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8.6
Definition (Continuous)
Let
![$f$](img13.gif)
be a complex function and let
![$p\in\mbox{{\rm dom}}(f)$](img618.gif)
. We say
is continuous
at ![$p$](img213.gif)
if
i.e., if
Let
![$B$](img258.gif)
be a subset of
![$S$](img427.gif)
. We say
is continuous on ![$B$](img258.gif)
if
![$f$](img13.gif)
is
continuous at
![$q$](img621.gif)
for all
![$q\in B$](img622.gif)
.
We say
![$f$](img13.gif)
is
continuous if
![$f$](img13.gif)
is
continuous on
![$\mbox{{\rm dom}}(f)$](img623.gif)
; i.e., if
![$f$](img13.gif)
is continuous at every point at which it is
defined.
8.7
Examples.
If
![$f(z)=z$](img624.gif)
for all
![$z\in\mbox{{\bf C}}$](img66.gif)
, then
![$f$](img13.gif)
is continuous. In this case
![$f\circ x=x$](img625.gif)
for
every sequence
![$x$](img428.gif)
so the condition for continuity at
![$p$](img213.gif)
is
If
![$a\in\mbox{{\bf C}}$](img86.gif)
, then the constant function
![$\tilde a$](img627.gif)
is continuous since for all
![$p\in\mbox{{\bf C}}$](img628.gif)
, and all complex sequences
![$x$](img428.gif)
,
Notice that
and
(Real part and imaginary part) are functions from
to
. In theorem 7.39 we showed if
is any complex sequence and
, then
and
Hence
![$\mbox{\rm Re}$](img630.gif)
and
![$\mbox{\rm Im}$](img631.gif)
are continuous functions on
![$\mbox{{\bf C}}$](img2.gif)
.
8.8
Theorem.
If
and
are functions from
to
defined by
then
and
are continuous.
Proof: Let
and let
be any sequence in
such that
;
i.e.,
is a null sequence. By the reverse triangle inequality,
and
so we have
and hence
It follows by the comparison theorem that
is a null sequence; i.e.,
. Hence
is continuous.
Since
, the comparison theorem shows that
i.e.,
is continuous.
8.9
Example.
If
then
![$f$](img13.gif)
is not continuous at
![$0$](img34.gif)
, since
but
Notice that to show that a function
![$f$](img13.gif)
is
not continuous at a point
![$a$](img590.gif)
in its
domain, it is sufficient to find
one sequence
![$\{x_n\}$](img387.gif)
in
![$\mbox{{\rm dom}}(f)$](img623.gif)
such that
![$\{x_n\}\to a$](img637.gif)
and either
![$\{f(x_n)\}$](img650.gif)
converges to a limit different from
![$f(a)$](img651.gif)
or
![$\{f(x_n)\}$](img650.gif)
diverges.
8.10
Theorem (Sum and Product theorems.)
Let
be complex functions, and let
. If
and
are continuous at
, then
,
, and
are continuous at
.
Proof: Let
be a sequence in domain
such that
. Then
for all
and
for all
, and by continuity of
and
at
, it follows that
By the sum theorem for sequences,
Hence
is continuous at
. The proofs of continuity for
and
are similar.
8.11
Theorem (Quotient theorem.)
Let
be complex functions and let
. If
and
are continuous at
, then
is continuous at
.
8.12
Exercise.
Prove the quotient theorem. Recall that
8.13
Theorem (Continuity of roots.)
Let
and let
for all
. Then
is continuous.
Proof: First we show
is continuous at
. Let
be a sequence in
such that
; i.e., such that
is a null sequence.
Then by the root theorem for null sequences (Theorem 7.19),
is a null sequence; i.e.,
, so
is
continuous at
.
Next we show that
is continuous at
. By the formula for a finite
geometric series (3.72 ), we have for all
![\begin{displaymath}
\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}](img667.gif) |
(8.14) |
If we replace
by
in (8.14), we get
,
i.e.,
Let
be a sequence in
. Then
so
Hence
is continuous at
.
Finally we show that
is continuous at arbitrary
.
Let
, and let
be a sequence n
.
Then
Thus
is continuous at
.
8.15
Definition (Composition of functions.)
Let
![$A,B,C,D$](img678.gif)
be sets, and let
![$f\colon A\to B$](img679.gif)
,
![$g\colon C\to D$](img680.gif)
be functions. We
define a function
![$g\circ f$](img681.gif)
by the rules:
8.16
Examples.
Let
![$f\colon\mbox{{\bf C}}\to\mbox{{\bf C}}$](img683.gif)
,
![$g\colon\mbox{{\bf C}}\to\mbox{{\bf C}}$](img684.gif)
be defined by
Then
and
If
and
are defined by
and
then
and
8.17
Theorem (Compositions of continuous functions.)
Let
be complex functions. If
is continuous at
, and
is
continuous at
, then
is continuous at
.
Proof: Let
be a sequence in
such that
.
Then for all
, we have
and
. By
continuity of
at
,
, and by continuity of
at
,
.
8.18
Example.
If
![$\displaystyle {f(x)=\sqrt{x^2+3}}$](img698.gif)
for all
![$x\in\mbox{{\bf R}}$](img699.gif)
, then
![$f$](img13.gif)
is continuous (i.e.,
![$f$](img13.gif)
is
continuous at
![$a$](img590.gif)
for all
![$a\in\mbox{{\bf R}}$](img700.gif)
.)
8.19
Exercise.
A
Let
![$f\colon\mbox{{\bf N}}\to\mbox{{\bf C}}$](img3.gif)
be defined
by
![$f(n)=n!$](img701.gif)
for all
![$n\in\mbox{{\bf N}}$](img9.gif)
. Is
![$f$](img13.gif)
continuous?
Next: 8.3 Limits
Up: 8. Continuity
Previous: 8.1 Compositions with Sequences
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