6. Continuity. Points of discontinuity.Uniform Continuity.
Answer:
Definition. We say that f is continues at points of x0 if f is defined on an open interval (a;b) continue x0 and we write (x)=f(x0).
Theorem: A function f is continues at x0 if and only if f is defined on an open interval (a;b) continue x0 and for every ε>0 there is δ>0 such that |f(x)-f(x0)|<ε, whenever |x-x0|<δ
Definition. A function f is continuous on [a;b], if :
(a)
(x)=f(x0)
exists for all x0
in[a;b)
(b)
(x)=f(x0-)
exists for all x0
in (a;b]
(c)
f(x0+)=f(x0-)=f(x0)
or
(x)=
(x)=f(x0)
For the function to be continuous at the point x0, necessary and sufficient condition is the following: (x)= (x)=f(x) (1)
Point of discontinuity:
| type and || type. | type: jump and removable.
If condition (1) is not true then point x0 of discontinuity.
If
(x)or
(x)=
, then x0
is a point of discontinuity (||type).
Theorem.If f and g are continuous on a set S , then so are f
=g, f-g, and fg. In addition is continuous at each x0 in S such that g(x0) 0.
Definition.
A function f is bounded below on a set S , if there is a
real number m such that f(x)
m
for all x
S.
In this
case, the set V={f(x) , x
S}
has a infimum
,
and we write α=inff(x) , x
S.
If there is a point x, in S such that f(x)=α we say α is the minimum of f an S, and we write α=minf(x)
6.2 Definition. F is bounded above on S, if there is a real number M, such that f(x) M for all xin S.
In this case V has supremum β , and we write β=supf(x) x S. If there is a point x2 in S such that f(x2)=β we say that β is maximum of f on S and write β=maxf(x)
If f is bounded above and below on a set S we say that f is bounded on S.
Uniform Continuity.
Definition.
A function f is uniformly continuous on a subset S of its
domain. If for every ε>0 there is δ>0 such that
|f(x)-f(x’)|<ε ,whenever |x-x’|<δ, and x, x’
Cantor’s theorem.If f is continuous on a closed and bounded interval [a;b] then f is uniformly continuous on [a;b]
Proof: Suppose that ε>0 , since f is continuous on [a;b] for each t in [a;b] there is a positive number δt such that:
|f(x)-f(t)|<
,
if |x-t|<2δt
(1) and x
[a;b].
if It=(t-δt;
t+δt),
the collection H={It|t
[a;b]}
is an open covering of [a;b] since [a;b] is compact , the
Heine-Borel theorem implies that there are finitely many points
t1,t2,…tn
in [a;b] such that It1,It2,…Itn
cover [a;b]. Now define δ=min(δt1,δt2,…,δtn).
We will show thatig |x-x’|<δ and x,x’
[a;b]
then |f(x) – f(x’)|<ε (2)
From the triangle inequality : |f(x) – f(x’)| = |(f(x) – f(tr)) + f(tr) – f(x’)| |f(x) – f(tr)| + |f(tr) – f(x’)| (3)
Since It1, It2,…, Itr cover [a;b] , x must be in one of these intervals. Suppose that x Itr ; that is: |x - tr|<δtr (4)
From (1) with t=tr , |f(x) – f(tr)|< (5)
From (2), (4) and the triangle inequality |x’ - tr|= |(x’ - x) + (x - tr)| |x’ - x| +|x - tr|< δ + δtr 2δtr ;
Therefore, (1) with t=t0 and x replaced by x’ implies that |f(x’) –f(tr)|<
This (3) and (5) imply that: |f(x) – f(x’)|<ε