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.










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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.










share|cite|improve this question



















  • 2




    Linking to copyrighted material is frowned upon. Posting the whole book goes way beyond fair use and is illegal in most jurisdictions.
    – Umberto P.
    Sep 5 at 11:25












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.










share|cite|improve this question















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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edited Sep 6 at 16:23









Aloizio Macedo♦

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asked Sep 5 at 11:17









Saaqib Mahmood

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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.
    – Umberto P.
    Sep 5 at 11:25












  • 2




    Linking to copyrighted material is frowned upon. Posting the whole book goes way beyond fair use and is illegal in most jurisdictions.
    – 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.
– 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.
– Umberto P.
Sep 5 at 11:25










2 Answers
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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,




enter image description here




and,




enter image description here




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

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      2 Answers
      2






      active

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      active

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      active

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      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,




      enter image description here




      and,




      enter image description here




      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 $(2)$.






      share|cite|improve this answer


























        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,




        enter image description here




        and,




        enter image description here




        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 $(2)$.






        share|cite|improve this answer
























          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,




          enter image description here




          and,




          enter image description here




          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 $(2)$.






          share|cite|improve this answer














          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,




          enter image description here




          and,




          enter image description here




          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 $(2)$.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Sep 6 at 8:25

























          answered Sep 5 at 11:24









          Calvin Khor

          8,81621133




          8,81621133




















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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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                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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                  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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                  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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                  answered Sep 5 at 11:34









                  Balaji sb

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