Is every monotonic additive function $f colon mathbbR to mathbbR$ continuous?
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Let a function $f colon mathbbR to mathbbR$ have the following two properties:
(1) For all $x_1, x_2 in mathbbR$ such that $x_1 < x_2$, we have
$$f left( x_1 right) leq f left( x_2 right). $$
(2) For all $x_1, x_2 in mathbbR$, we have
$$f left( x_1 + x_2 right) = f left( x_1 right) + f left( x_2 right). $$
Is such a function $f$ continuous at every point $c$ of $mathbbR$?
My Effort:
We can show that for every rational number $q$, we have
$$ f(q) = q f(1). $$
As $f$ is monotonic (increasing), so the set of points of discontinuity of $f$ is at most countable.
How to proceed from here?
If we could show that $f$ is continuous at every rational point $c in mathbbQ$, then $f$ would also be continuous at every point of $mathbbR$. Am I right?
What next?
Context:
Sec. 5.6 (in fact immediately after Theorem 5.6.4) in the book Introduction To Real Analysis by Robert G. Bartle & Donald R. Sherbert, 4th edition.
calculus real-analysis analysis continuity monotone-functions
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up vote
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Let a function $f colon mathbbR to mathbbR$ have the following two properties:
(1) For all $x_1, x_2 in mathbbR$ such that $x_1 < x_2$, we have
$$f left( x_1 right) leq f left( x_2 right). $$
(2) For all $x_1, x_2 in mathbbR$, we have
$$f left( x_1 + x_2 right) = f left( x_1 right) + f left( x_2 right). $$
Is such a function $f$ continuous at every point $c$ of $mathbbR$?
My Effort:
We can show that for every rational number $q$, we have
$$ f(q) = q f(1). $$
As $f$ is monotonic (increasing), so the set of points of discontinuity of $f$ is at most countable.
How to proceed from here?
If we could show that $f$ is continuous at every rational point $c in mathbbQ$, then $f$ would also be continuous at every point of $mathbbR$. Am I right?
What next?
Context:
Sec. 5.6 (in fact immediately after Theorem 5.6.4) in the book Introduction To Real Analysis by Robert G. Bartle & Donald R. Sherbert, 4th edition.
calculus real-analysis analysis continuity monotone-functions
2
Linking to copyrighted material is frowned upon. Posting the whole book goes way beyond fair use and is illegal in most jurisdictions.
â Umberto P.
Sep 5 at 11:25
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up vote
4
down vote
favorite
up vote
4
down vote
favorite
Let a function $f colon mathbbR to mathbbR$ have the following two properties:
(1) For all $x_1, x_2 in mathbbR$ such that $x_1 < x_2$, we have
$$f left( x_1 right) leq f left( x_2 right). $$
(2) For all $x_1, x_2 in mathbbR$, we have
$$f left( x_1 + x_2 right) = f left( x_1 right) + f left( x_2 right). $$
Is such a function $f$ continuous at every point $c$ of $mathbbR$?
My Effort:
We can show that for every rational number $q$, we have
$$ f(q) = q f(1). $$
As $f$ is monotonic (increasing), so the set of points of discontinuity of $f$ is at most countable.
How to proceed from here?
If we could show that $f$ is continuous at every rational point $c in mathbbQ$, then $f$ would also be continuous at every point of $mathbbR$. Am I right?
What next?
Context:
Sec. 5.6 (in fact immediately after Theorem 5.6.4) in the book Introduction To Real Analysis by Robert G. Bartle & Donald R. Sherbert, 4th edition.
calculus real-analysis analysis continuity monotone-functions
Let a function $f colon mathbbR to mathbbR$ have the following two properties:
(1) For all $x_1, x_2 in mathbbR$ such that $x_1 < x_2$, we have
$$f left( x_1 right) leq f left( x_2 right). $$
(2) For all $x_1, x_2 in mathbbR$, we have
$$f left( x_1 + x_2 right) = f left( x_1 right) + f left( x_2 right). $$
Is such a function $f$ continuous at every point $c$ of $mathbbR$?
My Effort:
We can show that for every rational number $q$, we have
$$ f(q) = q f(1). $$
As $f$ is monotonic (increasing), so the set of points of discontinuity of $f$ is at most countable.
How to proceed from here?
If we could show that $f$ is continuous at every rational point $c in mathbbQ$, then $f$ would also be continuous at every point of $mathbbR$. Am I right?
What next?
Context:
Sec. 5.6 (in fact immediately after Theorem 5.6.4) in the book Introduction To Real Analysis by Robert G. Bartle & Donald R. Sherbert, 4th edition.
calculus real-analysis analysis continuity monotone-functions
calculus real-analysis analysis continuity monotone-functions
edited Sep 6 at 16:23
Aloizio Macedoâ¦
22.8k23483
22.8k23483
asked Sep 5 at 11:17
Saaqib Mahmood
7,30842171
7,30842171
2
Linking to copyrighted material is frowned upon. Posting the whole book goes way beyond fair use and is illegal in most jurisdictions.
â Umberto P.
Sep 5 at 11:25
add a comment |Â
2
Linking to copyrighted material is frowned upon. Posting the whole book goes way beyond fair use and is illegal in most jurisdictions.
â Umberto P.
Sep 5 at 11:25
2
2
Linking to copyrighted material is frowned upon. Posting the whole book goes way beyond fair use and is illegal in most jurisdictions.
â Umberto P.
Sep 5 at 11:25
Linking to copyrighted material is frowned upon. Posting the whole book goes way beyond fair use and is illegal in most jurisdictions.
â Umberto P.
Sep 5 at 11:25
add a comment |Â
2 Answers
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8
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If $x$ is an irrational number, let $r_n$ be rationals that approach $x$ from below. Then as $x>r_n$, the monotonicity property $(1)$ of $f$ gives
$$f(x) ge f(r_n) = r_n f(1) to x f(1),$$
so we conclude $$f(x) ge x f(1),$$ and similarly, using rationals that approach $x$ from above, $f(x) le xf(1)$. Hence, $$f(x) = xf(1) quad forall xinmathbb R.$$
(edit) I note the relevant portions of the book,
and,
it seems that $(a)$ you misread the book, which says you only need to prove continuity at one
point, and $(b)$ you want a proof that first proves continuity, and then appeals to the known result(Exercise 5.2.12), i.e. that a function with property $(2)$ that is also continuous at one point is of the form $cx$. But as you showed, there are only countably many discontinuities, and $mathbb R$ is uncountable. So $f$ is continuous somewhere, and we're already done.
As a final remark, note that continuity at any one point $x$ immediately implies continuity at any other point $y$, since if $y_nto y$ then $x_n := y_n - y + x to x$ and the additive property $(2)$ implies that
$$ f(y_n) = f(x_n) + f(y) - f(x) to f(y).$$ Therefore the known result used above follows from the basic result for continuous functions with property $(2)$.
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Let $f$ be continuous at $0$. Since $f(x+y) = f(x)+f(y)$. Let $x_n rightarrow c$. Now $f(x_n-c) = f(x_n)+f(-c)$
Hence as $f$ is continuous at $0$, $lim_n rightarrow infty f(x_n-c) = f(0)$. Hence $lim_n rightarrow infty f(x_n-c) - f(-c)= lim_n rightarrow infty f(x_n) = f(0)-f(-c) = f(c)$. Hence sequentially continuous.
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
8
down vote
If $x$ is an irrational number, let $r_n$ be rationals that approach $x$ from below. Then as $x>r_n$, the monotonicity property $(1)$ of $f$ gives
$$f(x) ge f(r_n) = r_n f(1) to x f(1),$$
so we conclude $$f(x) ge x f(1),$$ and similarly, using rationals that approach $x$ from above, $f(x) le xf(1)$. Hence, $$f(x) = xf(1) quad forall xinmathbb R.$$
(edit) I note the relevant portions of the book,
and,
it seems that $(a)$ you misread the book, which says you only need to prove continuity at one
point, and $(b)$ you want a proof that first proves continuity, and then appeals to the known result(Exercise 5.2.12), i.e. that a function with property $(2)$ that is also continuous at one point is of the form $cx$. But as you showed, there are only countably many discontinuities, and $mathbb R$ is uncountable. So $f$ is continuous somewhere, and we're already done.
As a final remark, note that continuity at any one point $x$ immediately implies continuity at any other point $y$, since if $y_nto y$ then $x_n := y_n - y + x to x$ and the additive property $(2)$ implies that
$$ f(y_n) = f(x_n) + f(y) - f(x) to f(y).$$ Therefore the known result used above follows from the basic result for continuous functions with property $(2)$.
add a comment |Â
up vote
8
down vote
If $x$ is an irrational number, let $r_n$ be rationals that approach $x$ from below. Then as $x>r_n$, the monotonicity property $(1)$ of $f$ gives
$$f(x) ge f(r_n) = r_n f(1) to x f(1),$$
so we conclude $$f(x) ge x f(1),$$ and similarly, using rationals that approach $x$ from above, $f(x) le xf(1)$. Hence, $$f(x) = xf(1) quad forall xinmathbb R.$$
(edit) I note the relevant portions of the book,
and,
it seems that $(a)$ you misread the book, which says you only need to prove continuity at one
point, and $(b)$ you want a proof that first proves continuity, and then appeals to the known result(Exercise 5.2.12), i.e. that a function with property $(2)$ that is also continuous at one point is of the form $cx$. But as you showed, there are only countably many discontinuities, and $mathbb R$ is uncountable. So $f$ is continuous somewhere, and we're already done.
As a final remark, note that continuity at any one point $x$ immediately implies continuity at any other point $y$, since if $y_nto y$ then $x_n := y_n - y + x to x$ and the additive property $(2)$ implies that
$$ f(y_n) = f(x_n) + f(y) - f(x) to f(y).$$ Therefore the known result used above follows from the basic result for continuous functions with property $(2)$.
add a comment |Â
up vote
8
down vote
up vote
8
down vote
If $x$ is an irrational number, let $r_n$ be rationals that approach $x$ from below. Then as $x>r_n$, the monotonicity property $(1)$ of $f$ gives
$$f(x) ge f(r_n) = r_n f(1) to x f(1),$$
so we conclude $$f(x) ge x f(1),$$ and similarly, using rationals that approach $x$ from above, $f(x) le xf(1)$. Hence, $$f(x) = xf(1) quad forall xinmathbb R.$$
(edit) I note the relevant portions of the book,
and,
it seems that $(a)$ you misread the book, which says you only need to prove continuity at one
point, and $(b)$ you want a proof that first proves continuity, and then appeals to the known result(Exercise 5.2.12), i.e. that a function with property $(2)$ that is also continuous at one point is of the form $cx$. But as you showed, there are only countably many discontinuities, and $mathbb R$ is uncountable. So $f$ is continuous somewhere, and we're already done.
As a final remark, note that continuity at any one point $x$ immediately implies continuity at any other point $y$, since if $y_nto y$ then $x_n := y_n - y + x to x$ and the additive property $(2)$ implies that
$$ f(y_n) = f(x_n) + f(y) - f(x) to f(y).$$ Therefore the known result used above follows from the basic result for continuous functions with property $(2)$.
If $x$ is an irrational number, let $r_n$ be rationals that approach $x$ from below. Then as $x>r_n$, the monotonicity property $(1)$ of $f$ gives
$$f(x) ge f(r_n) = r_n f(1) to x f(1),$$
so we conclude $$f(x) ge x f(1),$$ and similarly, using rationals that approach $x$ from above, $f(x) le xf(1)$. Hence, $$f(x) = xf(1) quad forall xinmathbb R.$$
(edit) I note the relevant portions of the book,
and,
it seems that $(a)$ you misread the book, which says you only need to prove continuity at one
point, and $(b)$ you want a proof that first proves continuity, and then appeals to the known result(Exercise 5.2.12), i.e. that a function with property $(2)$ that is also continuous at one point is of the form $cx$. But as you showed, there are only countably many discontinuities, and $mathbb R$ is uncountable. So $f$ is continuous somewhere, and we're already done.
As a final remark, note that continuity at any one point $x$ immediately implies continuity at any other point $y$, since if $y_nto y$ then $x_n := y_n - y + x to x$ and the additive property $(2)$ implies that
$$ f(y_n) = f(x_n) + f(y) - f(x) to f(y).$$ Therefore the known result used above follows from the basic result for continuous functions with property $(2)$.
edited Sep 6 at 8:25
answered Sep 5 at 11:24
Calvin Khor
8,81621133
8,81621133
add a comment |Â
add a comment |Â
up vote
0
down vote
Let $f$ be continuous at $0$. Since $f(x+y) = f(x)+f(y)$. Let $x_n rightarrow c$. Now $f(x_n-c) = f(x_n)+f(-c)$
Hence as $f$ is continuous at $0$, $lim_n rightarrow infty f(x_n-c) = f(0)$. Hence $lim_n rightarrow infty f(x_n-c) - f(-c)= lim_n rightarrow infty f(x_n) = f(0)-f(-c) = f(c)$. Hence sequentially continuous.
add a comment |Â
up vote
0
down vote
Let $f$ be continuous at $0$. Since $f(x+y) = f(x)+f(y)$. Let $x_n rightarrow c$. Now $f(x_n-c) = f(x_n)+f(-c)$
Hence as $f$ is continuous at $0$, $lim_n rightarrow infty f(x_n-c) = f(0)$. Hence $lim_n rightarrow infty f(x_n-c) - f(-c)= lim_n rightarrow infty f(x_n) = f(0)-f(-c) = f(c)$. Hence sequentially continuous.
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Let $f$ be continuous at $0$. Since $f(x+y) = f(x)+f(y)$. Let $x_n rightarrow c$. Now $f(x_n-c) = f(x_n)+f(-c)$
Hence as $f$ is continuous at $0$, $lim_n rightarrow infty f(x_n-c) = f(0)$. Hence $lim_n rightarrow infty f(x_n-c) - f(-c)= lim_n rightarrow infty f(x_n) = f(0)-f(-c) = f(c)$. Hence sequentially continuous.
Let $f$ be continuous at $0$. Since $f(x+y) = f(x)+f(y)$. Let $x_n rightarrow c$. Now $f(x_n-c) = f(x_n)+f(-c)$
Hence as $f$ is continuous at $0$, $lim_n rightarrow infty f(x_n-c) = f(0)$. Hence $lim_n rightarrow infty f(x_n-c) - f(-c)= lim_n rightarrow infty f(x_n) = f(0)-f(-c) = f(c)$. Hence sequentially continuous.
answered Sep 5 at 11:34
Balaji sb
40325
40325
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2
Linking to copyrighted material is frowned upon. Posting the whole book goes way beyond fair use and is illegal in most jurisdictions.
â Umberto P.
Sep 5 at 11:25