is this function regular?












1












$begingroup$


I am learning nonsmooth analysis for discontinuous dynamical systems. An important concept in this field is the regularity of functions.



A function $f$ is regular at $x$ if its right directional derivative $f^{'}(x,v)$ is equal to its generalized directional derivative $f^{o}(x,v)$ at $x$ in any direction $v$.





The right directional derivative $f^{'}$ at $x$ in the direction $v$ is defined as
$$f^{'}(x;v)=lim_{hrightarrow 0^+}frac{f(x+hv)-f(x)}{h}$$ when this limit exists. The generalized directinal derivative $f^{o}$ at $x$ in the direction $v$ is defined as
$$begin{align}
f^{o}(:,v)&=limsup_{begin{array}{}yrightarrow x\hrightarrow0^+end{array}}frac{f(y+hv)-f(y)}{h}\
&=lim_{begin{array}{}deltarightarrow0^+\ epsilonrightarrow0^+end{array}}sup_{begin{array}{}yinmathcal{B}(x,delta)\ hin[0,epsilon)end{array}}frac{f(y+hv)-f(y)}{h}
end{align}$$





The definition of regular is not straightforward for beginners. Let's use an example to study it.



Define $$f_1(x)=begin{cases}x^2,&|x|<1,\1,& otherwise end{cases}$$
Then is function $f_1$ regular at $x=pm 1$ or
not?





Here is my calculation.
$$
begin{align}
f_1(1;v)^{'}&=lim_{hrightarrow 0^+}frac{f_1(1+hv)-f_1(1)}{h}\
&=begin{cases}0,&v>0,\2v,&v<0.end{cases}
end{align}
$$

and
$$begin{align}
f_1^{o}(1,v)&=limsup_{begin{array}{}yrightarrow x\hrightarrow0^+end{array}}frac{f_1(y+hv)-f_1(y)}{h}\
&=limsup_{hrightarrow 0^+}frac{f_1(1+kh+hv)-f_1(1+kh)}{h}\
&=begin{cases}
2v,&v>0,\
0,&v<0.
end{cases}
end{align}$$

Thus $f_1(x)$ is not regular at $x=1$. (Please point out errors if the result is incorrect).



More calculation steps:



Define $g(x)=begin{cases}2x,&|x|<1.\0,&x>1.end{cases}$, then by L'Hopital's Rule, one can obtain
$$begin{align}
f^{o}(1;v)&=limsup_{hrightarrow 0^+}frac{f(1+hv+hk)-f(1+hk)}{h}\
&=limsup_{hrightarrow 0^+}g(1+hv+kv)(v+k)-g(1+hk)k\
&=begin{cases}
sup&begin{cases}
0,&k>0,\
begin{cases}-2k,&v+k>0\2v,&v+k<0end{cases},&k<0,
end{cases},v>0,\
sup&begin{cases}
2v,&k<0,\
begin{cases}0,&2v+2k>0\2(v+k),&v+k<0end{cases},&k>0,
end{cases},v<0,
end{cases}\
&=begin{cases}2v,&v>0,\0,&v<0end{cases}
end{align}$$





I find a similar piecewise function is claimed to be regular in published papers. The analysis can be analogized as
if $v>0$, then
$$begin{align}
f_1^{o}(1,v)&=limsup_{begin{array}{} yrightarrow 1\trightarrow 0^{+} end{array}}frac{f_1(y+hv)-f_1(y)}{h}\
&leq limsup_{begin{array}{} y^{'}rightarrow 1\trightarrow 0^{+} end{array}}frac{f_1(1)-f_1(y^{'}-hv)}{h}\
&=-lim_{hrightarrow 0^+}frac{f_1(1)-f_1(1)}{h}\
&=0.
end{align}$$

And if $v<0$, then
$$begin{align}
f_1^{o}(1,v)&=limsup_{begin{array}{} yrightarrow 1\trightarrow 0^{+} end{array}}frac{f_1(y+hv)-f_1(y)}{h}\
&=limsup_{begin{array}{} yrightarrow 1\trightarrow 0^{+} end{array}}frac{f_1(y+hv)-f_1(1)}{h}\
&=2v
end{align}$$



The original version in the paper are
the piecewise potential function and regularity analysis.





Obviously, there is something wrong. Can anyone point it out?
Thank you.










share|cite|improve this question











$endgroup$

















    1












    $begingroup$


    I am learning nonsmooth analysis for discontinuous dynamical systems. An important concept in this field is the regularity of functions.



    A function $f$ is regular at $x$ if its right directional derivative $f^{'}(x,v)$ is equal to its generalized directional derivative $f^{o}(x,v)$ at $x$ in any direction $v$.





    The right directional derivative $f^{'}$ at $x$ in the direction $v$ is defined as
    $$f^{'}(x;v)=lim_{hrightarrow 0^+}frac{f(x+hv)-f(x)}{h}$$ when this limit exists. The generalized directinal derivative $f^{o}$ at $x$ in the direction $v$ is defined as
    $$begin{align}
    f^{o}(:,v)&=limsup_{begin{array}{}yrightarrow x\hrightarrow0^+end{array}}frac{f(y+hv)-f(y)}{h}\
    &=lim_{begin{array}{}deltarightarrow0^+\ epsilonrightarrow0^+end{array}}sup_{begin{array}{}yinmathcal{B}(x,delta)\ hin[0,epsilon)end{array}}frac{f(y+hv)-f(y)}{h}
    end{align}$$





    The definition of regular is not straightforward for beginners. Let's use an example to study it.



    Define $$f_1(x)=begin{cases}x^2,&|x|<1,\1,& otherwise end{cases}$$
    Then is function $f_1$ regular at $x=pm 1$ or
    not?





    Here is my calculation.
    $$
    begin{align}
    f_1(1;v)^{'}&=lim_{hrightarrow 0^+}frac{f_1(1+hv)-f_1(1)}{h}\
    &=begin{cases}0,&v>0,\2v,&v<0.end{cases}
    end{align}
    $$

    and
    $$begin{align}
    f_1^{o}(1,v)&=limsup_{begin{array}{}yrightarrow x\hrightarrow0^+end{array}}frac{f_1(y+hv)-f_1(y)}{h}\
    &=limsup_{hrightarrow 0^+}frac{f_1(1+kh+hv)-f_1(1+kh)}{h}\
    &=begin{cases}
    2v,&v>0,\
    0,&v<0.
    end{cases}
    end{align}$$

    Thus $f_1(x)$ is not regular at $x=1$. (Please point out errors if the result is incorrect).



    More calculation steps:



    Define $g(x)=begin{cases}2x,&|x|<1.\0,&x>1.end{cases}$, then by L'Hopital's Rule, one can obtain
    $$begin{align}
    f^{o}(1;v)&=limsup_{hrightarrow 0^+}frac{f(1+hv+hk)-f(1+hk)}{h}\
    &=limsup_{hrightarrow 0^+}g(1+hv+kv)(v+k)-g(1+hk)k\
    &=begin{cases}
    sup&begin{cases}
    0,&k>0,\
    begin{cases}-2k,&v+k>0\2v,&v+k<0end{cases},&k<0,
    end{cases},v>0,\
    sup&begin{cases}
    2v,&k<0,\
    begin{cases}0,&2v+2k>0\2(v+k),&v+k<0end{cases},&k>0,
    end{cases},v<0,
    end{cases}\
    &=begin{cases}2v,&v>0,\0,&v<0end{cases}
    end{align}$$





    I find a similar piecewise function is claimed to be regular in published papers. The analysis can be analogized as
    if $v>0$, then
    $$begin{align}
    f_1^{o}(1,v)&=limsup_{begin{array}{} yrightarrow 1\trightarrow 0^{+} end{array}}frac{f_1(y+hv)-f_1(y)}{h}\
    &leq limsup_{begin{array}{} y^{'}rightarrow 1\trightarrow 0^{+} end{array}}frac{f_1(1)-f_1(y^{'}-hv)}{h}\
    &=-lim_{hrightarrow 0^+}frac{f_1(1)-f_1(1)}{h}\
    &=0.
    end{align}$$

    And if $v<0$, then
    $$begin{align}
    f_1^{o}(1,v)&=limsup_{begin{array}{} yrightarrow 1\trightarrow 0^{+} end{array}}frac{f_1(y+hv)-f_1(y)}{h}\
    &=limsup_{begin{array}{} yrightarrow 1\trightarrow 0^{+} end{array}}frac{f_1(y+hv)-f_1(1)}{h}\
    &=2v
    end{align}$$



    The original version in the paper are
    the piecewise potential function and regularity analysis.





    Obviously, there is something wrong. Can anyone point it out?
    Thank you.










    share|cite|improve this question











    $endgroup$















      1












      1








      1





      $begingroup$


      I am learning nonsmooth analysis for discontinuous dynamical systems. An important concept in this field is the regularity of functions.



      A function $f$ is regular at $x$ if its right directional derivative $f^{'}(x,v)$ is equal to its generalized directional derivative $f^{o}(x,v)$ at $x$ in any direction $v$.





      The right directional derivative $f^{'}$ at $x$ in the direction $v$ is defined as
      $$f^{'}(x;v)=lim_{hrightarrow 0^+}frac{f(x+hv)-f(x)}{h}$$ when this limit exists. The generalized directinal derivative $f^{o}$ at $x$ in the direction $v$ is defined as
      $$begin{align}
      f^{o}(:,v)&=limsup_{begin{array}{}yrightarrow x\hrightarrow0^+end{array}}frac{f(y+hv)-f(y)}{h}\
      &=lim_{begin{array}{}deltarightarrow0^+\ epsilonrightarrow0^+end{array}}sup_{begin{array}{}yinmathcal{B}(x,delta)\ hin[0,epsilon)end{array}}frac{f(y+hv)-f(y)}{h}
      end{align}$$





      The definition of regular is not straightforward for beginners. Let's use an example to study it.



      Define $$f_1(x)=begin{cases}x^2,&|x|<1,\1,& otherwise end{cases}$$
      Then is function $f_1$ regular at $x=pm 1$ or
      not?





      Here is my calculation.
      $$
      begin{align}
      f_1(1;v)^{'}&=lim_{hrightarrow 0^+}frac{f_1(1+hv)-f_1(1)}{h}\
      &=begin{cases}0,&v>0,\2v,&v<0.end{cases}
      end{align}
      $$

      and
      $$begin{align}
      f_1^{o}(1,v)&=limsup_{begin{array}{}yrightarrow x\hrightarrow0^+end{array}}frac{f_1(y+hv)-f_1(y)}{h}\
      &=limsup_{hrightarrow 0^+}frac{f_1(1+kh+hv)-f_1(1+kh)}{h}\
      &=begin{cases}
      2v,&v>0,\
      0,&v<0.
      end{cases}
      end{align}$$

      Thus $f_1(x)$ is not regular at $x=1$. (Please point out errors if the result is incorrect).



      More calculation steps:



      Define $g(x)=begin{cases}2x,&|x|<1.\0,&x>1.end{cases}$, then by L'Hopital's Rule, one can obtain
      $$begin{align}
      f^{o}(1;v)&=limsup_{hrightarrow 0^+}frac{f(1+hv+hk)-f(1+hk)}{h}\
      &=limsup_{hrightarrow 0^+}g(1+hv+kv)(v+k)-g(1+hk)k\
      &=begin{cases}
      sup&begin{cases}
      0,&k>0,\
      begin{cases}-2k,&v+k>0\2v,&v+k<0end{cases},&k<0,
      end{cases},v>0,\
      sup&begin{cases}
      2v,&k<0,\
      begin{cases}0,&2v+2k>0\2(v+k),&v+k<0end{cases},&k>0,
      end{cases},v<0,
      end{cases}\
      &=begin{cases}2v,&v>0,\0,&v<0end{cases}
      end{align}$$





      I find a similar piecewise function is claimed to be regular in published papers. The analysis can be analogized as
      if $v>0$, then
      $$begin{align}
      f_1^{o}(1,v)&=limsup_{begin{array}{} yrightarrow 1\trightarrow 0^{+} end{array}}frac{f_1(y+hv)-f_1(y)}{h}\
      &leq limsup_{begin{array}{} y^{'}rightarrow 1\trightarrow 0^{+} end{array}}frac{f_1(1)-f_1(y^{'}-hv)}{h}\
      &=-lim_{hrightarrow 0^+}frac{f_1(1)-f_1(1)}{h}\
      &=0.
      end{align}$$

      And if $v<0$, then
      $$begin{align}
      f_1^{o}(1,v)&=limsup_{begin{array}{} yrightarrow 1\trightarrow 0^{+} end{array}}frac{f_1(y+hv)-f_1(y)}{h}\
      &=limsup_{begin{array}{} yrightarrow 1\trightarrow 0^{+} end{array}}frac{f_1(y+hv)-f_1(1)}{h}\
      &=2v
      end{align}$$



      The original version in the paper are
      the piecewise potential function and regularity analysis.





      Obviously, there is something wrong. Can anyone point it out?
      Thank you.










      share|cite|improve this question











      $endgroup$




      I am learning nonsmooth analysis for discontinuous dynamical systems. An important concept in this field is the regularity of functions.



      A function $f$ is regular at $x$ if its right directional derivative $f^{'}(x,v)$ is equal to its generalized directional derivative $f^{o}(x,v)$ at $x$ in any direction $v$.





      The right directional derivative $f^{'}$ at $x$ in the direction $v$ is defined as
      $$f^{'}(x;v)=lim_{hrightarrow 0^+}frac{f(x+hv)-f(x)}{h}$$ when this limit exists. The generalized directinal derivative $f^{o}$ at $x$ in the direction $v$ is defined as
      $$begin{align}
      f^{o}(:,v)&=limsup_{begin{array}{}yrightarrow x\hrightarrow0^+end{array}}frac{f(y+hv)-f(y)}{h}\
      &=lim_{begin{array}{}deltarightarrow0^+\ epsilonrightarrow0^+end{array}}sup_{begin{array}{}yinmathcal{B}(x,delta)\ hin[0,epsilon)end{array}}frac{f(y+hv)-f(y)}{h}
      end{align}$$





      The definition of regular is not straightforward for beginners. Let's use an example to study it.



      Define $$f_1(x)=begin{cases}x^2,&|x|<1,\1,& otherwise end{cases}$$
      Then is function $f_1$ regular at $x=pm 1$ or
      not?





      Here is my calculation.
      $$
      begin{align}
      f_1(1;v)^{'}&=lim_{hrightarrow 0^+}frac{f_1(1+hv)-f_1(1)}{h}\
      &=begin{cases}0,&v>0,\2v,&v<0.end{cases}
      end{align}
      $$

      and
      $$begin{align}
      f_1^{o}(1,v)&=limsup_{begin{array}{}yrightarrow x\hrightarrow0^+end{array}}frac{f_1(y+hv)-f_1(y)}{h}\
      &=limsup_{hrightarrow 0^+}frac{f_1(1+kh+hv)-f_1(1+kh)}{h}\
      &=begin{cases}
      2v,&v>0,\
      0,&v<0.
      end{cases}
      end{align}$$

      Thus $f_1(x)$ is not regular at $x=1$. (Please point out errors if the result is incorrect).



      More calculation steps:



      Define $g(x)=begin{cases}2x,&|x|<1.\0,&x>1.end{cases}$, then by L'Hopital's Rule, one can obtain
      $$begin{align}
      f^{o}(1;v)&=limsup_{hrightarrow 0^+}frac{f(1+hv+hk)-f(1+hk)}{h}\
      &=limsup_{hrightarrow 0^+}g(1+hv+kv)(v+k)-g(1+hk)k\
      &=begin{cases}
      sup&begin{cases}
      0,&k>0,\
      begin{cases}-2k,&v+k>0\2v,&v+k<0end{cases},&k<0,
      end{cases},v>0,\
      sup&begin{cases}
      2v,&k<0,\
      begin{cases}0,&2v+2k>0\2(v+k),&v+k<0end{cases},&k>0,
      end{cases},v<0,
      end{cases}\
      &=begin{cases}2v,&v>0,\0,&v<0end{cases}
      end{align}$$





      I find a similar piecewise function is claimed to be regular in published papers. The analysis can be analogized as
      if $v>0$, then
      $$begin{align}
      f_1^{o}(1,v)&=limsup_{begin{array}{} yrightarrow 1\trightarrow 0^{+} end{array}}frac{f_1(y+hv)-f_1(y)}{h}\
      &leq limsup_{begin{array}{} y^{'}rightarrow 1\trightarrow 0^{+} end{array}}frac{f_1(1)-f_1(y^{'}-hv)}{h}\
      &=-lim_{hrightarrow 0^+}frac{f_1(1)-f_1(1)}{h}\
      &=0.
      end{align}$$

      And if $v<0$, then
      $$begin{align}
      f_1^{o}(1,v)&=limsup_{begin{array}{} yrightarrow 1\trightarrow 0^{+} end{array}}frac{f_1(y+hv)-f_1(y)}{h}\
      &=limsup_{begin{array}{} yrightarrow 1\trightarrow 0^{+} end{array}}frac{f_1(y+hv)-f_1(1)}{h}\
      &=2v
      end{align}$$



      The original version in the paper are
      the piecewise potential function and regularity analysis.





      Obviously, there is something wrong. Can anyone point it out?
      Thank you.







      real-analysis derivatives control-theory






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Jan 9 at 16:41







      wjxjmj

















      asked Jan 9 at 8:13









      wjxjmjwjxjmj

      62




      62






















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