Weierstrass Function is Continuous
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Theorem
Let $a \in \openint 0 1$.
Let $b$ be a strictly positive odd integer such that:
- $\ds a b > 1 + \frac 3 2 \pi$
Let $f: \R \to \R$ be a real function defined by:
- $\ds \map f x = \sum_{n \mathop = 0}^\infty a^n \map \cos {b^n \pi x}$
for each $x \in \R$.
Then $f$ is well-defined and continuous.
Proof
Note that:
- $\ds \sup_{x \mathop \in \R} \size {a^n \map \cos {b^n \pi x} } = a^n$
Since $a \in \openint 0 1$:
- $\ds \sum_{n \mathop = 0}^\infty a^n$ converges.
So, by the Weierstrass M-Test:
- $\ds \sum_{n \mathop = 0}^\infty a^n \map \cos {b^n \pi x}$ converges uniformly on $\R$.
That is, $f$ is well-defined.
Further, from Uniformly Convergent Series of Continuous Functions Converges to Continuous Function: Corollary:
- $f$ is continuous.
$\blacksquare$