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Next: 8.3 Limits Up: 8. Continuity Previous: 8.1 Compositions with Sequences   Index

8.2 Continuity

8.6   Definition (Continuous) Let $f$ be a complex function and let $p\in\mbox{{\rm dom}}(f)$. We say $f$ is continuous at $p$ if

\begin{displaymath}\mbox{ for every sequence } x \mbox{ in } \mbox{{\rm dom}}(f)...{1ex}\Longrightarrow\hspace{1ex}$}f\circ x\to

i.e., if

\begin{displaymath}\mbox{ for every sequence } \{x_n\}\mbox{ in } \mbox{{\rm dom...
...ace{1ex}\Longrightarrow\hspace{1ex}$}\{f(x_n)\}\to f(p)\right).\end{displaymath}

Let $B$ be a subset of $S$. We say $f$ is continuous on $B$ if $f$ is continuous at $q$ for all $q\in B$. We say $f$ is continuous if $f$ is continuous on $\mbox{{\rm dom}}(f)$; i.e., if $f$ is continuous at every point at which it is defined.

8.7   Examples. If $f(z)=z$ for all $z\in\mbox{{\bf C}}$, then $f$ is continuous. In this case $f\circ x=x$ for every sequence $x$ so the condition for continuity at $p$ is

\begin{displaymath}x\to p\mbox{$\hspace{1ex}\Longrightarrow\hspace{1ex}$}x\to p.\end{displaymath}

If $a\in\mbox{{\bf C}}$, then the constant function $\tilde a$ is continuous since for all $p\in\mbox{{\bf C}}$, and all complex sequences $x$,

\begin{displaymath}x\to p\mbox{$\hspace{1ex}\Longrightarrow\hspace{1ex}$}\tilde a\circ x=\tilde a\to a=\tilde a(p).\end{displaymath}

Notice that $\mbox{\rm Re}$ and $\mbox{\rm Im}$ (Real part and imaginary part) are functions from $\mbox{{\bf C}}$ to $\mbox{{\bf R}}$. In theorem 7.39 we showed if $x$ is any complex sequence and $L\in\mbox{{\bf C}}$, then

\begin{displaymath}x\to L\mbox{$\hspace{1ex}\Longrightarrow\hspace{1ex}$}\mbox{\rm Re}(x)\to \mbox{\rm Re}(L)\end{displaymath}


\begin{displaymath}x\to L\mbox{$\hspace{1ex}\Longrightarrow\hspace{1ex}$}\mbox{\rm Im}(x)\to\mbox{\rm Im}(L).\end{displaymath}

Hence $\mbox{\rm Re}$ and $\mbox{\rm Im}$ are continuous functions on $\mbox{{\bf C}}$.

8.8   Theorem. If $\mbox{\rm abs}$ and $\mbox{\rm conj}$ are functions from $\mbox{{\bf C}}$ to $\mbox{{\bf C}}$ defined by

\mbox{\rm abs}(z)&=&\vert z\vert \mbox{ for all } z\in\mbox{{\...
...r all } z\in\mbox{{\bf C}},\glossary{$\mbox{\rm conj}(z) = z^*$}

then $\mbox{\rm abs}$ and $\mbox{\rm conj}$ are continuous.

Proof: Let $a\in\mbox{{\bf C}}$ and let $x$ be any sequence in $\mbox{{\bf C}}$ such that $\{x_n\}\to a$; i.e., $\{x_n-a\}$ is a null sequence. By the reverse triangle inequality,

\begin{displaymath}\vert x_n-a\vert\geq \vert x_n\vert-\vert a\vert\end{displaymath}


\begin{displaymath}\vert x_n-a\vert=\vert a-x_n\vert\geq \vert a\vert-\vert x_n\vert,\end{displaymath}

so we have

\begin{displaymath}-\vert x_n-a\vert\leq \vert x_n\vert-\vert a\vert\leq \vert x_n-a\vert\end{displaymath}

and hence

\begin{displaymath}\left\vert \vert x_n\vert-\vert a\vert\right\vert\leq\vert x_n-a\vert.\end{displaymath}

It follows by the comparison theorem that $\{\vert x_n\vert-\vert a\vert\}$ is a null sequence; i.e., $\{\vert x_n\vert\}\to\vert a\vert$. Hence $\mbox{\rm abs}$ is continuous.

Since $\vert x_n^*-a^*\vert=\vert(x_n-a)^*\vert=\vert x_n-a\vert$, the comparison theorem shows that

\begin{displaymath}\{x_n\}\to a\mbox{$\hspace{1ex}\Longrightarrow\hspace{1ex}$}\{x_n^*\}\to a^*;\end{displaymath}

i.e., $\mbox{\rm conj}$ is continuous. $\mid\!\mid\!\mid$

8.9   Example. If

\begin{displaymath}f(z)=\cases{ z &for $z\in\mbox{{\bf C}}\backslash\{0\}$, \cr
1 &for $z=0$,\cr}\end{displaymath}

then $f$ is not continuous at $0$, since

\left\{{1\over n}\right\}\to 0 \end{displaymath}


\begin{displaymath}\left\{f\left({1\over n}\right)\right\}=\left\{{1\over n}\right\}\to 0\neq f(0).

Notice that to show that a function $f$ is not continuous at a point $a$ in its domain, it is sufficient to find one sequence $\{x_n\}$ in $\mbox{{\rm dom}}(f)$ such that $\{x_n\}\to a$ and either $\{f(x_n)\}$ converges to a limit different from $f(a)$ or $\{f(x_n)\}$ diverges.

8.10   Theorem (Sum and Product theorems.) Let $f,g$ be complex functions, and let $a\in\mbox{{\rm dom}}(f)\cap\mbox{{\rm dom}}(g)$. If $f$ and $g$ are continuous at $a$, then $f+g$, $f-g$, and $f\cdot g$ are continuous at $a$.

Proof: Let $\{x_n\}$ be a sequence in domain $(f+g)$ such that $\{x_n\}\to a$. Then $x_n\in\mbox{{\rm dom}}(f)$ for all $n$ and $x_n\in\mbox{{\rm dom}}(g)$ for all $n$, and by continuity of $f$ and $g$ at $a$, it follows that

\begin{displaymath}\{f(x_n)\}\to f(a) \mbox{ and } \{g(x_n)\}\to g(a).\end{displaymath}

By the sum theorem for sequences,

\begin{displaymath}\{(f+g)(x_n)\}=\{f(x_n)+g(x_n)\}\to f(a)+g(a)=(f+g)(a).\end{displaymath}

Hence $f+g$ is continuous at $a$. The proofs of continuity for $f-g$ and $f\cdot g$ are similar.

8.11   Theorem (Quotient theorem.) Let $f,g$ be complex functions and let $\displaystyle {a\in\mbox{{\rm dom}}\left({f\over g}\right)}$. If $f$ and $g$ are continuous at $a$, then $\displaystyle {{f\over g}}$ is continuous at $a$.

8.12   Exercise. Prove the quotient theorem. Recall that

\begin{displaymath}\mbox{{\rm dom}}\left({f\over
g}\right)=\left(\mbox{{\rm dom}...
...}(g)\right)\backslash\{z\in\mbox{{\rm dom}}(g)\colon g(z)= 0\}.\end{displaymath}

8.13   Theorem (Continuity of roots.) Let $p\in\mbox{{\bf Z}}_{\geq
1}$ and let $\displaystyle {f_p(x)=x^{{1\over p}}}$ for all $x\in[0,\infty)$. Then $f_p$ is continuous.

Proof: First we show $f_p$ is continuous at $0$. Let $\{x_n\}$ be a sequence in $[0,\infty)$ such that $\{x_n\}\to 0$; i.e., such that $\{x_n\}$ is a null sequence. Then by the root theorem for null sequences (Theorem 7.19), $\displaystyle {\left\{x_n^{{1\over p}}\right\}}$ is a null sequence; i.e., $\displaystyle {\{f_p(x_n)\}=\left\{x_n^{{1\over p}}\right\}\to 0=f_p(0)}$, so $f_p$ is continuous at $0$.

Next we show that $f_p$ is continuous at $1$. By the formula for a finite geometric series (3.72 ), we have for all $x\in[0,\infty)$

\vert x^p-1\vert = \left\vert(x-1)\sum_{j=0}^{p-1}x^j\right\vert = \vert x-1\vert \sum_{j=0}^{p-1}x^j \geq
\vert x-1\vert.
\end{displaymath} (8.14)

If we replace $x$ by $y^{1\over p}$ in (8.14), we get $ \vert y-1\vert = \vert (y^{1\over p})^p - 1\vert \geq \vert y^{1\over p}-1\vert$, i.e.,

\vert y^{1\over p} - 1\vert \leq \vert y-1\vert \mbox{ for all }y\in[0,\infty).\end{displaymath}

Let $\{y_n\}$ be a sequence in $[0,\infty)$. Then

\begin{displaymath}\vert(y_n)^{1\over p} -1\vert \leq \vert y_n-1\vert \mbox{ for all }n\in \mbox{{\bf N}},\end{displaymath}


\{ y_n\} \to 1 & \mbox{$\Longrightarrow$}& \vert y_n-1\vert \t...
...\} \to 1\\
&\mbox{$\Longrightarrow$}& \{ f_p(y_n)\} \to f_p(1).

Hence $f_p$ is continuous at $1$.

Finally we show that $f_p$ is continuous at arbitrary $a \in (0,\infty)$. Let $a\in [0,\infty)$, and let $\{z_n\}$ be a sequence n $[0,\infty)$. Then

\{z_n\}\to a &\mbox{$\Longrightarrow$}& {1\over a}\{z_n\} \to ...
...over p} \\
&\mbox{$\Longrightarrow$}& \{ f_p(z_n)\} \to f_p(a).

Thus $f_p$ is continuous at $a$. $\mid\!\mid\!\mid$

8.15   Definition (Composition of functions.) Let $A,B,C,D$ be sets, and let $f\colon A\to B$, $g\colon C\to D$ be functions. We define a function $g\circ f$ by the rules:

\mbox{ domain }(g\circ f)&=&\{x\in\mbox{{\rm dom}}f\colon f(x)...
...eft(f(x)\right) \mbox{ for all } x\in\mbox{{\rm dom}}(g\circ f).

8.16   Examples. Let $f\colon\mbox{{\bf C}}\to\mbox{{\bf C}}$, $g\colon\mbox{{\bf C}}\to\mbox{{\bf C}}$ be defined by

f(z)&=&z^2+1\mbox{ for all } z\in\mbox{{\bf C}}\\
g(z)&=&(1+z^*)\mbox{ for all } z\in\mbox{{\bf C}}.


(f\circ g)(z)&=&f\left(g(z)\right)=(1+z^*)^2+1=1+2z^*+(z^*)^2+1 \\


\begin{displaymath}(g\circ f)(z)=g\left(f(z)\right)=1+(z^2+1)^*=1+(z^*)^2+1=2+(z^*)^2.\end{displaymath}

If $f\colon\mbox{{\bf R}}\to\mbox{{\bf R}}$ and $g\colon[-1,\infty)\to\mbox{{\bf R}}$ are defined by

\begin{displaymath}f(x)=x^2-1 \mbox{ for all } x\in\mbox{{\bf R}}\end{displaymath}


\begin{displaymath}g(x)=\sqrt{1+x} \mbox{ for all } x\in[-1,\infty),\end{displaymath}


(f\circ g)(x)&=&(\sqrt{1+x})^2-1 \mbox{ for all } x\in[-1,\inf...
... all } x\in[-1,\infty) \\
&=&x \mbox{ for all } x\in[-1,\infty)


\begin{displaymath}(g\circ f)(x)=\sqrt{1+(x^2-1)}=\sqrt{x^2}=\vert x\vert \mbox{ for all }x\in\mbox{{\bf R}}.\end{displaymath}

8.17   Theorem (Compositions of continuous functions.) Let $f,g$ be complex functions. If $f$ is continuous at $a\in\mbox{{\bf C}}$, and $g$ is continuous at $f(a)$, then $g\circ f$ is continuous at $a$.

Proof: Let $\{x_n\}$ be a sequence in $\mbox{{\rm dom}}(g\circ f)$ such that $\{x_n\}\to a$. Then for all $n\in\mbox{{\bf N}}$, we have $x_n\in\mbox{{\rm dom}}(f)$ and $f(x_n)\in\mbox{{\rm dom}}(g)$. By continuity of $f$ at $a$, $\{f(x_n)\}\to f(a)$, and by continuity of $g$ at $f(a)$, $\{g\left(f(x_n)\right)\}\to g\left(f(a)\right)$. $\mid\!\mid\!\mid$

8.18   Example. If $\displaystyle {f(x)=\sqrt{x^2+3}}$ for all $x\in\mbox{{\bf R}}$, then $f$ is continuous (i.e., $f$ is continuous at $a$ for all $a\in\mbox{{\bf R}}$.)

8.19   Exercise. A Let $f\colon\mbox{{\bf N}}\to\mbox{{\bf C}}$ be defined by $f(n)=n!$ for all $n\in\mbox{{\bf N}}$. Is $f$ continuous?

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Next: 8.3 Limits Up: 8. Continuity Previous: 8.1 Compositions with Sequences   Index