Proof of Continuity of Complex-Valued Functions, and the Modulus of a Power

Routine facts about real and complex numbers, and the values of the pointwise defined functions $f+g$, $fg$, $\lambda f$, $1/g$, $h\circ f$, $\operatorname{Re}f$, $\operatorname{Im}f$, $|f|$ and $\overline{f}$, are used without comment (Routine Facts, Continued: Typed Arithmetic, Maps and Predicativity); the continuity results cited below, stated for the earlier definitions of metric spaces and continuity, are read in the current ones by The Earlier and the New Versions of the Definitions of Metric Spaces, Convergence, Continuity and Compactness Agree (i) and The Earlier and the New Versions of the Definitions of Metric Spaces, Convergence, Continuity and Compactness Agree (vii). Distances in $\mathbb{C}$ are moduli of differences (Metric Spaces: Balls, the Metric Topology, and the Real and Complex Numbers as Metric Spaces (v)) and distances in $\mathbb{R}$ are absolute values of differences (Metric Spaces: Balls, the Metric Topology, and the Real and Complex Numbers as Metric Spaces (iv)).

The modulus of a power. We prove (i) by induction on $n$. For $n=0$ we have $z^{0}=1$ and $|z|^{0}=1$ (Powers with Exponents in the Natural Numbers with Zero (ii)), and the modulus of the real number $1$ is its absolute value $1$ (The Modulus of a Complex Number (ii)); so both sides equal $1$. Suppose now that $|z^{n}|=|z|^{n}$. By the laws of exponents, $z^{n+1}=z^{n}z$ (Powers in a Commutative Ring, a Field and an Ordered Field: Exponent Laws, Factorisation, Geometric Sums, Monotonicity and Bernoulli's Inequality (i), Powers in a Commutative Ring, a Field and an Ordered Field: Exponent Laws, Factorisation, Geometric Sums, Monotonicity and Bernoulli's Inequality (ii)), and the modulus is multiplicative (Conjugation and Modulus of Complex Numbers: Field Automorphism, Real and Imaginary Parts, Multiplicativity, Inverses, Component Bounds, Triangle Inequalities and Finite Sums (iv)); hence

$$|z^{n+1}|=|z^{n}|\,|z|=|z|^{n}\,|z|=|z|^{n+1},$$

which completes the induction.

Constants and the identity. We prove (ii). Let $\varepsilon>0$. If $f$ is the constant function with value $c$, then for every $x\in A$ the distance of $f(x)=c$ and $f(a)=c$ is $|c-c|=0<\varepsilon$, so every $\delta>0$ meets the requirement of Continuous Map Between Metric Spaces (i). If $(X,d)=(\mathbb{C},d_{\mathbb{C}})$ and $f$ is the identity $w\mapsto w$ on $A$, then the distance of the values $f(w)$ and $f(a)$ is $d_{\mathbb{C}}(w,a)$, the very distance of the points; so $\delta=\varepsilon$ works.

Sums, products and scalar multiples. We prove (iii); let $f$ and $g$ be continuous at $a$ relative to $A$. For $x\in A$, the triangle inequality (Conjugation and Modulus of Complex Numbers: Field Automorphism, Real and Imaginary Parts, Multiplicativity, Inverses, Component Bounds, Triangle Inequalities and Finite Sums (vii)) and multiplicativity of the modulus (Conjugation and Modulus of Complex Numbers: Field Automorphism, Real and Imaginary Parts, Multiplicativity, Inverses, Component Bounds, Triangle Inequalities and Finite Sums (iv)) give

$$|(f+g)(x)-(f+g)(a)|\le|f(x)-f(a)|+|g(x)-g(a)|,\qquad|\lambda f(x)-\lambda f(a)|=|\lambda|\,|f(x)-f(a)|,$$

and, writing $f(x)g(x)-f(a)g(a)=f(x)\bigl(g(x)-g(a)\bigr)+g(a)\bigl(f(x)-f(a)\bigr)$,

$$|f(x)g(x)-f(a)g(a)|\le|f(x)|\,|g(x)-g(a)|+|g(a)|\,|f(x)-f(a)|.$$

Let $\varepsilon>0$. For the product we choose in this order. First, by continuity of $f$, choose $\delta_{1}>0$ such that $|f(x)-f(a)|<1$ whenever $x\in A$ and $d(a,x)<\delta_{1}$; for such $x$ the triangle inequality gives $|f(x)|<|f(a)|+1$. Next choose $\delta_{2}>0$ from the continuity of $f$ for the tolerance $\varepsilon/(2(|g(a)|+1))$, and $\delta_{3}>0$ from the continuity of $g$ for the tolerance $\varepsilon/(2(|f(a)|+1))$. Finally let $\delta=\min(\delta_{1},\delta_{2},\delta_{3})$. For $x\in A$ with $d(a,x)<\delta$ the second display gives

$$|f(x)g(x)-f(a)g(a)|<(|f(a)|+1)\frac{\varepsilon}{2(|f(a)|+1)}+|g(a)|\frac{\varepsilon}{2(|g(a)|+1)}<\varepsilon,$$

so $fg$ is continuous at $a$ relative to $A$. For $f+g$, take the minimum of the $\delta$'s that the continuity of $f$ and of $g$ provide for $\varepsilon/2$; for $\lambda f$, take the $\delta$ that the continuity of $f$ provides for $\varepsilon/(|\lambda|+1)$, since $|\lambda|\,\varepsilon/(|\lambda|+1)<\varepsilon$. In each case the first display gives the required bound.

Reciprocals. We prove (iv); let $g$ be continuous at $a$ relative to $A$ and nowhere zero on $A$. Since $g(a)\neq0$, we have $|g(a)|>0$ (Conjugation and Modulus of Complex Numbers: Field Automorphism, Real and Imaginary Parts, Multiplicativity, Inverses, Component Bounds, Triangle Inequalities and Finite Sums (iii)). Let $\varepsilon>0$; we choose in this order. First choose $\delta_{1}>0$ such that $|g(x)-g(a)|<|g(a)|/2$ whenever $x\in A$ and $d(a,x)<\delta_{1}$; for such $x$ the reverse triangle inequality (Conjugation and Modulus of Complex Numbers: Field Automorphism, Real and Imaginary Parts, Multiplicativity, Inverses, Component Bounds, Triangle Inequalities and Finite Sums (vii)) gives $|g(x)|\ge|g(a)|-|g(x)-g(a)|\ge|g(a)|/2$, and so, by multiplicativity of the modulus and the modulus of an inverse (Conjugation and Modulus of Complex Numbers: Field Automorphism, Real and Imaginary Parts, Multiplicativity, Inverses, Component Bounds, Triangle Inequalities and Finite Sums (iv), Conjugation and Modulus of Complex Numbers: Field Automorphism, Real and Imaginary Parts, Multiplicativity, Inverses, Component Bounds, Triangle Inequalities and Finite Sums (v)),

$$\Bigl|\frac{1}{g(x)}-\frac{1}{g(a)}\Bigr|=\frac{|g(a)-g(x)|}{|g(x)|\,|g(a)|}\le\frac{2\,|g(x)-g(a)|}{|g(a)|^{2}}.$$

Next choose $\delta_{2}>0$ from the continuity of $g$ for the tolerance $\varepsilon|g(a)|^{2}/2$. Finally let $\delta=\min(\delta_{1},\delta_{2})$: for $x\in A$ with $d(a,x)<\delta$ the display is less than $\varepsilon$, so $1/g$ is continuous at $a$ relative to $A$.

Composition. Assertion (v) is the composition rule Continuity of Identity Maps, Compositions and Reciprocals (ii) applied with $(X,d_{X})=(X,d)$, $(Y,d_{Y})=(Z,d_{Z})=(\mathbb{C},d_{\mathbb{C}})$, the subsets $A\subseteq X$ and $B\subseteq\mathbb{C}$, the inner map $f$ and the outer map $h$.

Parts, modulus and conjugate. We prove (vi). For $x\in A$ put $u=f(x)-f(a)$. Additivity of the real and imaginary parts (Conjugation and Modulus of Complex Numbers: Field Automorphism, Real and Imaginary Parts, Multiplicativity, Inverses, Component Bounds, Triangle Inequalities and Finite Sums (ii)), the component bounds (Conjugation and Modulus of Complex Numbers: Field Automorphism, Real and Imaginary Parts, Multiplicativity, Inverses, Component Bounds, Triangle Inequalities and Finite Sums (vi)), the reverse triangle inequality (Conjugation and Modulus of Complex Numbers: Field Automorphism, Real and Imaginary Parts, Multiplicativity, Inverses, Component Bounds, Triangle Inequalities and Finite Sums (vii)), additivity of conjugation (Conjugation and Modulus of Complex Numbers: Field Automorphism, Real and Imaginary Parts, Multiplicativity, Inverses, Component Bounds, Triangle Inequalities and Finite Sums (i)) and $|\overline{u}|=|u|$ (Conjugation and Modulus of Complex Numbers: Field Automorphism, Real and Imaginary Parts, Multiplicativity, Inverses, Component Bounds, Triangle Inequalities and Finite Sums (iii)) give

$$|\operatorname{Re}f(x)-\operatorname{Re}f(a)|=|\operatorname{Re}u|\le|u|,\quad|\operatorname{Im}f(x)-\operatorname{Im}f(a)|\le|u|,\quad\bigl||f(x)|-|f(a)|\bigr|\le|u|,\quad|\overline{f(x)}-\overline{f(a)}|=|u|.$$

Hence, if $f$ is continuous at $a$ relative to $A$ and $\varepsilon>0$, the $\delta$ that the continuity of $f$ provides for $\varepsilon$ serves $\operatorname{Re}f$, $\operatorname{Im}f$, $|f|$ and $\overline{f}$ at once. Conversely, suppose that $\operatorname{Re}f$ and $\operatorname{Im}f$ are continuous at $a$ relative to $A$. Since $|u|\le|\operatorname{Re}u|+|\operatorname{Im}u|$ (Conjugation and Modulus of Complex Numbers: Field Automorphism, Real and Imaginary Parts, Multiplicativity, Inverses, Component Bounds, Triangle Inequalities and Finite Sums (vi)), given $\varepsilon>0$ the minimum of the $\delta$'s that the continuity of $\operatorname{Re}f$ and of $\operatorname{Im}f$ provide for $\varepsilon/2$ serves $f$.