Continuity and Uniform Continuity - Continuity and Differentiability, CSIR-NET Mathematical Sciences Notes | Study Mathematics for IIT JAM, CSIR NET, UGC NET - Mathematics

Mathematics: Continuity and Uniform Continuity - Continuity and Differentiability, CSIR-NET Mathematical Sciences Notes | Study Mathematics for IIT JAM, CSIR NET, UGC NET - Mathematics

The document Continuity and Uniform Continuity - Continuity and Differentiability, CSIR-NET Mathematical Sciences Notes | Study Mathematics for IIT JAM, CSIR NET, UGC NET - Mathematics is a part of the Mathematics Course Mathematics for IIT JAM, CSIR NET, UGC NET.
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 Page 1


Continuity and Uniform Continuity
521
May 12, 2010
1. ThroughoutS will denote a subset of the real numbers R andf :S! R
will be a real valued function dened on S. The set S may be bounded like
S = (0; 5) =fx2 R : 0<x< 5g
or innite like
S = (0;1) =fx2 R : 0<xg:
It may even be all of R. The valuef(x) of the functionf at the pointx2S
will be dened by a formula (or formulas).
Denition 2. The function f is said to be continuous on S i
8x
0
2S8"> 09> 08x2S

jxx
0
j< =) jf(x)f(x
0
)j<"

:
Hence f is not continuous
1
on S i
9x
0
2S9"> 08> 09x2S

jxx
0
j< andjf(x)f(x
0
)j"

:
Denition 3. The functionf is said to be uniformly continuous on S i
8"> 09> 08x
0
2S8x2S

jxx
0
j< =) jf(x)f(x
0
)j<"

:
Hence f is not uniformly continuous on S i
9"> 08> 09x
0
2S9x2S

jxx
0
j< andjf(x)f(x
0
)j"

:
1
For an example of a function which is not continuous see Example 22 below.
Page 2


Continuity and Uniform Continuity
521
May 12, 2010
1. ThroughoutS will denote a subset of the real numbers R andf :S! R
will be a real valued function dened on S. The set S may be bounded like
S = (0; 5) =fx2 R : 0<x< 5g
or innite like
S = (0;1) =fx2 R : 0<xg:
It may even be all of R. The valuef(x) of the functionf at the pointx2S
will be dened by a formula (or formulas).
Denition 2. The function f is said to be continuous on S i
8x
0
2S8"> 09> 08x2S

jxx
0
j< =) jf(x)f(x
0
)j<"

:
Hence f is not continuous
1
on S i
9x
0
2S9"> 08> 09x2S

jxx
0
j< andjf(x)f(x
0
)j"

:
Denition 3. The functionf is said to be uniformly continuous on S i
8"> 09> 08x
0
2S8x2S

jxx
0
j< =) jf(x)f(x
0
)j<"

:
Hence f is not uniformly continuous on S i
9"> 08> 09x
0
2S9x2S

jxx
0
j< andjf(x)f(x
0
)j"

:
1
For an example of a function which is not continuous see Example 22 below.
4. The only dierence between the two denitions is the order of the quan-
tiers. When you prove f is continuous your proof will have the form
Choose x
0
2S. Choose "> 0. Let  =(x
0
;"). Choose x2S.
Assumejxx
0
j<. Thereforejf(x)f(x
0
)j<".
The expression for(x
0
;") can involve bothx
0
and" but must be independent
ofx. The order of the quaniers in the denition signals this; in the proof x
has not yet been chosen at the point where  is dened so the denition of
 must not involve x. (The represent the proof thatjf(x)f(x
0
)j < "
follows from the earlier steps in the proof.) When you prove f is uniformly
continuous your proof will have the form
Choose "> 0. Let  =("). Choose x
0
2S. Choose x2S.
Assumejxx
0
j<. Thereforejf(x)f(x
0
)j<".
so the expression for  can only involve " and must not involve either x or
x
0
.
It is obvious that a uniformly continuous function is continuous: if we can
nd a  which works for all x
0
, we can nd one (the same one) which works
for any particular x
0
. We will see below that there are continuous functions
which are not uniformly continuous.
Example 5. LetS = R andf(x) = 3x + 7. Thenf is uniformly continuous
on S.
Proof. Choose "> 0. Let  ="=3. Choose x
0
2 R. Choose x2 R. Assume
jxx
0
j<. Then
jf(x)f(x
0
)j =j(3x + 7) (3x
0
+ 7)j = 3jxx
0
j< 3 =":
Example 6. Let S =fx2 R : 0 < x < 4g and f(x) = x
2
. Then f is
uniformly continuous on S.
Proof. Choose " > 0. Let  = "=8. Choose x
0
2 S. Choose x2 S. Thus
0<x
0
< 4 and 0<x< 4 so 0<x +x
0
< 8. Assumejxx
0
j<. Then
jf(x)f(x
0
)j =jx
2
x
2
0
j = (x +x
0
)jxx
0
j< (4 + 4) =":
Page 3


Continuity and Uniform Continuity
521
May 12, 2010
1. ThroughoutS will denote a subset of the real numbers R andf :S! R
will be a real valued function dened on S. The set S may be bounded like
S = (0; 5) =fx2 R : 0<x< 5g
or innite like
S = (0;1) =fx2 R : 0<xg:
It may even be all of R. The valuef(x) of the functionf at the pointx2S
will be dened by a formula (or formulas).
Denition 2. The function f is said to be continuous on S i
8x
0
2S8"> 09> 08x2S

jxx
0
j< =) jf(x)f(x
0
)j<"

:
Hence f is not continuous
1
on S i
9x
0
2S9"> 08> 09x2S

jxx
0
j< andjf(x)f(x
0
)j"

:
Denition 3. The functionf is said to be uniformly continuous on S i
8"> 09> 08x
0
2S8x2S

jxx
0
j< =) jf(x)f(x
0
)j<"

:
Hence f is not uniformly continuous on S i
9"> 08> 09x
0
2S9x2S

jxx
0
j< andjf(x)f(x
0
)j"

:
1
For an example of a function which is not continuous see Example 22 below.
4. The only dierence between the two denitions is the order of the quan-
tiers. When you prove f is continuous your proof will have the form
Choose x
0
2S. Choose "> 0. Let  =(x
0
;"). Choose x2S.
Assumejxx
0
j<. Thereforejf(x)f(x
0
)j<".
The expression for(x
0
;") can involve bothx
0
and" but must be independent
ofx. The order of the quaniers in the denition signals this; in the proof x
has not yet been chosen at the point where  is dened so the denition of
 must not involve x. (The represent the proof thatjf(x)f(x
0
)j < "
follows from the earlier steps in the proof.) When you prove f is uniformly
continuous your proof will have the form
Choose "> 0. Let  =("). Choose x
0
2S. Choose x2S.
Assumejxx
0
j<. Thereforejf(x)f(x
0
)j<".
so the expression for  can only involve " and must not involve either x or
x
0
.
It is obvious that a uniformly continuous function is continuous: if we can
nd a  which works for all x
0
, we can nd one (the same one) which works
for any particular x
0
. We will see below that there are continuous functions
which are not uniformly continuous.
Example 5. LetS = R andf(x) = 3x + 7. Thenf is uniformly continuous
on S.
Proof. Choose "> 0. Let  ="=3. Choose x
0
2 R. Choose x2 R. Assume
jxx
0
j<. Then
jf(x)f(x
0
)j =j(3x + 7) (3x
0
+ 7)j = 3jxx
0
j< 3 =":
Example 6. Let S =fx2 R : 0 < x < 4g and f(x) = x
2
. Then f is
uniformly continuous on S.
Proof. Choose " > 0. Let  = "=8. Choose x
0
2 S. Choose x2 S. Thus
0<x
0
< 4 and 0<x< 4 so 0<x +x
0
< 8. Assumejxx
0
j<. Then
jf(x)f(x
0
)j =jx
2
x
2
0
j = (x +x
0
)jxx
0
j< (4 + 4) =":
7. In both of the preceeding proofs the function f satised an inequality of
form
jf(x
1
)f(x
2
)jMjx
1
x
2
j (1)
for x
1
;x
2
2S. In Example 5 we had
j(3x
1
+ 7) (3x
2
+ 7)j 3jx
1
x
2
j
and in Example 6 we had
jx
2
1
x
2
2
j 8jx
1
x
2
j
for 0<x
1
;x
2
< 4. An inequality of form (1) is called a Lipschitz inequality
and the constant M is called the corresponding Lipschitz constant.
Theorem 8. Iff satises (1) for x
1
;x
2
2S, thenf is uniformly continuous
on S.
Proof. Choose "> 0. Let  ="=M. Choose x
0
2S. Choose x2S. Assume
thatjxx
0
j<. Then
jf(x)f(x
0
)jMjxx
0
j<M =":
9. The Lipschitz constant depend might depend on the interval. For example,
jx
2
1
x
2
2
j = (x
1
+x
2
)jx
1
x
2
j 2ajx
1
x
2
j
for 0 < x
1
;x
2
< a but the function f(x) = x
2
does not satisfy a Lipschitz
inequality on the whole interval (0;1) since
jx
2
1
x
2
2
j = (x
1
+x
2
)jx
1
x
2
j>Mjx
1
x
2
j
if x
1
=M and x
2
=x
1
+ 1. In fact,
Example 10. The function f(x) =x
2
is continuous but not uniformly con-
tinuous on the interval S = (0;1).
Proof. We show f is continuous on S, i.e.
8x
0
2S8"> 09> 08x2S

jxx
0
j< =) jx
2
x
2
0
j<"

:
Page 4


Continuity and Uniform Continuity
521
May 12, 2010
1. ThroughoutS will denote a subset of the real numbers R andf :S! R
will be a real valued function dened on S. The set S may be bounded like
S = (0; 5) =fx2 R : 0<x< 5g
or innite like
S = (0;1) =fx2 R : 0<xg:
It may even be all of R. The valuef(x) of the functionf at the pointx2S
will be dened by a formula (or formulas).
Denition 2. The function f is said to be continuous on S i
8x
0
2S8"> 09> 08x2S

jxx
0
j< =) jf(x)f(x
0
)j<"

:
Hence f is not continuous
1
on S i
9x
0
2S9"> 08> 09x2S

jxx
0
j< andjf(x)f(x
0
)j"

:
Denition 3. The functionf is said to be uniformly continuous on S i
8"> 09> 08x
0
2S8x2S

jxx
0
j< =) jf(x)f(x
0
)j<"

:
Hence f is not uniformly continuous on S i
9"> 08> 09x
0
2S9x2S

jxx
0
j< andjf(x)f(x
0
)j"

:
1
For an example of a function which is not continuous see Example 22 below.
4. The only dierence between the two denitions is the order of the quan-
tiers. When you prove f is continuous your proof will have the form
Choose x
0
2S. Choose "> 0. Let  =(x
0
;"). Choose x2S.
Assumejxx
0
j<. Thereforejf(x)f(x
0
)j<".
The expression for(x
0
;") can involve bothx
0
and" but must be independent
ofx. The order of the quaniers in the denition signals this; in the proof x
has not yet been chosen at the point where  is dened so the denition of
 must not involve x. (The represent the proof thatjf(x)f(x
0
)j < "
follows from the earlier steps in the proof.) When you prove f is uniformly
continuous your proof will have the form
Choose "> 0. Let  =("). Choose x
0
2S. Choose x2S.
Assumejxx
0
j<. Thereforejf(x)f(x
0
)j<".
so the expression for  can only involve " and must not involve either x or
x
0
.
It is obvious that a uniformly continuous function is continuous: if we can
nd a  which works for all x
0
, we can nd one (the same one) which works
for any particular x
0
. We will see below that there are continuous functions
which are not uniformly continuous.
Example 5. LetS = R andf(x) = 3x + 7. Thenf is uniformly continuous
on S.
Proof. Choose "> 0. Let  ="=3. Choose x
0
2 R. Choose x2 R. Assume
jxx
0
j<. Then
jf(x)f(x
0
)j =j(3x + 7) (3x
0
+ 7)j = 3jxx
0
j< 3 =":
Example 6. Let S =fx2 R : 0 < x < 4g and f(x) = x
2
. Then f is
uniformly continuous on S.
Proof. Choose " > 0. Let  = "=8. Choose x
0
2 S. Choose x2 S. Thus
0<x
0
< 4 and 0<x< 4 so 0<x +x
0
< 8. Assumejxx
0
j<. Then
jf(x)f(x
0
)j =jx
2
x
2
0
j = (x +x
0
)jxx
0
j< (4 + 4) =":
7. In both of the preceeding proofs the function f satised an inequality of
form
jf(x
1
)f(x
2
)jMjx
1
x
2
j (1)
for x
1
;x
2
2S. In Example 5 we had
j(3x
1
+ 7) (3x
2
+ 7)j 3jx
1
x
2
j
and in Example 6 we had
jx
2
1
x
2
2
j 8jx
1
x
2
j
for 0<x
1
;x
2
< 4. An inequality of form (1) is called a Lipschitz inequality
and the constant M is called the corresponding Lipschitz constant.
Theorem 8. Iff satises (1) for x
1
;x
2
2S, thenf is uniformly continuous
on S.
Proof. Choose "> 0. Let  ="=M. Choose x
0
2S. Choose x2S. Assume
thatjxx
0
j<. Then
jf(x)f(x
0
)jMjxx
0
j<M =":
9. The Lipschitz constant depend might depend on the interval. For example,
jx
2
1
x
2
2
j = (x
1
+x
2
)jx
1
x
2
j 2ajx
1
x
2
j
for 0 < x
1
;x
2
< a but the function f(x) = x
2
does not satisfy a Lipschitz
inequality on the whole interval (0;1) since
jx
2
1
x
2
2
j = (x
1
+x
2
)jx
1
x
2
j>Mjx
1
x
2
j
if x
1
=M and x
2
=x
1
+ 1. In fact,
Example 10. The function f(x) =x
2
is continuous but not uniformly con-
tinuous on the interval S = (0;1).
Proof. We show f is continuous on S, i.e.
8x
0
2S8"> 09> 08x2S

jxx
0
j< =) jx
2
x
2
0
j<"

:
Choose x
0
. Let a = x
0
+ 1 and  = min(1;"=2a). (Note that  depends on
x
0
since a does.) Choose x2S. Assumejxx
0
j<. Thenjxx
0
j< 1 so
x<x
0
+ 1 =a so x;x
0
<a so
jx
2
x
2
0
j = (x +x
0
)jxx
0
j 2ajxx
0
j< 2a 2a
"
2a
="
as required.
We show that f is not uniformly continuous on S, i.e.
9"> 08> 09x
0
2S9x2S

jxx
0
j< andjx
2
x
2
0
j"

:
Let " = 1. Choose  > 0. Let x
0
= 1= and x =x
0
+=2. Thenjxx
0
j =
=2< but
jx
2
x
2
0
j =






1

+

2

2


1


2





= 1 +

2
4
> 1 ="
as required. (Note that x
0
is large when  is small.)
11. According to the Mean Value Theorem from calculus for a dierentiable
function f we have
f(x
1
)f(x
2
) =f
0
(c)(x
2
x
1
):
for some c between x
1
and x
2
. (The slope (f(x
1
)f(x
2
))=(x
1
x
2
) of the
secant line joining the two points (x
1
;f(x
1
)) and (x
2
;f(x
2
)) on the graph is
the same as the slope f
0
(c) of the tangent point at the intermediate point
(c;f(c)).) If x
1
and x
2
lie in some interval S andjf
0
(c)j M for all c2 S
we conclude that the Lipschitz inequality (1) holds on S. We don't want to
use the Mean Value Theorem without rst proving it, but we certainly can
use it to guess an appropriate value of M and then prove the inequality by
other means.
12. For example, consider the function f(x) = x
1
dened on the interval
S = (a;1) where a> 0. For x
1
;x
2
2S the Mean Value Theorem says that
x
1
1
x
1
2
=c
2
(x
1
x
2
) where c is between x
1
and x
2
. If x
1
;x
2
2S then
c2 S (as c is between x
1
and x
2
) and hence c > a so c
2
< a
2
. We can
prove the inequality
jx
1
1
x
1
2
ja
2
jx
2
x
2
j
Page 5


Continuity and Uniform Continuity
521
May 12, 2010
1. ThroughoutS will denote a subset of the real numbers R andf :S! R
will be a real valued function dened on S. The set S may be bounded like
S = (0; 5) =fx2 R : 0<x< 5g
or innite like
S = (0;1) =fx2 R : 0<xg:
It may even be all of R. The valuef(x) of the functionf at the pointx2S
will be dened by a formula (or formulas).
Denition 2. The function f is said to be continuous on S i
8x
0
2S8"> 09> 08x2S

jxx
0
j< =) jf(x)f(x
0
)j<"

:
Hence f is not continuous
1
on S i
9x
0
2S9"> 08> 09x2S

jxx
0
j< andjf(x)f(x
0
)j"

:
Denition 3. The functionf is said to be uniformly continuous on S i
8"> 09> 08x
0
2S8x2S

jxx
0
j< =) jf(x)f(x
0
)j<"

:
Hence f is not uniformly continuous on S i
9"> 08> 09x
0
2S9x2S

jxx
0
j< andjf(x)f(x
0
)j"

:
1
For an example of a function which is not continuous see Example 22 below.
4. The only dierence between the two denitions is the order of the quan-
tiers. When you prove f is continuous your proof will have the form
Choose x
0
2S. Choose "> 0. Let  =(x
0
;"). Choose x2S.
Assumejxx
0
j<. Thereforejf(x)f(x
0
)j<".
The expression for(x
0
;") can involve bothx
0
and" but must be independent
ofx. The order of the quaniers in the denition signals this; in the proof x
has not yet been chosen at the point where  is dened so the denition of
 must not involve x. (The represent the proof thatjf(x)f(x
0
)j < "
follows from the earlier steps in the proof.) When you prove f is uniformly
continuous your proof will have the form
Choose "> 0. Let  =("). Choose x
0
2S. Choose x2S.
Assumejxx
0
j<. Thereforejf(x)f(x
0
)j<".
so the expression for  can only involve " and must not involve either x or
x
0
.
It is obvious that a uniformly continuous function is continuous: if we can
nd a  which works for all x
0
, we can nd one (the same one) which works
for any particular x
0
. We will see below that there are continuous functions
which are not uniformly continuous.
Example 5. LetS = R andf(x) = 3x + 7. Thenf is uniformly continuous
on S.
Proof. Choose "> 0. Let  ="=3. Choose x
0
2 R. Choose x2 R. Assume
jxx
0
j<. Then
jf(x)f(x
0
)j =j(3x + 7) (3x
0
+ 7)j = 3jxx
0
j< 3 =":
Example 6. Let S =fx2 R : 0 < x < 4g and f(x) = x
2
. Then f is
uniformly continuous on S.
Proof. Choose " > 0. Let  = "=8. Choose x
0
2 S. Choose x2 S. Thus
0<x
0
< 4 and 0<x< 4 so 0<x +x
0
< 8. Assumejxx
0
j<. Then
jf(x)f(x
0
)j =jx
2
x
2
0
j = (x +x
0
)jxx
0
j< (4 + 4) =":
7. In both of the preceeding proofs the function f satised an inequality of
form
jf(x
1
)f(x
2
)jMjx
1
x
2
j (1)
for x
1
;x
2
2S. In Example 5 we had
j(3x
1
+ 7) (3x
2
+ 7)j 3jx
1
x
2
j
and in Example 6 we had
jx
2
1
x
2
2
j 8jx
1
x
2
j
for 0<x
1
;x
2
< 4. An inequality of form (1) is called a Lipschitz inequality
and the constant M is called the corresponding Lipschitz constant.
Theorem 8. Iff satises (1) for x
1
;x
2
2S, thenf is uniformly continuous
on S.
Proof. Choose "> 0. Let  ="=M. Choose x
0
2S. Choose x2S. Assume
thatjxx
0
j<. Then
jf(x)f(x
0
)jMjxx
0
j<M =":
9. The Lipschitz constant depend might depend on the interval. For example,
jx
2
1
x
2
2
j = (x
1
+x
2
)jx
1
x
2
j 2ajx
1
x
2
j
for 0 < x
1
;x
2
< a but the function f(x) = x
2
does not satisfy a Lipschitz
inequality on the whole interval (0;1) since
jx
2
1
x
2
2
j = (x
1
+x
2
)jx
1
x
2
j>Mjx
1
x
2
j
if x
1
=M and x
2
=x
1
+ 1. In fact,
Example 10. The function f(x) =x
2
is continuous but not uniformly con-
tinuous on the interval S = (0;1).
Proof. We show f is continuous on S, i.e.
8x
0
2S8"> 09> 08x2S

jxx
0
j< =) jx
2
x
2
0
j<"

:
Choose x
0
. Let a = x
0
+ 1 and  = min(1;"=2a). (Note that  depends on
x
0
since a does.) Choose x2S. Assumejxx
0
j<. Thenjxx
0
j< 1 so
x<x
0
+ 1 =a so x;x
0
<a so
jx
2
x
2
0
j = (x +x
0
)jxx
0
j 2ajxx
0
j< 2a 2a
"
2a
="
as required.
We show that f is not uniformly continuous on S, i.e.
9"> 08> 09x
0
2S9x2S

jxx
0
j< andjx
2
x
2
0
j"

:
Let " = 1. Choose  > 0. Let x
0
= 1= and x =x
0
+=2. Thenjxx
0
j =
=2< but
jx
2
x
2
0
j =






1

+

2

2


1


2





= 1 +

2
4
> 1 ="
as required. (Note that x
0
is large when  is small.)
11. According to the Mean Value Theorem from calculus for a dierentiable
function f we have
f(x
1
)f(x
2
) =f
0
(c)(x
2
x
1
):
for some c between x
1
and x
2
. (The slope (f(x
1
)f(x
2
))=(x
1
x
2
) of the
secant line joining the two points (x
1
;f(x
1
)) and (x
2
;f(x
2
)) on the graph is
the same as the slope f
0
(c) of the tangent point at the intermediate point
(c;f(c)).) If x
1
and x
2
lie in some interval S andjf
0
(c)j M for all c2 S
we conclude that the Lipschitz inequality (1) holds on S. We don't want to
use the Mean Value Theorem without rst proving it, but we certainly can
use it to guess an appropriate value of M and then prove the inequality by
other means.
12. For example, consider the function f(x) = x
1
dened on the interval
S = (a;1) where a> 0. For x
1
;x
2
2S the Mean Value Theorem says that
x
1
1
x
1
2
=c
2
(x
1
x
2
) where c is between x
1
and x
2
. If x
1
;x
2
2S then
c2 S (as c is between x
1
and x
2
) and hence c > a so c
2
< a
2
. We can
prove the inequality
jx
1
1
x
1
2
ja
2
jx
2
x
2
j
for x
1
;x
2
a as follows. First a
2
x
1
x
2
since ax
1
and ax
2
. Then
jx
1
1
x
1
2
j =
jx
1
x
2
j
x
1
x
2

jx
1
x
2
j
a
2
(2)
where we have used the fact that 
1
< 
1
if 0 <  < . It follows that
that the function f(x) is uniformly continuous on any interval (a;1) where
a > 0. Notice however that the Lipschitz constant M = a
2
depends on
the interval. In fact, the function f(x) = x
1
does not satisfy a Lipshitz
inequality on the interval (0;1).
13. We can discover a Lipscitz inequality for the square root functionf(x) =
p
x in much the same way. Consider the function f(x) =
p
x dened on the
interval S = (a;1) where a > 0. For x
1
;x
2
2 S the Mean Value Theorem
says that
p
x
1

p
x
2
= (x
1
x
2
)=(2
p
c) where c is between x
1
and x
2
. If
x
1
;x
2
2 S then c 2 S (as c is between x
1
and x
2
) and hence c > a so
(2
p
c)
1
< (2
p
a)
1
. We can prove the inequality
j
p
x
1

p
x
2
j
jx
1
x
2
j
2
p
a
(3)
for x
1
;x
2
a as follows: Divide the equation
(
p
x
1

p
x
2
)(
p
x
1
+
p
x
2
) = ((
p
x
1
)
2
 (
p
x
2
)
2
) =x
1
x
2
by (
p
x
1
+
p
x
2
), take absolute values, and use (
p
x
1
+
p
x
2
) 2
p
a. Again the
Lipschitz constant M = (2
p
a)
1
depends on the interval and the function
does not satisfy a Lipschitz inequality on the interval (0;1).
Example 14. The function f(x) = x
1
is continuous but not uniformly
continuous on the interval S = (0;1).
Proof. We show f is continuous on S, i.e.
8x
0
2S8"> 09> 08x2S

jxx
0
j< =)




1
x

1
x
0




<"

:
Choose x
0
. Let a =x
0
=2 and  = min(x
0
a;a
2
"). Choose x2S. Assume
jxx
0
j < . Then x
0
xjxx
0
j < x
0
a sox <a so a < x so
x;x
0
<a so by (2)




1
x

1
x
0





jx
1
x
2
j
a
2
<

a
2

a
2
"
a
2
="
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