Weierstrass Approximation Theorem/Complex Case

Theorem

Let $\Bbb I = \closedint a b$ be a closed real interval.

Let $f: \Bbb I \to \C$ be a continuous complex function.

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:

\(\ds \norm {p - f}_\infty\)

\(=\)

\(\ds \sup_{x \mathop \in \Bbb I} \size {\map p x - \map f x}\)

Definition of Supremum Norm

\(\ds \)

\(=\)

\(\ds \sup_{x \mathop \in \Bbb I} \size {\map u x + i \map v x - \map \Re {\map f x} - i \map \Im {\map f x} }\)

Definition of Complex Number

\(\ds \)

\(\le\)

\(\ds \sup_{x \mathop \in \Bbb I} \size {\map u x - \map \Re {\map f x} } + \sup_{x \mathop \in \Bbb I} \size i \size {\map v x - \map \Im {\map f x} }\)

Triangle Inequality for Complex Numbers

\(\ds \)

\(=\)

\(\ds \norm {u - \Re f}_\infty + \norm {v - \Im f}_\infty\)

Definition of Supremum Norm, and complex modulus of $i$

\(\ds \)

\(<\)

\(\ds \dfrac \epsilon 2 + \dfrac \epsilon 2\)

\(\ds \)

\(=\)

\(\ds \epsilon\)

$\blacksquare$