Differentiation and Uniformly Convergent Sequences of Functions

# Differentiation and Uniformly Convergent Sequences of Functions

We will now investigate differentiation with regards to uniformly convergent sequences of functions.

 Theorem 1: Let [$(f_n(x))_{n=1}^{\infty}$ be a sequence of real-valued functions such that $f_n’$ exists on $(a, b)$ for each $n \in \mathbb{N}$. Let $(f_n’(x))_{n=1}^{\infty}$ be the sequence of derivative functions, and suppose that $(f_n’(x))_{n=1}^{\infty}$ uniformly converges to a limit function $g$ on $(a, b)$. Assume that there exists an $x_0 \in (a, b)$ such that the numerical sequence $(f_n(x_0))_{n=1}^{\infty}$ converges. Then: a) $(f_n(x))_{n=1}^{\infty}$ uniformly converges to some function $f(x)$ on $(a, b)$. b) $f$ is differentiable on $(a, b)$ and $f’(x) = g(x)$ for all $x \in (a, b)$.
• Proof of a) Let $(f_n(x))_{n=1}^{\infty}$ be a sequence of real-valued differentiable functions on $(a, b)$ for which $(f_n’(x))_{n=1}^{\infty}$ uniformly converges to some function $g(x)$ on $(a, b)$, and assume that the numerical sequence $(f_n(x_0))_{n=1}^{\infty}$ converges for some $x_0 \in (a, b)$. We will show that then $(f_n(x))_{n=1}^{\infty}$ uniformly converges to some function $f(x)$ on $(a, b)$ by using Cauchy’s criterion for uniform convergence presented on the Uniformly Cauchy Sequences of Functions page.
• Let $m, n \in \mathbb{N}$ and define a function $F_{m,n}$ for all $x \in (a, b)$ by:
(1)
\begin{align} \quad F_{m,n}(x) = f_m(x) - f_n(x) \end{align}
• Since $f_m$ and $f_n$ are both differentiable on $(a, b)$, $F_{m,n}$ is differentiable on $(a, b)$ and hence continuous on $(a, b)$. Moreover, consider the interval whose endpoints are $x$ and $x_0$. Without loss of generality, assume this interval is of the form $[x_0, x]$. Then $F$ is continuous on $[x_0, x]$ and differentiable on $(x_0, x)$, so by the Mean Value theorem there exists a $\xi$ between $x_0$ and $x$ such that:
(2)
\begin{align} \quad F_{m,n}(x) - F_{m,n}(x_0) &= F_{m,n}’(\xi) (x - x_0) \\ \quad F_{m,n}(x) &= F_{m,n}(x_0) + F_{m,n}’(\xi) (x - x_0) \\ \quad f_m(x) - f_n(x) &= [f_m(x_0) - f_n(x_0)] + [f_m’(\xi) + f_n’(\xi)](x - x_0) \\ \end{align}
• Taking the absolute values of both sides and applying the triangle inequality gives us:
(3)
\begin{align} \quad \mid f_m(x) - f_n(x) \mid & \leq \mid f_m(x_0) - f_n(x_0) \mid + \mid f_m’(\xi) - f_n’(\xi) \mid \mid x - x_0 \mid \\ \quad \mid f_m(x) - f_n(x) \mid & \leq \mid f_m(x_0) - f_n(x_0) \mid + \mid f_m’(\xi) - f_n’(\xi) \mid (b - a) \end{align}
• Let $\epsilon > 0$ be given. Since $(f_n(x_0))_{n=1}^{\infty}$ converges, by the Cauchy criterion for numerical sequences, for $\epsilon_1 = \frac{\epsilon}{2} > 0$ there exists an $N_1 \in \mathbb{N}$ such that if $m, n \geq N_1$ we have that:
(4)
\begin{align} \quad \mid f_m(x_0) - f_n(x_0) \mid < \epsilon_1 = \frac{\epsilon}{2} \quad (*) \\ \end{align}
• Furthermore, since $(f_n’(x))_{n=1}^{\infty}$ converges uniformly, it satisfies the Cauchy criterion and so for $\epsilon_2 = \frac{\epsilon}{2(b - a)} > 0$ there exists an $N_2 \in \mathbb{N}$ such that for all $m, n \geq N_2$ and for all $x \in (a, b)$ we have that:
(5)
\begin{align} \quad \mid f_m’(x) - f_n’(x) \mid < \epsilon_2 = \frac{\epsilon}{2(b - a)} \quad (**) \end{align}
• Notice that $\xi \in (a, b)$. Let $N = \max \{ N_1, N_2 \}$. Then for all $m, n \geq N$ we have that $(*)$ and $(**)$ hold simultaneously and:
(6)
\begin{align} \quad \mid f_m(x) - f_n(x) \mid < \epsilon_1 + \epsilon_2(b - a) = \frac{\epsilon}{2} + \frac{\epsilon}{2(b - a)} (b - a) = \epsilon \end{align}
• So Cauchy’s criterion holds for the sequence $(f_n(x))_{n=1}^{\infty}$ and thus $(f_n(x))_{n=1}^{\infty}$ uniformly converges on $(a, b)$ to some function $f(x)$. [[$\blacksquare]] • Proof of b) We now establish that the limit function$f$is differentiable on$(a, b)$and that$f'(x) = g(x)$for each$x \in (a, b)$. Let$c \in (a, b). Then we have that: (7) \begin{align} \quad f'(c) = \lim_{x \to c} \frac{f(x) - f(c)}{x - c} \end{align} • Define a new functionG(x)$for each$x \in (a, b)by: (8) \begin{align} \quad G(x) = \left\{\begin{matrix} \frac{f(x)-f(c)}{x - c} & \mathrm{if} \: x \neq c\\ g(c) & \mathrm{if} \: x = c \end{matrix}\right. \end{align} • For eachn \in \mathbb{N}$define the function$G_n(x)$for$x \in (a, b)by: (9) \begin{align} \quad G_n(x) = \left\{\begin{matrix} \frac{f_n(x)-f_n(c)}{x - c} & \mathrm{if} \: x \neq c\\ f_n'(c) & \mathrm{if} \: x = c \end{matrix}\right. \end{align} • ThenG_n$is continuous at$c$since$\displaystyle{f_n'(c) = \lim_{n \to \infty} \frac{f_n(x) - f(c)}{x - c}}$exists. We claim that$(G_n(X))_{n=1}^{\infty}$uniformly converges to$G(x)$on$(a, b)$. We will use Cauchy's criterion again. Let$m, n \in \mathbb{N}, so that: (10) \begin{align} \quad G_m(x) - G_n(x) = \left\{\begin{matrix} \frac{[f_m(x)-f_n(x)] - [f_m(c) - f_n(c)]}{x - c} & \mathrm{if} \: x \neq c\\ f_m'(c) - f_n'(c) & \mathrm{if} \: x = c \end{matrix}\right. \end{align} • By the Mean Value Theorem there exists a\xi$between$x$and$csuch that: (11) \begin{align} \quad \mid [f_m(x)-f_n(x)] - [f_m(c) - f_n(c)] \mid = \mid f_m'(\xi) - f_n'(\xi) \mid \mid x - c \mid \end{align} • Therefore: (12) \begin{align} \quad G_m(x) - G_n(x) = \left\{\begin{matrix} f_m'(\xi) - f_n'(\xi) & \mathrm{if} \: x \neq c\\ f_m'(c) - f_n'(c) & \mathrm{if} \: x = c \end{matrix}\right. \end{align} • Now since(f_n'(x))_{n=1}^{\infty}$uniformly converges to$g(x)$on$(a, b)$we have that for all$\epsilon > 0$there exists an$N \in \mathbb{N}$such that for all$m, n \geq N$and for all$x \in (a, b)$we have that$\mid G_m(x) - G_n(x) \mid < \epsilon$. So, since$(G_n(x))_{n=1}^{\infty}$uniformly converges on$(a, b)$we have that$G$is continuous at$c$, and so$f'(c) = g(c)$for all$c \in (a, b)$.$\blacksquare\$