Sequences of Functions Review

# Sequences of Functions Review

We will now review some of the recent material regarding sequences of functions. Let $(f_n(x))_{n=1}^{\infty}$ be a sequence of functions with common domain $X$.

• On the Sequences of Functions we defined a Sequence of Functions denoted $(f_n(x))_{n=1}^{\infty}$ where $f_n(x)$ is a function for each $n \in \mathbb{N}$. We looked at a couple examples of sequences of functions such as the sequence $(f_n(x))_{n=1}^{\infty} = (nx)_{n=1}^{\infty} = (x, 2x, 3x, ...)$ of diagonal lines with increasing slope.
• On the Pointwise Convergence of Sequences of Functions we said that $(f_n(x))_{n=1}^{\infty}$ is Pointwise Convergent to the function $f(x)$ denoted $\displaystyle{\lim_{n \to \infty} f_n(x) = f(x) \: \mathit{pointwise}}$ if for all $x \in X$ and for all $\epsilon > 0$ there exists an $N \in \mathbb{N}$ such that if $n \geq N$ then:
(1)
\begin{align} \quad \mid f_n(x) - f(x) \mid < \epsilon \end{align}
• Similarly on the Uniform Convergence of Sequences of Functions page we said that $(f_n(x))_{n=1}^{\infty}$ is Uniformly Convergent to the function $f(x)$ denoted $\displaystyle{\lim_{n \to \infty} f_n(x) = f(x) \: \mathit{uniformly}}$ if for all $\epsilon > 0$ there exists an $N \in \mathbb{N}$ such that if $n \geq N$ then for all $x \in X$ we have that:
(2)
\begin{align} \quad \mid f_n(x) - f(x) \mid < \epsilon \end{align}
• On the Pointwise Cauchy Sequences of Functions page we said that $(f_n(x))_{n=1}^{\infty}$ is Pointwise Cauchy if for all $x \in X$ and for all $\epsilon > 0$ there exists an $N \in \mathbb{N}$ such that if $m, n \geq N$ we have that:
(3)
\begin{align} \quad \mid f_m(x) - f_n(x) \mid < \epsilon \end{align}
• We then proved that a sequence of functions $(f_n(x))_{n=1}^{\infty}$ is pointwise Cauchy if and only if it is pointwise convergent.
• Similarly on the Uniformly Cauchy Sequences of Functions page we said that $(f_n(x))_{n=1}^{\infty}$ if Uniformly Cauchy if for all $\epsilon > 0$ there exists an $N \in \mathbb{N}$ such that if $m, n \geq N$ then for all $x \in X$ we have that:
(4)
\begin{align} \quad \mid f_m(x) - f_n(x) \mid < \epsilon \end{align}
• We then proved that a sequence of functions $(f_n(x))_{n=1}^{\infty}$ is uniformly Cauchy if and only if it is uniformly convergent.
• On the Differentiation and Uniformly Convergent Sequences of Functions page we saw that if $(f_n(x))_{n=1}^{\infty}$ is a sequence of differentiable functions on $(a, b)$ where $(f_n'(x))_{n=1}^{\infty}$ uniformly converges to some function $g$ on $(a, b)$ and if there exists an $x_0 \in (a, b)$ such that the numerical sequence $(f_n(x_0))_{n=1}^{\infty}$ converges then $(f_n(x))_{n=1}^{\infty}$ uniformly converges on $(a, b)$ to some function $f(x)$ and moreover, $f$ is differentiable with $f'(x) = g(x)$ for all $x \in (a, b)$.
(5)
\begin{align} \quad \int_a^b f(x) \: dx = \int_a^b \lim_{n \to \infty} f_n(x) \: dx = \lim_{n \to \infty} \int_a^b f_n(x) \: dx \end{align}
(6)
\begin{align} \quad \int_a^b f(x) \: d \alpha(x) = \int_a^b \lim_{n \to \infty} f_n(x) \: d \alpha (x) = \lim_{n \to \infty} \int_a^b f_n(x) \: d \alpha (x \end{align}