## Friday, December 16, 2011

### Weierstrass M-test (Real Analysis)

In mathematics, the Weierstrass M-test is a criterion for determining whether an infinite series of functions converges uniformly. It applies to series whose terms are functions with real or complex values, and is analogous to the comparison test for determining the convergence of series of real or complex numbers.

Statement

Conditions Suppose {fn} is a sequence of real- or complex-valued functions defined on a set A, and that there exist positive constants Mn such that
$|f_n(x)|\leq M_n$
for all n1 and all $x \in A$, where the series
$\sum_{n=1}^{\infty} M_n$
converges.
Conclusion Then, the series
$\sum_{n=1}^{\infty} f_n (x)$
converges uniformly on A.

Remark: The result is often used in combination with the uniform limit theorem. Together they say that if, in addition to the above conditions, the set A is a topological space and the functions fn are continuous on A,
, then the series converges to a continuous function.

Generalization

A more general version of the Weierstrass M-test holds if the codomain of the functions {fn} is any Banach space, in which case the statement
$|f_n|\leq M_n$
may be replaced by
$\|f_n\|\leq M_n$,
where $\|\cdot\|$ is the norm on the Banach space. For an example of the use of this test on a Banach space, see the article Fréchet derivative.

Proof
Suppose that the conditions stated above hold. We must show that $\sum_{n=1}^{\infty}f_n(x)$ converges uniformly on A.
First note that, since $\sum_{n=1}^{\infty}M_n$ converges, the comparison test gives that $\sum_{n=1}^{\infty}f_n(x)$ is absolutely convergent. Therefore, $\sum_{n=1}^{\infty}f_n(x)$ converges pointwise on A to some s(x).
Again because $\sum_{n=1}^{\infty}M_n$ converges, for every ε > 0 there exists an $N \in \mathbb{N}$ such that
$\left|\sum_{n=1}^{\infty}M_n - \sum_{n=1}^{n_0}M_n\right| = \left|\sum_{n=n_0+1}^{\infty}M_n \right| = \sum_{n=n_0+1}^{\infty}M_n < \varepsilon$ whenever $n_0 \ge N$.
Thus, if $s_{n_0}(x)$ represents the sum of the first n0 terms of (fn(x)),
$|s_{n_0}(x)-s(x)| = \left|\sum_{n=n_0+1}^{\infty} f_n(x)\right| \le \sum_{n=n_0+1}^{\infty}|f_n(x)| \le \sum_{n=n_0+1}^{\infty}M_n < \varepsilon$, for all $x \in A$ whenever $n_0 \ge N.$
Since this shows that the tail of s(x) is less than ε independently of the value of $x \in A$, uniform convergence is proved.

Exponential function convergence using Weirstrass M-test
The series expansion of the exponential function can be shown to be uniformly convergent on any bounded subset S of $\mathbb{C}$ using the Weierstrass M-test.
Here is the series:
$\sum_{n=0}^{\infty}\frac{z^n}{n!}.$
Any bounded subset is a subset of some disc DR of radius R, centered on the origin in the complex plane. The Weierstrass M-test requires us to find an upper bound Mn on the terms of the series, with Mn independent of the position in the disc:
$\left| \frac{z^n}{n!}\right|\le M_n , \forall z\in D_R.$
This is trivial:
$\left| \frac{z^n}{n!}\right| \le \frac{\left| z\right|^n}{n!} \le \frac{R^n}{n!}$
$\Rightarrow M_n=\frac{R^n}{n!}.$
If $\sum_{n=0}^{\infty}M_n$ is convergent, then the M-test asserts that the original series is uniformly convergent.
The ratio test can be used here:
$\lim_{n \to \infty}\frac{M_{n+1}}{M_n}=\lim_{n \to \infty}\frac{R^{n+1}}{R^n}\frac{n!}{(n+1)!}=\lim_{n \to \infty}\frac{R}{n+1}=0$
which means the series over Mn is convergent. Thus the original series converges uniformly for all $z\in D_R$, and since $S\subset D_R$, the series is also uniformly convergent on S.