__1.png)
and
an example
The continuity of
and
an example
Suppose that
__2.png)
is uniformly continuous on
![$[A,B]$](graphics/x_f(x)__3.png)
and bounded by
__4.png)
;
that is,
__5.png)
is UC and
__6.png)
for all
![$x\in \lbrack A,B]$](graphics/x_f(x)__7.png)
.
Let
__8.png)
.
__9.png)
(Here
we have used first the triangle inequality, then the fact that
__10.png)
,
and finally that
__11.png)
and
__12.png)
.)
Now suppose that we are given
__13.png)
;
we must make the above expression
__14.png)
by making
__15.png)
sufficiently small
(__16.png)
some
__17.png)
).
We will make each of the two added pieces
__18.png)
.
Since
__19.png)
is UC and
__20.png)
,
we can find a modulus of continuity
__21.png)
such that
__22.png)
when
__23.png)
(all on
![$[A,B]$](graphics/x_f(x)__24.png)
of course). Also,
__25.png)
will be less than
__26.png)
when
__27.png)
.
Since we want both conditions to hold, we
let__28.png)
Thus,
when
__29.png)
we'll have
__30.png)
and
__31.png)
,
so we
get__32.png)
Done.
Now let's work out an actual example:
__33.png)
on the interval
![$[-1,3]$](graphics/x_f(x)__34.png)
;
we suppose that we are given
__35.png)
.
We will need a modulus of continuity
__36.png)
for
__37.png)
on
![$[1,3]$](graphics/x_f(x)__38.png)
as well as a bound
__39.png)
.
Let's do
__40.png)
first:__41.png)
thus,
__42.png)
so we can take
__43.png)
.
(The actual "best" bound is
__44.png)
.)
Now let's find a modulus of continuity:
__45.png)
To
make
__46.png)
it suffices to make
__47.png)
,
so we can take
__48.png)
.
Finally, note that
__49.png)
.
Now we
compute:__50.png)
Thus,
if
__51.png)
on
![$[-1,3]$](graphics/x_f(x)__52.png)
,
then
__53.png)
__54.png)
__55.png)