Can a cube of discontinuous function be continuous?












14














Can a cube (meaning $g(x) = f(x)^3 = f(x) cdot f(x) cdot f(x)$) of discontinuous function $f: D to mathbb{R}$ ($D$ is subset of $mathbb{R}$) be continuous? I think it can't, since $x^3$ is injective, but I am not able to prove it or find a counterexample.






continuity







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  • 10




    The point is not that $x longmapsto x^3$ is injective, as much as $x longmapsto x^{1/3}$ is continuous.
    – Mindlack
    yesterday






  • 3




    What is your domain? It matters really quite a lot.
    – user3482749
    yesterday








  • 2




    I think the injectivity is very much to the point. $f(x) =sqrt x$ is also continuous on $[0,∞)$ but there are any number of discontinuous functions $g$ on that interval with $g(x)cdot g(x)$ continuous.
    – MJD
    yesterday












  • No. But the cube of a non-differentiable function can be differentiable : $|x|^3$
    – Eric Duminil
    yesterday








  • 3




    I misspoke: it's the range that matters. If it's $mathbb{C}$ and your domain is, say, connected, it's false (send some subset of your domain to $1$, and the rest to one of the other cube roots of unity).
    – user3482749
    18 hours ago
















14














Can a cube (meaning $g(x) = f(x)^3 = f(x) cdot f(x) cdot f(x)$) of discontinuous function $f: D to mathbb{R}$ ($D$ is subset of $mathbb{R}$) be continuous? I think it can't, since $x^3$ is injective, but I am not able to prove it or find a counterexample.






continuity







share|cite|improve this question




















  • 10




    The point is not that $x longmapsto x^3$ is injective, as much as $x longmapsto x^{1/3}$ is continuous.
    – Mindlack
    yesterday






  • 3




    What is your domain? It matters really quite a lot.
    – user3482749
    yesterday








  • 2




    I think the injectivity is very much to the point. $f(x) =sqrt x$ is also continuous on $[0,∞)$ but there are any number of discontinuous functions $g$ on that interval with $g(x)cdot g(x)$ continuous.
    – MJD
    yesterday












  • No. But the cube of a non-differentiable function can be differentiable : $|x|^3$
    – Eric Duminil
    yesterday








  • 3




    I misspoke: it's the range that matters. If it's $mathbb{C}$ and your domain is, say, connected, it's false (send some subset of your domain to $1$, and the rest to one of the other cube roots of unity).
    – user3482749
    18 hours ago














14












14








14


1





Can a cube (meaning $g(x) = f(x)^3 = f(x) cdot f(x) cdot f(x)$) of discontinuous function $f: D to mathbb{R}$ ($D$ is subset of $mathbb{R}$) be continuous? I think it can't, since $x^3$ is injective, but I am not able to prove it or find a counterexample.






continuity







share|cite|improve this question















Can a cube (meaning $g(x) = f(x)^3 = f(x) cdot f(x) cdot f(x)$) of discontinuous function $f: D to mathbb{R}$ ($D$ is subset of $mathbb{R}$) be continuous? I think it can't, since $x^3$ is injective, but I am not able to prove it or find a counterexample.






continuity




continuity






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share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited yesterday







J. Abraham

















asked yesterday









J. AbrahamJ. Abraham

528316




528316








  • 10




    The point is not that $x longmapsto x^3$ is injective, as much as $x longmapsto x^{1/3}$ is continuous.
    – Mindlack
    yesterday






  • 3




    What is your domain? It matters really quite a lot.
    – user3482749
    yesterday








  • 2




    I think the injectivity is very much to the point. $f(x) =sqrt x$ is also continuous on $[0,∞)$ but there are any number of discontinuous functions $g$ on that interval with $g(x)cdot g(x)$ continuous.
    – MJD
    yesterday












  • No. But the cube of a non-differentiable function can be differentiable : $|x|^3$
    – Eric Duminil
    yesterday








  • 3




    I misspoke: it's the range that matters. If it's $mathbb{C}$ and your domain is, say, connected, it's false (send some subset of your domain to $1$, and the rest to one of the other cube roots of unity).
    – user3482749
    18 hours ago














  • 10




    The point is not that $x longmapsto x^3$ is injective, as much as $x longmapsto x^{1/3}$ is continuous.
    – Mindlack
    yesterday






  • 3




    What is your domain? It matters really quite a lot.
    – user3482749
    yesterday








  • 2




    I think the injectivity is very much to the point. $f(x) =sqrt x$ is also continuous on $[0,∞)$ but there are any number of discontinuous functions $g$ on that interval with $g(x)cdot g(x)$ continuous.
    – MJD
    yesterday












  • No. But the cube of a non-differentiable function can be differentiable : $|x|^3$
    – Eric Duminil
    yesterday








  • 3




    I misspoke: it's the range that matters. If it's $mathbb{C}$ and your domain is, say, connected, it's false (send some subset of your domain to $1$, and the rest to one of the other cube roots of unity).
    – user3482749
    18 hours ago








10




10




The point is not that $x longmapsto x^3$ is injective, as much as $x longmapsto x^{1/3}$ is continuous.
– Mindlack
yesterday




The point is not that $x longmapsto x^3$ is injective, as much as $x longmapsto x^{1/3}$ is continuous.
– Mindlack
yesterday




3




3




What is your domain? It matters really quite a lot.
– user3482749
yesterday






What is your domain? It matters really quite a lot.
– user3482749
yesterday






2




2




I think the injectivity is very much to the point. $f(x) =sqrt x$ is also continuous on $[0,∞)$ but there are any number of discontinuous functions $g$ on that interval with $g(x)cdot g(x)$ continuous.
– MJD
yesterday






I think the injectivity is very much to the point. $f(x) =sqrt x$ is also continuous on $[0,∞)$ but there are any number of discontinuous functions $g$ on that interval with $g(x)cdot g(x)$ continuous.
– MJD
yesterday














No. But the cube of a non-differentiable function can be differentiable : $|x|^3$
– Eric Duminil
yesterday






No. But the cube of a non-differentiable function can be differentiable : $|x|^3$
– Eric Duminil
yesterday






3




3




I misspoke: it's the range that matters. If it's $mathbb{C}$ and your domain is, say, connected, it's false (send some subset of your domain to $1$, and the rest to one of the other cube roots of unity).
– user3482749
18 hours ago




I misspoke: it's the range that matters. If it's $mathbb{C}$ and your domain is, say, connected, it's false (send some subset of your domain to $1$, and the rest to one of the other cube roots of unity).
– user3482749
18 hours ago










2 Answers
2






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oldest

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16














If a function $f(x)$ is continuous, then its cube root $g(x) = f(x)^{1/3}$ is also continuous.



So the contrapositive is also true, which is:



If a function $g(x)$ is not continuous, then its cube $f(x) = g(x)^3$ is not continuous either.



(Strictly speaking, the contrapositive is actually "if the cube root $f(x)^{1/3}$ of a function $f(x)$ is not continuous, then the function $f(x)$ is not continuous either". But this is equivalent.)






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    35














    Since $phi : mathbb{R} to mathbb{R}, phi(x) = x^3$, is a homeomorphism, you see that $f$ is continuous iff $phi circ f$ is continuous.






    share|cite|improve this answer



















    • 8




      To lower the level of this answer, note that $phi^{-1}(x) = sqrt[3] x$ is also a continuous map ("homeomorphism" means a continuous, invertible map whose inverse is also continuous). So if $phicirc f$ is continuous, so is $f = phi^{-1}circ phi circ f$.
      – Paul Sinclair
      15 hours ago











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    2 Answers
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    2 Answers
    2






    active

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    active

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    active

    oldest

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    16














    If a function $f(x)$ is continuous, then its cube root $g(x) = f(x)^{1/3}$ is also continuous.



    So the contrapositive is also true, which is:



    If a function $g(x)$ is not continuous, then its cube $f(x) = g(x)^3$ is not continuous either.



    (Strictly speaking, the contrapositive is actually "if the cube root $f(x)^{1/3}$ of a function $f(x)$ is not continuous, then the function $f(x)$ is not continuous either". But this is equivalent.)






    share|cite|improve this answer


























      16














      If a function $f(x)$ is continuous, then its cube root $g(x) = f(x)^{1/3}$ is also continuous.



      So the contrapositive is also true, which is:



      If a function $g(x)$ is not continuous, then its cube $f(x) = g(x)^3$ is not continuous either.



      (Strictly speaking, the contrapositive is actually "if the cube root $f(x)^{1/3}$ of a function $f(x)$ is not continuous, then the function $f(x)$ is not continuous either". But this is equivalent.)






      share|cite|improve this answer
























        16












        16








        16






        If a function $f(x)$ is continuous, then its cube root $g(x) = f(x)^{1/3}$ is also continuous.



        So the contrapositive is also true, which is:



        If a function $g(x)$ is not continuous, then its cube $f(x) = g(x)^3$ is not continuous either.



        (Strictly speaking, the contrapositive is actually "if the cube root $f(x)^{1/3}$ of a function $f(x)$ is not continuous, then the function $f(x)$ is not continuous either". But this is equivalent.)






        share|cite|improve this answer












        If a function $f(x)$ is continuous, then its cube root $g(x) = f(x)^{1/3}$ is also continuous.



        So the contrapositive is also true, which is:



        If a function $g(x)$ is not continuous, then its cube $f(x) = g(x)^3$ is not continuous either.



        (Strictly speaking, the contrapositive is actually "if the cube root $f(x)^{1/3}$ of a function $f(x)$ is not continuous, then the function $f(x)$ is not continuous either". But this is equivalent.)







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered yesterday









        Tanner SwettTanner Swett

        4,1961639




        4,1961639























            35














            Since $phi : mathbb{R} to mathbb{R}, phi(x) = x^3$, is a homeomorphism, you see that $f$ is continuous iff $phi circ f$ is continuous.






            share|cite|improve this answer



















            • 8




              To lower the level of this answer, note that $phi^{-1}(x) = sqrt[3] x$ is also a continuous map ("homeomorphism" means a continuous, invertible map whose inverse is also continuous). So if $phicirc f$ is continuous, so is $f = phi^{-1}circ phi circ f$.
              – Paul Sinclair
              15 hours ago
















            35














            Since $phi : mathbb{R} to mathbb{R}, phi(x) = x^3$, is a homeomorphism, you see that $f$ is continuous iff $phi circ f$ is continuous.






            share|cite|improve this answer



















            • 8




              To lower the level of this answer, note that $phi^{-1}(x) = sqrt[3] x$ is also a continuous map ("homeomorphism" means a continuous, invertible map whose inverse is also continuous). So if $phicirc f$ is continuous, so is $f = phi^{-1}circ phi circ f$.
              – Paul Sinclair
              15 hours ago














            35












            35








            35






            Since $phi : mathbb{R} to mathbb{R}, phi(x) = x^3$, is a homeomorphism, you see that $f$ is continuous iff $phi circ f$ is continuous.






            share|cite|improve this answer














            Since $phi : mathbb{R} to mathbb{R}, phi(x) = x^3$, is a homeomorphism, you see that $f$ is continuous iff $phi circ f$ is continuous.







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited yesterday

























            answered yesterday









            Paul FrostPaul Frost

            9,6953732




            9,6953732








            • 8




              To lower the level of this answer, note that $phi^{-1}(x) = sqrt[3] x$ is also a continuous map ("homeomorphism" means a continuous, invertible map whose inverse is also continuous). So if $phicirc f$ is continuous, so is $f = phi^{-1}circ phi circ f$.
              – Paul Sinclair
              15 hours ago














            • 8




              To lower the level of this answer, note that $phi^{-1}(x) = sqrt[3] x$ is also a continuous map ("homeomorphism" means a continuous, invertible map whose inverse is also continuous). So if $phicirc f$ is continuous, so is $f = phi^{-1}circ phi circ f$.
              – Paul Sinclair
              15 hours ago








            8




            8




            To lower the level of this answer, note that $phi^{-1}(x) = sqrt[3] x$ is also a continuous map ("homeomorphism" means a continuous, invertible map whose inverse is also continuous). So if $phicirc f$ is continuous, so is $f = phi^{-1}circ phi circ f$.
            – Paul Sinclair
            15 hours ago




            To lower the level of this answer, note that $phi^{-1}(x) = sqrt[3] x$ is also a continuous map ("homeomorphism" means a continuous, invertible map whose inverse is also continuous). So if $phicirc f$ is continuous, so is $f = phi^{-1}circ phi circ f$.
            – Paul Sinclair
            15 hours ago


















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