Weierstrass M-Test

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Theorem

Let $f_n$ be a sequence of real functions defined on a domain $D \subseteq \R$.

Let $\displaystyle \sup_{x \mathop \in D} \left|{f_n \left({x}\right)}\right| \le M_n$ for each integer $n$ and some constants $M_n$

Let $\displaystyle \sum_{i \mathop = 1}^\infty M_i < \infty$.


Then $\displaystyle \sum_{i \mathop = 1}^\infty f_i$ converges uniformly on $D$.


Proof

Let:

$\displaystyle S_n = \sum_{i \mathop = 1}^n f_i$

Let:

$\displaystyle f = \lim_{n \to \infty} S_n$

To show the partial sums converge uniformly to $f$, we must show that:

$\displaystyle \lim_{n \to \infty}\sup_{x \mathop \in D} \left|{f - S_n}\right| = 0$

But:

\(\displaystyle \sup_{x \mathop \in D} \left\vert{f - S_n}\right\vert\) \(=\) \(\displaystyle \sup_{x \mathop \in D} \left\vert{\left({f_1 + f_2 + \cdots}\right) - \left({f_1 + f_2 + \cdots + f_n}\right)}\right\vert\)
\(\displaystyle \) \(=\) \(\displaystyle \sup_{x \mathop \in D} \left\vert{f_{n+1} + f_{n+2} + \ldots}\right\vert\)


By the Triangle Inequality, this value is less than or equal to:

$\displaystyle \sum_{i \mathop = n+1}^\infty \sup_{x \mathop \in D} \left|{f_i \left({x}\right)}\right| \le \sum_{i \mathop = n+1}^\infty M_i$

We have that:

$\displaystyle 0 \le \sum_{i \mathop = 1}^\infty M_n < \infty$

It follows from Tail of Convergent Series tends to Zero:

$\displaystyle 0 \le \lim_{n \to \infty} \sum_{i \mathop = n+1}^\infty \sup_{x \mathop \in D} \left|{f_i \left({x}\right)}\right| \le \lim_{n \to \infty} \sum_{i \mathop = n+1}^\infty M_i = 0$


So:

$\displaystyle \lim_{n \to \infty} \sup_{x \mathop \in D} \left|{f - S_n}\right| = 0$


Hence the series converges uniformly on the domain.

$\blacksquare$



Also known as

Some sources do not use the hyphen: Weierstrass $M$ test.


Source of Name

This entry was named for Karl Theodor Wilhelm Weierstrass.


Historical Note

The Weierstrass M-Test was developed by Karl Weierstrass during his investigation of power series.


Sources