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











up vote
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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
    2 hours ago















up vote
6
down vote

favorite
1












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
    2 hours ago













up vote
6
down vote

favorite
1









up vote
6
down vote

favorite
1






1





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 2 hours ago









Glorfindel

3,38481730




3,38481730










asked 2 hours ago









Jiu

42119




42119












  • 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
    2 hours ago


















  • 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
    2 hours ago
















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
2 hours ago




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
2 hours ago










1 Answer
1






active

oldest

votes

















up vote
7
down vote



accepted










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





















  • Thanks for your answer! What if we suppose that one of the functions is continuous? Or should I ask this in another question?
    – Jiu
    2 hours ago










  • I'm not sure whether it is true or not in that case. Sorry
    – Olof Rubin
    2 hours ago






  • 1




    I just found the same question, see my edit!
    – Jiu
    2 hours ago










  • Interesting, makes sense that I couldn’t think of a counterexample
    – Olof Rubin
    1 hour ago











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1 Answer
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1 Answer
1






active

oldest

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active

oldest

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active

oldest

votes








up vote
7
down vote



accepted










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





















  • Thanks for your answer! What if we suppose that one of the functions is continuous? Or should I ask this in another question?
    – Jiu
    2 hours ago










  • I'm not sure whether it is true or not in that case. Sorry
    – Olof Rubin
    2 hours ago






  • 1




    I just found the same question, see my edit!
    – Jiu
    2 hours ago










  • Interesting, makes sense that I couldn’t think of a counterexample
    – Olof Rubin
    1 hour ago















up vote
7
down vote



accepted










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





















  • Thanks for your answer! What if we suppose that one of the functions is continuous? Or should I ask this in another question?
    – Jiu
    2 hours ago










  • I'm not sure whether it is true or not in that case. Sorry
    – Olof Rubin
    2 hours ago






  • 1




    I just found the same question, see my edit!
    – Jiu
    2 hours ago










  • Interesting, makes sense that I couldn’t think of a counterexample
    – Olof Rubin
    1 hour ago













up vote
7
down vote



accepted







up vote
7
down vote



accepted






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










answered 2 hours ago









Olof Rubin

771213




771213












  • Thanks for your answer! What if we suppose that one of the functions is continuous? Or should I ask this in another question?
    – Jiu
    2 hours ago










  • I'm not sure whether it is true or not in that case. Sorry
    – Olof Rubin
    2 hours ago






  • 1




    I just found the same question, see my edit!
    – Jiu
    2 hours ago










  • Interesting, makes sense that I couldn’t think of a counterexample
    – Olof Rubin
    1 hour ago


















  • Thanks for your answer! What if we suppose that one of the functions is continuous? Or should I ask this in another question?
    – Jiu
    2 hours ago










  • I'm not sure whether it is true or not in that case. Sorry
    – Olof Rubin
    2 hours ago






  • 1




    I just found the same question, see my edit!
    – Jiu
    2 hours ago










  • Interesting, makes sense that I couldn’t think of a counterexample
    – Olof Rubin
    1 hour ago
















Thanks for your answer! What if we suppose that one of the functions is continuous? Or should I ask this in another question?
– Jiu
2 hours ago




Thanks for your answer! What if we suppose that one of the functions is continuous? Or should I ask this in another question?
– Jiu
2 hours ago












I'm not sure whether it is true or not in that case. Sorry
– Olof Rubin
2 hours ago




I'm not sure whether it is true or not in that case. Sorry
– Olof Rubin
2 hours ago




1




1




I just found the same question, see my edit!
– Jiu
2 hours ago




I just found the same question, see my edit!
– Jiu
2 hours ago












Interesting, makes sense that I couldn’t think of a counterexample
– Olof Rubin
1 hour ago




Interesting, makes sense that I couldn’t think of a counterexample
– Olof Rubin
1 hour ago


















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