|f_n - f\|_A - 0| < \epsilon$ so $\displaystyle\lim_{n \to \infty} \| f_n - f \|_A = 0$

($\Leftarrow$) for a given $\epsilon > 0$ there exists some $N \in \mathbb N$ such that if $n \geq N$, $| \|f_n - f\|_A - 0| < \epsilon$, i.e. $\|f_n - f\|_A < \epsilon$, i.e.

$$\sup \{|f_n(x) - f(x)| : x \in A \} < \epsilon$$

i.e. $|f_n(x) - f(x)| < \epsilon$ for all $x \in A$ i.e. $f_n$ converges uniformly to $f$ on $A$.

2Cauchy Criterion¶

Theorem: Let $(f_n)$ be a sequence of bounded functions on $A \subseteq \mathbb R$ to $\mathbb R$. $(f_n)$ converges uniformly to a bounded function $f$ on $A$ $\iff$ for a given $\epsilon > 0$, there exists some $N \in \mathbb N$ such that if $n, m \geq N$ then $\|f_n - f_m\|_A < \epsilon$.

Proof: ($\Rightarrow$) $\epsilon > 0$ given, there exists $N \in \mathbb N$ such that if $n \geq N$, $|f_n(x) - f(x)| < \epsilon / 4$ for all $x \in A$. Thus, if $n, m \geq N$

$$\begin{align*} |f_n(x) - f_m(x)| &= |f_n(x) - f(x) + f(x) - f_m(x)| \\ &\leq |f_n(x) - f(x)| + |f_m(x) - f(x)| \\ &< \frac{\epsilon}{4} + \frac{\epsilon}{4} = \frac{\epsilon}{2} \end{align*}$$

thus $\sup \{|f_n(x) - f_m(x)| : x \in A \} = \| f_n - f_m \|_A \leq \epsilon/2 < \epsilon$.

($\Leftarrow$) if $\epsilon > 0$ is given, there exists some $N \in \mathbb N$ such that if $n, m \geq N$, $\|f_n - f_m\|_A < \epsilon / 2$. Thus for $x \in A$, $|f_n(x) - f_m(x)| < \epsilon / 2$, i.e. $(f_n(x))$ is a Cauchy sequence in $\mathbb R$. Thus $f_n(x)$ converges in $\mathbb R$. Call its limit $f(x)$ - we must show that $f_n$ converges uniformly to $f$. If we let $m \to \infty$ we can obtain

$$|f_n(x) - f(x)| \leq \frac{\epsilon}{2} < \epsilon$$

by the squeeze theorem. Thus $f_n \rightrightarrows f$. It remains to show that $f$ is bounded. $f_N$ is bounded, i.e. $|f_N(x)| \leq B$ for all $x \in A$ for some $B \in \mathbb R$. Thus since $|f_N(x) - f(x)| < \epsilon$ for all $x \in A$, $|f(x)| < \epsilon + |f_N(x)| \leq B + \epsilon$ for all $x \in A$.

3Continuity of functions formed by uniformly convergent sequences¶

Theorem: Let $(f_n)$ be a sequence of continuous functions on $A \subseteq \mathbb R$ and suppose $f_n \rightrightarrows f$ on $A$ to a function $f : A \to \mathbb R$. Then $f$ is continuous on $A$.

Proof: $f_n \rightrightarrows f$ on $A$ thus for a given $\epsilon > 0$, there exists some $N \in \mathbb N$ such that if $n \geq N$ then $|f_n(x) - f(x)| < \epsilon/3$ for all $x \in A$. In particular, $|f_N(x) - f(x)| < \epsilon/3$ for all $x \in A$. $f_N(x)$ is continuous on $A$. Let $c \in A$, $\epsilon > 0$ given, $\exists \delta > 0$ such that if $|x - c| < \delta$ then $|f_N(x) - f_N(c)| < \epsilon / 3$. Thus if $|x - c| < \delta$,

$$\begin{align*} |f(x) - f(c)| &= |f(x) - f_N(x) + f_N(x) - f_N(c) + f_N(c) - f(c)| \\ &\leq |f(x) - f_N(x)| + |f_N(x) - f_N(c)| + |f_N(c) - f(c)| \\ &< \frac{\epsilon}{3} + \frac{\epsilon}{3} + \frac{\epsilon}{3} = \epsilon \end{align*}$$

i.e. $f$ is continuous at $c$.