Weierstrass Approximation Theorem/Complex Case

Theorem
Let $f : \Bbb I \to \C$ be a continuous complex function, where $\Bbb I = \closedint a b$ is a closed real interval.

Let $\epsilon \in \R_{>0}$.

Then there exists a complex polynomial function $p : \Bbb I \to \C$ such that:


 * $\norm { p - f }_\infty < \epsilon$

where $\norm { f }_\infty$ denotes the supremum norm of $f$ on $\Bbb I$.

Proof
Let $\Re f : \Bbb \to \R$ be the real function defined by $\map {\Re f}{x} = \map {\Re }{ \map f x }$, where $\map \Re z$ denotes the real part of a complex number $z \in \C$.

Let $\Im f : \Bbb \to \R$ be the real function defined by $\map {\Im f}{x} = \map {\Im }{ \map f x }$, where $\map \Im z$ denotes the imaginary part of $z \in \C$.

From Real and Imaginary Part Projections are Continuous and Composite of Continuous Mappings is Continuous, it follows that $\Re f$ and $\Im f$ are continuous functions.

By the Weierstrass Approximation Theorem, there exist two real polynomial functions $u ,v : \Bbb I \to \R$ such that:


 * $\norm { u - \Re f }_\infty < \dfrac \epsilon 2$, $\norm { v - \Im f }_\infty < \dfrac \epsilon 2$

Set $p := u + iv$, so $p$ becomes a complex polynomial function.

Then: