Differentiation and Uniformly Convergent Series of Functions

# Differentiation and Uniformly Convergent Series of Functions

On the Differentiation and Uniformly Convergent Sequences of Functions page we saw that if $(f_n(x))_{n=1}^{\infty}$ is a sequence of functions that are differentiable on $(a, b)$, $(f_n'(x))_{n=1}^{\infty}$ is uniformly convergent to a function $g(x)$ 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 we can conclude that $(f_n(x))_{n=1}^{\infty}$ uniformly converges to some function $f(x)$ and that $f'(x) - g(x)$.

We will now state and prove an analogous theorem for series of differentiable functions.

Theorem 1: Let $\displaystyle{\sum_{n=1}^{\infty} f_n(x)}$ be a series of differentiable functions on $(a, b)$. Suppose that $\displaystyle{\sum_{n=1}^{\infty} f_n'(x)}$ is uniformly convergent to $g(x)$ on $(a, b)$ and that there exists an $x_0 \in (a, b)$ such that the numerical series $\displaystyle{\sum_{n=1}^{\infty} f_n(x_0)}$ converges. Then $\displaystyle{\sum_{n=1}^{\infty} f_n(x)}$ is uniformly convergent to some function $f(x)$ on $(a, b)$ and $f'(x) = g(x)$. |

**Proof:**For all $n \in \mathbb{N}$ we have that the sequence of partial sums, $(s_n(x))_{n=1}^{\infty}$ for $\displaystyle{\sum_{n=1}^{\infty} f_n(x)}$ has general term:

\begin{align} \quad s_n(x) = \sum_{k=1}^{n} f_k(x) = f_1(x) + f_2(x) + ... + f_n(x) \end{align}

- Each $s_n$ is differentiable (as a finite sum of differentiable functions is differentiable) and taking the derivative of both sides gives us:

\begin{align} \quad s_n'(x) = \sum_{k=1}^{n} f_k'(x) = f_1'(x) + f_2'(x) + ... + f_n'(x) \end{align}

- Notice that this is simply the general term for the sequence of partial sums, $(s_n'(x))_{n=1}^{\infty}$ for the series $\displaystyle{\sum_{n=1}^{\infty} f_n'(x)}$, which is uniformly convergent to $g(x)$ on $(a, b)$.

- Now, since $\displaystyle{\sum_{n=1}^{\infty} f_n(x_0)}$ converges, the numerical sequence of partial sums $(s_n(x_0))_{n=1}^{\infty}$ converges.

- So $(s_n(x))_{n=1}^{\infty}$ is a sequence of functions that are differentiable, $(s_n'(x))_{n=1}^{\infty}$ uniformly converges to $g(x)$ on $(a, b)$, and there exists an $x_0 \in (a, b)$ such that the numerical sequence $(s_n(x_0))_{n=1}^{\infty}$ converges. So by the theorem presented on the Differentiation and Uniformly Convergent Sequences of Functions we have that $(s_n(x))_{n=1}^{\infty}$ is uniformly convergent on $(a, b)$ to some limit function $f(x)$, i.e., $\displaystyle{\sum_{n=1}^{\infty} f_n(x)}$ converges uniformly to $f(x)$ on $(a, b)$, and moreover, $f'(x) = g(x)$. $\blacksquare$