A Lipschitz function must has a set of intervals s.t. $cup_kI_k=mathbb R$ and $f$ is either convex or concave...












0












$begingroup$


Let $f:mathbb Rtomathbb R$ be a continuous function.



$f$ is said to be "regular" if there exists a set of intervals ${I_k}_{kin K}$ (indexed with $k$ where $K$ is an arbitrary set), such that $cup_kI_k=mathbb R$ and $f$ is either convex or concave on $I_k$ $forall k$.



Claim: $f$ is regular iff $f$ is absolute continuous.



If the claim is wrong, then what is the sufficient and necessary condition for the property “regular"?



It seems that Lipschitz is necessary and second differentiable is sufficient.



What theorems might be useful for me to explore the answer?






real-analysis analysis functions continuity lipschitz-functions







share|cite|improve this question











$endgroup$

















    0












    $begingroup$


    Let $f:mathbb Rtomathbb R$ be a continuous function.



    $f$ is said to be "regular" if there exists a set of intervals ${I_k}_{kin K}$ (indexed with $k$ where $K$ is an arbitrary set), such that $cup_kI_k=mathbb R$ and $f$ is either convex or concave on $I_k$ $forall k$.



    Claim: $f$ is regular iff $f$ is absolute continuous.



    If the claim is wrong, then what is the sufficient and necessary condition for the property “regular"?



    It seems that Lipschitz is necessary and second differentiable is sufficient.



    What theorems might be useful for me to explore the answer?






    real-analysis analysis functions continuity lipschitz-functions







    share|cite|improve this question











    $endgroup$















      0












      0








      0





      $begingroup$


      Let $f:mathbb Rtomathbb R$ be a continuous function.



      $f$ is said to be "regular" if there exists a set of intervals ${I_k}_{kin K}$ (indexed with $k$ where $K$ is an arbitrary set), such that $cup_kI_k=mathbb R$ and $f$ is either convex or concave on $I_k$ $forall k$.



      Claim: $f$ is regular iff $f$ is absolute continuous.



      If the claim is wrong, then what is the sufficient and necessary condition for the property “regular"?



      It seems that Lipschitz is necessary and second differentiable is sufficient.



      What theorems might be useful for me to explore the answer?






      real-analysis analysis functions continuity lipschitz-functions







      share|cite|improve this question











      $endgroup$




      Let $f:mathbb Rtomathbb R$ be a continuous function.



      $f$ is said to be "regular" if there exists a set of intervals ${I_k}_{kin K}$ (indexed with $k$ where $K$ is an arbitrary set), such that $cup_kI_k=mathbb R$ and $f$ is either convex or concave on $I_k$ $forall k$.



      Claim: $f$ is regular iff $f$ is absolute continuous.



      If the claim is wrong, then what is the sufficient and necessary condition for the property “regular"?



      It seems that Lipschitz is necessary and second differentiable is sufficient.



      What theorems might be useful for me to explore the answer?






      real-analysis analysis functions continuity lipschitz-functions




      real-analysis analysis functions continuity lipschitz-functions






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Dec 26 '18 at 15:27









      David C. Ullrich

      61.7k44095




      61.7k44095










      asked Dec 26 '18 at 8:11









      High GPAHigh GPA

      914422




      914422






















          2 Answers
          2






          active

          oldest

          votes


















          1












          $begingroup$

          No, a regular function need not be Lipschitz. Consider $f(t)=t^2$. Of course that's "locally Lipschitz", meaning Lipschitz when restricted to any compact interval. If $f(t)=|t|^{1/2}$ then $f$ is regular but not even locally Lipschitz.



          Presumably in the definition you mean that each interval contains more than one point, because if ${a}=[a,a]$ counts as an "interval" then every function is regular. Assuming so, then no, $f''$ continuous does not imply that $f$ is regular: Let $$g(t)=begin{cases}tsin(1/t),&(tne0),
          \0,&(t=0).end{cases}$$           Then $g$ is continuous, there exists $f$ with $f''=g$, but $f$ is not concave or convex on any interval containing the origin, because $f''$ changes sign on any such interval.



          I doubt that the sort of characterization you want exists.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Thank you Daivd for your intuitive answer. Is monotonicity (quasilinearty) a sufficient condition? My intuition is no but I can't tell why.
            $endgroup$
            – High GPA
            Dec 26 '18 at 22:11








          • 1




            $begingroup$
            @HighGPA Montonicity is certainly not sufficient.. I don't know what you mean by "monotonicity (quasilinearty)".
            $endgroup$
            – David C. Ullrich
            Dec 27 '18 at 1:14



















          0












          $begingroup$

          An absolutely continuous function is differentiable a.e. but only once. A convex function in $I_k$ has an increasing derivative at all but countably many points and so it is differentiable twice a.e. So a "regular" function must be differentiable at all but countably many points in the interior of the $I_k$ and the second derivative has to exist a.e. in the interior of the $I_k$.
          So the equivalence is false. I don't know what family of functions would be equivalent to regular.
          Definitely polynomials and very smooth functions, but it is not an iff.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Thank you for your kind answer! Define $int$ as interior, then $cup_k int(I_k)$ could be the complement of fat-Cantor set so the Lebesgue measure of $cup_k int(I_k)$ could be arbitrarily close to zero. That is, a regular function is not necessarily differentiable almost every where. Note the $K$ is an arbitrary set not a countable set. Not sure if my understanding is correct. Feel free to teach me anything. Thank you again for your kin interests.
            $endgroup$
            – High GPA
            Dec 26 '18 at 21:43










          • $begingroup$
            Yes, what I wrote is valid only in the interior of the intervals $I_k$. I edited my answer.
            $endgroup$
            – Gio67
            Dec 27 '18 at 7:21












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          2 Answers
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          1












          $begingroup$

          No, a regular function need not be Lipschitz. Consider $f(t)=t^2$. Of course that's "locally Lipschitz", meaning Lipschitz when restricted to any compact interval. If $f(t)=|t|^{1/2}$ then $f$ is regular but not even locally Lipschitz.



          Presumably in the definition you mean that each interval contains more than one point, because if ${a}=[a,a]$ counts as an "interval" then every function is regular. Assuming so, then no, $f''$ continuous does not imply that $f$ is regular: Let $$g(t)=begin{cases}tsin(1/t),&(tne0),
          \0,&(t=0).end{cases}$$           Then $g$ is continuous, there exists $f$ with $f''=g$, but $f$ is not concave or convex on any interval containing the origin, because $f''$ changes sign on any such interval.



          I doubt that the sort of characterization you want exists.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Thank you Daivd for your intuitive answer. Is monotonicity (quasilinearty) a sufficient condition? My intuition is no but I can't tell why.
            $endgroup$
            – High GPA
            Dec 26 '18 at 22:11








          • 1




            $begingroup$
            @HighGPA Montonicity is certainly not sufficient.. I don't know what you mean by "monotonicity (quasilinearty)".
            $endgroup$
            – David C. Ullrich
            Dec 27 '18 at 1:14
















          1












          $begingroup$

          No, a regular function need not be Lipschitz. Consider $f(t)=t^2$. Of course that's "locally Lipschitz", meaning Lipschitz when restricted to any compact interval. If $f(t)=|t|^{1/2}$ then $f$ is regular but not even locally Lipschitz.



          Presumably in the definition you mean that each interval contains more than one point, because if ${a}=[a,a]$ counts as an "interval" then every function is regular. Assuming so, then no, $f''$ continuous does not imply that $f$ is regular: Let $$g(t)=begin{cases}tsin(1/t),&(tne0),
          \0,&(t=0).end{cases}$$           Then $g$ is continuous, there exists $f$ with $f''=g$, but $f$ is not concave or convex on any interval containing the origin, because $f''$ changes sign on any such interval.



          I doubt that the sort of characterization you want exists.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Thank you Daivd for your intuitive answer. Is monotonicity (quasilinearty) a sufficient condition? My intuition is no but I can't tell why.
            $endgroup$
            – High GPA
            Dec 26 '18 at 22:11








          • 1




            $begingroup$
            @HighGPA Montonicity is certainly not sufficient.. I don't know what you mean by "monotonicity (quasilinearty)".
            $endgroup$
            – David C. Ullrich
            Dec 27 '18 at 1:14














          1












          1








          1





          $begingroup$

          No, a regular function need not be Lipschitz. Consider $f(t)=t^2$. Of course that's "locally Lipschitz", meaning Lipschitz when restricted to any compact interval. If $f(t)=|t|^{1/2}$ then $f$ is regular but not even locally Lipschitz.



          Presumably in the definition you mean that each interval contains more than one point, because if ${a}=[a,a]$ counts as an "interval" then every function is regular. Assuming so, then no, $f''$ continuous does not imply that $f$ is regular: Let $$g(t)=begin{cases}tsin(1/t),&(tne0),
          \0,&(t=0).end{cases}$$           Then $g$ is continuous, there exists $f$ with $f''=g$, but $f$ is not concave or convex on any interval containing the origin, because $f''$ changes sign on any such interval.



          I doubt that the sort of characterization you want exists.






          share|cite|improve this answer









          $endgroup$



          No, a regular function need not be Lipschitz. Consider $f(t)=t^2$. Of course that's "locally Lipschitz", meaning Lipschitz when restricted to any compact interval. If $f(t)=|t|^{1/2}$ then $f$ is regular but not even locally Lipschitz.



          Presumably in the definition you mean that each interval contains more than one point, because if ${a}=[a,a]$ counts as an "interval" then every function is regular. Assuming so, then no, $f''$ continuous does not imply that $f$ is regular: Let $$g(t)=begin{cases}tsin(1/t),&(tne0),
          \0,&(t=0).end{cases}$$           Then $g$ is continuous, there exists $f$ with $f''=g$, but $f$ is not concave or convex on any interval containing the origin, because $f''$ changes sign on any such interval.



          I doubt that the sort of characterization you want exists.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 26 '18 at 15:44









          David C. UllrichDavid C. Ullrich

          61.7k44095




          61.7k44095












          • $begingroup$
            Thank you Daivd for your intuitive answer. Is monotonicity (quasilinearty) a sufficient condition? My intuition is no but I can't tell why.
            $endgroup$
            – High GPA
            Dec 26 '18 at 22:11








          • 1




            $begingroup$
            @HighGPA Montonicity is certainly not sufficient.. I don't know what you mean by "monotonicity (quasilinearty)".
            $endgroup$
            – David C. Ullrich
            Dec 27 '18 at 1:14


















          • $begingroup$
            Thank you Daivd for your intuitive answer. Is monotonicity (quasilinearty) a sufficient condition? My intuition is no but I can't tell why.
            $endgroup$
            – High GPA
            Dec 26 '18 at 22:11








          • 1




            $begingroup$
            @HighGPA Montonicity is certainly not sufficient.. I don't know what you mean by "monotonicity (quasilinearty)".
            $endgroup$
            – David C. Ullrich
            Dec 27 '18 at 1:14
















          $begingroup$
          Thank you Daivd for your intuitive answer. Is monotonicity (quasilinearty) a sufficient condition? My intuition is no but I can't tell why.
          $endgroup$
          – High GPA
          Dec 26 '18 at 22:11






          $begingroup$
          Thank you Daivd for your intuitive answer. Is monotonicity (quasilinearty) a sufficient condition? My intuition is no but I can't tell why.
          $endgroup$
          – High GPA
          Dec 26 '18 at 22:11






          1




          1




          $begingroup$
          @HighGPA Montonicity is certainly not sufficient.. I don't know what you mean by "monotonicity (quasilinearty)".
          $endgroup$
          – David C. Ullrich
          Dec 27 '18 at 1:14




          $begingroup$
          @HighGPA Montonicity is certainly not sufficient.. I don't know what you mean by "monotonicity (quasilinearty)".
          $endgroup$
          – David C. Ullrich
          Dec 27 '18 at 1:14











          0












          $begingroup$

          An absolutely continuous function is differentiable a.e. but only once. A convex function in $I_k$ has an increasing derivative at all but countably many points and so it is differentiable twice a.e. So a "regular" function must be differentiable at all but countably many points in the interior of the $I_k$ and the second derivative has to exist a.e. in the interior of the $I_k$.
          So the equivalence is false. I don't know what family of functions would be equivalent to regular.
          Definitely polynomials and very smooth functions, but it is not an iff.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Thank you for your kind answer! Define $int$ as interior, then $cup_k int(I_k)$ could be the complement of fat-Cantor set so the Lebesgue measure of $cup_k int(I_k)$ could be arbitrarily close to zero. That is, a regular function is not necessarily differentiable almost every where. Note the $K$ is an arbitrary set not a countable set. Not sure if my understanding is correct. Feel free to teach me anything. Thank you again for your kin interests.
            $endgroup$
            – High GPA
            Dec 26 '18 at 21:43










          • $begingroup$
            Yes, what I wrote is valid only in the interior of the intervals $I_k$. I edited my answer.
            $endgroup$
            – Gio67
            Dec 27 '18 at 7:21
















          0












          $begingroup$

          An absolutely continuous function is differentiable a.e. but only once. A convex function in $I_k$ has an increasing derivative at all but countably many points and so it is differentiable twice a.e. So a "regular" function must be differentiable at all but countably many points in the interior of the $I_k$ and the second derivative has to exist a.e. in the interior of the $I_k$.
          So the equivalence is false. I don't know what family of functions would be equivalent to regular.
          Definitely polynomials and very smooth functions, but it is not an iff.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Thank you for your kind answer! Define $int$ as interior, then $cup_k int(I_k)$ could be the complement of fat-Cantor set so the Lebesgue measure of $cup_k int(I_k)$ could be arbitrarily close to zero. That is, a regular function is not necessarily differentiable almost every where. Note the $K$ is an arbitrary set not a countable set. Not sure if my understanding is correct. Feel free to teach me anything. Thank you again for your kin interests.
            $endgroup$
            – High GPA
            Dec 26 '18 at 21:43










          • $begingroup$
            Yes, what I wrote is valid only in the interior of the intervals $I_k$. I edited my answer.
            $endgroup$
            – Gio67
            Dec 27 '18 at 7:21














          0












          0








          0





          $begingroup$

          An absolutely continuous function is differentiable a.e. but only once. A convex function in $I_k$ has an increasing derivative at all but countably many points and so it is differentiable twice a.e. So a "regular" function must be differentiable at all but countably many points in the interior of the $I_k$ and the second derivative has to exist a.e. in the interior of the $I_k$.
          So the equivalence is false. I don't know what family of functions would be equivalent to regular.
          Definitely polynomials and very smooth functions, but it is not an iff.






          share|cite|improve this answer











          $endgroup$



          An absolutely continuous function is differentiable a.e. but only once. A convex function in $I_k$ has an increasing derivative at all but countably many points and so it is differentiable twice a.e. So a "regular" function must be differentiable at all but countably many points in the interior of the $I_k$ and the second derivative has to exist a.e. in the interior of the $I_k$.
          So the equivalence is false. I don't know what family of functions would be equivalent to regular.
          Definitely polynomials and very smooth functions, but it is not an iff.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Dec 27 '18 at 7:20

























          answered Dec 26 '18 at 8:50









          Gio67Gio67

          12.8k1627




          12.8k1627












          • $begingroup$
            Thank you for your kind answer! Define $int$ as interior, then $cup_k int(I_k)$ could be the complement of fat-Cantor set so the Lebesgue measure of $cup_k int(I_k)$ could be arbitrarily close to zero. That is, a regular function is not necessarily differentiable almost every where. Note the $K$ is an arbitrary set not a countable set. Not sure if my understanding is correct. Feel free to teach me anything. Thank you again for your kin interests.
            $endgroup$
            – High GPA
            Dec 26 '18 at 21:43










          • $begingroup$
            Yes, what I wrote is valid only in the interior of the intervals $I_k$. I edited my answer.
            $endgroup$
            – Gio67
            Dec 27 '18 at 7:21


















          • $begingroup$
            Thank you for your kind answer! Define $int$ as interior, then $cup_k int(I_k)$ could be the complement of fat-Cantor set so the Lebesgue measure of $cup_k int(I_k)$ could be arbitrarily close to zero. That is, a regular function is not necessarily differentiable almost every where. Note the $K$ is an arbitrary set not a countable set. Not sure if my understanding is correct. Feel free to teach me anything. Thank you again for your kin interests.
            $endgroup$
            – High GPA
            Dec 26 '18 at 21:43










          • $begingroup$
            Yes, what I wrote is valid only in the interior of the intervals $I_k$. I edited my answer.
            $endgroup$
            – Gio67
            Dec 27 '18 at 7:21
















          $begingroup$
          Thank you for your kind answer! Define $int$ as interior, then $cup_k int(I_k)$ could be the complement of fat-Cantor set so the Lebesgue measure of $cup_k int(I_k)$ could be arbitrarily close to zero. That is, a regular function is not necessarily differentiable almost every where. Note the $K$ is an arbitrary set not a countable set. Not sure if my understanding is correct. Feel free to teach me anything. Thank you again for your kin interests.
          $endgroup$
          – High GPA
          Dec 26 '18 at 21:43




          $begingroup$
          Thank you for your kind answer! Define $int$ as interior, then $cup_k int(I_k)$ could be the complement of fat-Cantor set so the Lebesgue measure of $cup_k int(I_k)$ could be arbitrarily close to zero. That is, a regular function is not necessarily differentiable almost every where. Note the $K$ is an arbitrary set not a countable set. Not sure if my understanding is correct. Feel free to teach me anything. Thank you again for your kin interests.
          $endgroup$
          – High GPA
          Dec 26 '18 at 21:43












          $begingroup$
          Yes, what I wrote is valid only in the interior of the intervals $I_k$. I edited my answer.
          $endgroup$
          – Gio67
          Dec 27 '18 at 7:21




          $begingroup$
          Yes, what I wrote is valid only in the interior of the intervals $I_k$. I edited my answer.
          $endgroup$
          – Gio67
          Dec 27 '18 at 7:21


















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