Series of Functions

The uniform convergence of sequence of functions can be naturally extended to uniform convergence of series of functions just like what we did for real number series.

Definition. Let $(f_n: X \to \mathbf{R})$ be a sequence of functions. Then the series $\sum_{n=1}^\infty f_n$ converges at a point $x \in X$ if the sequence of partial sums

\[g_n = \sum_{j=1}^n f_j\]

converges at $x$. The series converges uniformly if $(g_n)$ converges uniformly.

Definition. The series $\sum f_n$ converges absolutely at $x$ if $\sum \vert f_n \vert$ converges at $x$. The series $\sum f_n$ converges absolutely uniformly if $\sum \vert f_n \vert$ converges uniformly.

Similar to real number series, we have the following result.

Proposition. If $\sum f_n$ converges absolutely uniformly, then $\sum f_n$ converges uniformly.

Proof.

Let $g_n = \sum_{j=1}^n f_j$ and $h_n = \sum_{j=1}^n \vert f_j \vert$. Then for $n > m$,

\[\vert g_n - g_m \vert = \left| \sum_{j=m+1}^n f_j \right| \le \sum_{j=m+1}^n \vert f_j \vert = \vert h_n - h_m \vert\]

Since $\sup \vert h_n - h_m \vert \to 0$, $\sup \vert g_n - g_m \vert \to 0$ so $\sum f_n$ converges uniformly.

Proposition. [Weierstrass M-test] Let $(f_n: X \to \mathbf{R}$ be a sequence of functions. If, for each $n$, there is a constant $M_n$ such that

\[\sup_{x \in X} \vert f_n(x) \vert \le M_n\]

and $\sum M_n$ converges, then $\sum f_n$ converges (absolutely and) uniformly on $X$.

Proof.

Given $\varepsilon > 0$. Since $\sum M_n$ converges, there exists $N$ such that for all $N < m < n$,

\[\vert M_n - M_m \vert \le \vert M_{m+1} \vert + \cdots + \vert M_n \vert < \varepsilon\]

Let $g_n = \sum_{j=1}^n \vert f_n \vert$. For the same $N$, we have for all $N < m < n$,

\[\sup \vert g_n - g_m \vert \le \sup(\vert f_{m+1} \vert + \cdots + \vert f_n \vert) \le \vert M_{m+1} \vert + \cdots + \vert M_n \vert < \varepsilon\]

Proposition. [Dirichlet Test] Let $(a_n: X \to \mathbf{R})$ and $(f_n: X \to \mathbf{R})$ be sequences of functions. If

  • $s_n = \sum a_j$ is uniformly bounded on $X$, i.e. there is $M$ such that $\vert s_n \vert < M$ for all $n$ and $x \in X$;

  • for each $x$, $(f_n(x))$ is a monotonic sequence;

  • $f_n \to 0$ uniformly on $X$,

then $\sum a_n f_n$ converges uniformly on $X$.

Proposition. [Abel Test] Let $(a_n: X \to \mathbf{R})$ and $(f_n: X \to \mathbf{R})$ be sequences of functions. If

  • the series $\sum a_j$ converges uniformly on $X$;

  • for each $x$, $(f_n(x))$ is a monotonic sequence;

  • $f_n(x)$ is uniformly bounded on $X$,

then $\sum a_n f_n$ converges uniformly on $X$.

Power Series

With uniformity we can prove some deeper results about power series.

Proposition. If the power series $\sum a_n z^n$ has radius of convergence $R$ and $0 < r < R$, then $\sum a_n z^n$ converges uniformly for $\vert z \vert \le r$.

Proof.

When $\vert z \vert \le r$, we have $\vert a_n z^n \vert \le \vert a_n \vert r^n$ in which the series $\sum a_n r^n$ converges since $0 < r < R$. By Weierstrass M-test, $\sum a_n z^n$ converges uniformly for $\vert z \vert \le r$.

Corollary. The sum of a power series is continuous within its circle of convergence.

Proof.

Let $z_0$ be any point such that $\vert z_0 \vert < R$. If $\vert z_0 < r < R$, $\sum a_n z^n$ converges uniformly for $\vert z \vert \le r$ so $\sum a_n z^n$ is continuous at $z_0$.

Proposition. [Principle of equating coefficients] If $\sum a_n z^n = \sum b_n z^n$ for a sequence of non-zero values of $z$ tending to $0$ as limit, then $a_n = b_n$ for all $n$.

Proof.

Let $(z_k)$ be the non-zero complex sequence satisfying the conditions. Assume the contrary that $a_m, b_m$ being the first pair of unequal coefficients. Then

\[z_m \sum_{n = 0}^\infty (a_{m+n} - b_{m+n}) z^n = 0\]

and therefore

\[\sum_{n = 0}^\infty (a_{m+n} - b_{m+n}) z^n = 0\]

for all non-zero $z_k$. The last power series clearly has a positive radius of convergence so it is continuous at $z = 0$. Hence, by substituting $z = 0$ we have $a_m - b_m = 0$ which is a contradiction.

Proposition. [Termwise differentiation] Suppose that $\sum a_n x^n$ is a real power series converges to $f(x)$ with radius of convergence $R > 0$. Then $f(x)$ is differentiable in $(-R, R)$ where $f’(x) = \sum n a_n x^{n-1}$ has the same radius of convergence $R$.

Proof.

Let $R’$ be the radius of convergence for $g(x) = \sum n a_n x^{n-1}$. We have

\[\vert a_n x^n \vert = \vert a_n x^{n-1} \vert \vert x \vert \le \vert n a_n x^{n-1} \vert \vert x \vert\]

so by comparison test, if $g(x)$ converges absolutely for some $x$ then $f(x)$ also converges and $R’ \le R$.

If $R’ < R$, then there are $R’ < r’ < r < R$ such that $\sum n a_n (r’)^{n-1}$ diverges but $\sum a_n r^n$ converges. But

\[{ \vert a_n r^n \vert \over \vert n a_n (r')^{n-1} \vert } = \left( {r \over r'} \right)^{n-1} {r \over n} \ge 1\]

so $\vert n a_n (r’)^{n-1} \vert \le \vert a_n r^n \vert$ for sufficiently large $n$ so we must have $R’ = R$.

Let $f_n(x) = \sum_{j=0}^n a_j x^j$ and $f_n’(x) = \sum_{j=1}^n j a_j x^j$. $f_n(x)$ is obviously convergent at a point and $f_n’(x)$ is uniformly convergent for $\vert x \vert \le r < R$. So on any closed subinterval of $(-R, R)$, $f(x) = \lim_{n \to \infty} f_n(x)$ is differentiable with $f’(x) = \lim_{n \to \infty} f_n’(x)$.

References