Does the sum of two functions satisfying the intermediate value property also have this property?












10














If functions $f$ and $g$ both satisfy the intermediate value property, does their sum also satisfy this property? If not, what if I suppose in addition that $f$ is continuous?



Thanks in advance!



Edit: I found the second part of my question here:
Is the sum of a Darboux function and a continuous function Darboux?






calculus analysis continuity







share|cite|improve this question
























  • The best thing you can aim for is to show that the image of $f+g$ is an interval. (In particular it takes all the values in-between but not necessarily in the right order)
    – Yanko
    Dec 17 at 9:57
















10














If functions $f$ and $g$ both satisfy the intermediate value property, does their sum also satisfy this property? If not, what if I suppose in addition that $f$ is continuous?



Thanks in advance!



Edit: I found the second part of my question here:
Is the sum of a Darboux function and a continuous function Darboux?






calculus analysis continuity







share|cite|improve this question
























  • The best thing you can aim for is to show that the image of $f+g$ is an interval. (In particular it takes all the values in-between but not necessarily in the right order)
    – Yanko
    Dec 17 at 9:57














10












10








10


2





If functions $f$ and $g$ both satisfy the intermediate value property, does their sum also satisfy this property? If not, what if I suppose in addition that $f$ is continuous?



Thanks in advance!



Edit: I found the second part of my question here:
Is the sum of a Darboux function and a continuous function Darboux?






calculus analysis continuity







share|cite|improve this question















If functions $f$ and $g$ both satisfy the intermediate value property, does their sum also satisfy this property? If not, what if I suppose in addition that $f$ is continuous?



Thanks in advance!



Edit: I found the second part of my question here:
Is the sum of a Darboux function and a continuous function Darboux?






calculus analysis continuity




calculus analysis continuity






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 17 at 10:32









Glorfindel

3,41981830




3,41981830










asked Dec 17 at 9:47









Jiu

435111




435111












  • The best thing you can aim for is to show that the image of $f+g$ is an interval. (In particular it takes all the values in-between but not necessarily in the right order)
    – Yanko
    Dec 17 at 9:57


















  • The best thing you can aim for is to show that the image of $f+g$ is an interval. (In particular it takes all the values in-between but not necessarily in the right order)
    – Yanko
    Dec 17 at 9:57
















The best thing you can aim for is to show that the image of $f+g$ is an interval. (In particular it takes all the values in-between but not necessarily in the right order)
– Yanko
Dec 17 at 9:57




The best thing you can aim for is to show that the image of $f+g$ is an interval. (In particular it takes all the values in-between but not necessarily in the right order)
– Yanko
Dec 17 at 9:57










1 Answer
1






active

oldest

votes


















14














Consider the functions, $f:[0,1]rightarrow [-1,1]$ and $g:[0,1]rightarrow[-1,1]$ where



$$f = begin{cases}sinfrac{1}{x},& x>0 \ 0, & x = 0end{cases}$$
and
$$g = begin{cases}-sinfrac{1}{x},& x>0 \ 1, & x = 0end{cases}$$






share|cite|improve this answer



















  • 1




    Thanks for your answer! What if we suppose that one of the functions is continuous? Or should I ask this in another question?
    – Jiu
    Dec 17 at 10:05






  • 1




    I'm not sure whether it is true or not in that case. Sorry
    – Olof Rubin
    Dec 17 at 10:18






  • 2




    I just found the same question, see my edit!
    – Jiu
    Dec 17 at 10:29






  • 1




    Interesting, makes sense that I couldn’t think of a counterexample
    – Olof Rubin
    Dec 17 at 11:15











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1 Answer
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1 Answer
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active

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14














Consider the functions, $f:[0,1]rightarrow [-1,1]$ and $g:[0,1]rightarrow[-1,1]$ where



$$f = begin{cases}sinfrac{1}{x},& x>0 \ 0, & x = 0end{cases}$$
and
$$g = begin{cases}-sinfrac{1}{x},& x>0 \ 1, & x = 0end{cases}$$






share|cite|improve this answer



















  • 1




    Thanks for your answer! What if we suppose that one of the functions is continuous? Or should I ask this in another question?
    – Jiu
    Dec 17 at 10:05






  • 1




    I'm not sure whether it is true or not in that case. Sorry
    – Olof Rubin
    Dec 17 at 10:18






  • 2




    I just found the same question, see my edit!
    – Jiu
    Dec 17 at 10:29






  • 1




    Interesting, makes sense that I couldn’t think of a counterexample
    – Olof Rubin
    Dec 17 at 11:15
















14














Consider the functions, $f:[0,1]rightarrow [-1,1]$ and $g:[0,1]rightarrow[-1,1]$ where



$$f = begin{cases}sinfrac{1}{x},& x>0 \ 0, & x = 0end{cases}$$
and
$$g = begin{cases}-sinfrac{1}{x},& x>0 \ 1, & x = 0end{cases}$$






share|cite|improve this answer



















  • 1




    Thanks for your answer! What if we suppose that one of the functions is continuous? Or should I ask this in another question?
    – Jiu
    Dec 17 at 10:05






  • 1




    I'm not sure whether it is true or not in that case. Sorry
    – Olof Rubin
    Dec 17 at 10:18






  • 2




    I just found the same question, see my edit!
    – Jiu
    Dec 17 at 10:29






  • 1




    Interesting, makes sense that I couldn’t think of a counterexample
    – Olof Rubin
    Dec 17 at 11:15














14












14








14






Consider the functions, $f:[0,1]rightarrow [-1,1]$ and $g:[0,1]rightarrow[-1,1]$ where



$$f = begin{cases}sinfrac{1}{x},& x>0 \ 0, & x = 0end{cases}$$
and
$$g = begin{cases}-sinfrac{1}{x},& x>0 \ 1, & x = 0end{cases}$$






share|cite|improve this answer














Consider the functions, $f:[0,1]rightarrow [-1,1]$ and $g:[0,1]rightarrow[-1,1]$ where



$$f = begin{cases}sinfrac{1}{x},& x>0 \ 0, & x = 0end{cases}$$
and
$$g = begin{cases}-sinfrac{1}{x},& x>0 \ 1, & x = 0end{cases}$$







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Dec 17 at 15:46









Community♦

1




1










answered Dec 17 at 9:56









Olof Rubin

991315




991315








  • 1




    Thanks for your answer! What if we suppose that one of the functions is continuous? Or should I ask this in another question?
    – Jiu
    Dec 17 at 10:05






  • 1




    I'm not sure whether it is true or not in that case. Sorry
    – Olof Rubin
    Dec 17 at 10:18






  • 2




    I just found the same question, see my edit!
    – Jiu
    Dec 17 at 10:29






  • 1




    Interesting, makes sense that I couldn’t think of a counterexample
    – Olof Rubin
    Dec 17 at 11:15














  • 1




    Thanks for your answer! What if we suppose that one of the functions is continuous? Or should I ask this in another question?
    – Jiu
    Dec 17 at 10:05






  • 1




    I'm not sure whether it is true or not in that case. Sorry
    – Olof Rubin
    Dec 17 at 10:18






  • 2




    I just found the same question, see my edit!
    – Jiu
    Dec 17 at 10:29






  • 1




    Interesting, makes sense that I couldn’t think of a counterexample
    – Olof Rubin
    Dec 17 at 11:15








1




1




Thanks for your answer! What if we suppose that one of the functions is continuous? Or should I ask this in another question?
– Jiu
Dec 17 at 10:05




Thanks for your answer! What if we suppose that one of the functions is continuous? Or should I ask this in another question?
– Jiu
Dec 17 at 10:05




1




1




I'm not sure whether it is true or not in that case. Sorry
– Olof Rubin
Dec 17 at 10:18




I'm not sure whether it is true or not in that case. Sorry
– Olof Rubin
Dec 17 at 10:18




2




2




I just found the same question, see my edit!
– Jiu
Dec 17 at 10:29




I just found the same question, see my edit!
– Jiu
Dec 17 at 10:29




1




1




Interesting, makes sense that I couldn’t think of a counterexample
– Olof Rubin
Dec 17 at 11:15




Interesting, makes sense that I couldn’t think of a counterexample
– Olof Rubin
Dec 17 at 11:15


















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