Legendre transformation of a piecewise function












0












$begingroup$


Let $f:mathbb{R}tomathbb{R}$ be defined by
$$f(x)=begin{cases}
1 & text{ if } x<1 \
3x-1 & text{ if } xgeq 1
end{cases}$$
How do I determine the Legendre transform $f^*$ of $f$, and the Legendre transform $(f^{*})^*$ of $f^*$?



Given a function $g:Ato mathbb{R}$, where $Asubseteq mathbb{R}^k$, the definition of $g^*$ is defined by
$$g^*(p)=sup{xp-g(x)mid xin A}$$
for $pin mathbb{R}^k$, where the supremum on the right side is finite.



Clearly, we have
$$
sup{xp-f(x)mid x <1}=-1;\
sup{xp-f(x)mid x geq 1}=1.
$$
Determining the maximum of them, we get that $f^*(p)=1$ and the domain of $f^*$ is ${3}$. Am I understanding this correctly? The Legendre transform has different names, such as a Legendre-Fenchel or a conjugate function.






real-analysis convex-analysis supremum-and-infimum







share|cite|improve this question











$endgroup$

















    0












    $begingroup$


    Let $f:mathbb{R}tomathbb{R}$ be defined by
    $$f(x)=begin{cases}
    1 & text{ if } x<1 \
    3x-1 & text{ if } xgeq 1
    end{cases}$$
    How do I determine the Legendre transform $f^*$ of $f$, and the Legendre transform $(f^{*})^*$ of $f^*$?



    Given a function $g:Ato mathbb{R}$, where $Asubseteq mathbb{R}^k$, the definition of $g^*$ is defined by
    $$g^*(p)=sup{xp-g(x)mid xin A}$$
    for $pin mathbb{R}^k$, where the supremum on the right side is finite.



    Clearly, we have
    $$
    sup{xp-f(x)mid x <1}=-1;\
    sup{xp-f(x)mid x geq 1}=1.
    $$
    Determining the maximum of them, we get that $f^*(p)=1$ and the domain of $f^*$ is ${3}$. Am I understanding this correctly? The Legendre transform has different names, such as a Legendre-Fenchel or a conjugate function.






    real-analysis convex-analysis supremum-and-infimum







    share|cite|improve this question











    $endgroup$















      0












      0








      0





      $begingroup$


      Let $f:mathbb{R}tomathbb{R}$ be defined by
      $$f(x)=begin{cases}
      1 & text{ if } x<1 \
      3x-1 & text{ if } xgeq 1
      end{cases}$$
      How do I determine the Legendre transform $f^*$ of $f$, and the Legendre transform $(f^{*})^*$ of $f^*$?



      Given a function $g:Ato mathbb{R}$, where $Asubseteq mathbb{R}^k$, the definition of $g^*$ is defined by
      $$g^*(p)=sup{xp-g(x)mid xin A}$$
      for $pin mathbb{R}^k$, where the supremum on the right side is finite.



      Clearly, we have
      $$
      sup{xp-f(x)mid x <1}=-1;\
      sup{xp-f(x)mid x geq 1}=1.
      $$
      Determining the maximum of them, we get that $f^*(p)=1$ and the domain of $f^*$ is ${3}$. Am I understanding this correctly? The Legendre transform has different names, such as a Legendre-Fenchel or a conjugate function.






      real-analysis convex-analysis supremum-and-infimum







      share|cite|improve this question











      $endgroup$




      Let $f:mathbb{R}tomathbb{R}$ be defined by
      $$f(x)=begin{cases}
      1 & text{ if } x<1 \
      3x-1 & text{ if } xgeq 1
      end{cases}$$
      How do I determine the Legendre transform $f^*$ of $f$, and the Legendre transform $(f^{*})^*$ of $f^*$?



      Given a function $g:Ato mathbb{R}$, where $Asubseteq mathbb{R}^k$, the definition of $g^*$ is defined by
      $$g^*(p)=sup{xp-g(x)mid xin A}$$
      for $pin mathbb{R}^k$, where the supremum on the right side is finite.



      Clearly, we have
      $$
      sup{xp-f(x)mid x <1}=-1;\
      sup{xp-f(x)mid x geq 1}=1.
      $$
      Determining the maximum of them, we get that $f^*(p)=1$ and the domain of $f^*$ is ${3}$. Am I understanding this correctly? The Legendre transform has different names, such as a Legendre-Fenchel or a conjugate function.






      real-analysis convex-analysis supremum-and-infimum




      real-analysis convex-analysis supremum-and-infimum






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Jan 12 at 19:58







      UnknownW

















      asked Jan 10 at 15:48









      UnknownWUnknownW

      1,045922




      1,045922






















          1 Answer
          1






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          1












          $begingroup$

          Hint:



          $$f^*(p) = max Big{ sup_{x<1} big(y x -1big), sup_{xgeq 1} big( y x - 3x+1big)Big}$$



          Now let's evaluate the first supremum in the max.



          If $y<0$, then that supremum is $+infty$ (by letting $xto-infty$).

          If $y geq 0$, then that supremum is $lim_{xto 1^-} (yx-1)=y-1$.



          Now argue similarly for the second supremum and put it all together.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Wow, that's easier than I thought. I've not thought the way you have just shown. Thank you for that. I have another similar question about finding the supremum of ${xy-frac{x^2}{2}mid xin [-1,1]}$. Do you have time to help me with this one? Or can you confirm this: What I found out is that the supremum is $y^2/2$ if $yin[-1,1]$, and $y-1/2$ if $y>1$ and $-y-1/2$ if $y<-1$.
            $endgroup$
            – UnknownW
            Jan 14 at 19:25












          • $begingroup$
            Please post this as a separate question, the comments are not the right medium for that. The procedure is similar.
            $endgroup$
            – max_zorn
            Jan 15 at 0:38










          • $begingroup$
            I have evaluated the second supremum, which I got as follows: If $y>3$ then the supremum is infinite, while if $yleq 3$ then the supremum is $y-2$. So, when we are finding $f^*$, its domain must be $[0,3]$, and so the result is $max{y-1,y-2}=y-1$ for all $yin [0,3]$. Do you agree?
            $endgroup$
            – UnknownW
            Jan 15 at 4:16














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          1 Answer
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          active

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          1 Answer
          1






          active

          oldest

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          active

          oldest

          votes






          active

          oldest

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          1












          $begingroup$

          Hint:



          $$f^*(p) = max Big{ sup_{x<1} big(y x -1big), sup_{xgeq 1} big( y x - 3x+1big)Big}$$



          Now let's evaluate the first supremum in the max.



          If $y<0$, then that supremum is $+infty$ (by letting $xto-infty$).

          If $y geq 0$, then that supremum is $lim_{xto 1^-} (yx-1)=y-1$.



          Now argue similarly for the second supremum and put it all together.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Wow, that's easier than I thought. I've not thought the way you have just shown. Thank you for that. I have another similar question about finding the supremum of ${xy-frac{x^2}{2}mid xin [-1,1]}$. Do you have time to help me with this one? Or can you confirm this: What I found out is that the supremum is $y^2/2$ if $yin[-1,1]$, and $y-1/2$ if $y>1$ and $-y-1/2$ if $y<-1$.
            $endgroup$
            – UnknownW
            Jan 14 at 19:25












          • $begingroup$
            Please post this as a separate question, the comments are not the right medium for that. The procedure is similar.
            $endgroup$
            – max_zorn
            Jan 15 at 0:38










          • $begingroup$
            I have evaluated the second supremum, which I got as follows: If $y>3$ then the supremum is infinite, while if $yleq 3$ then the supremum is $y-2$. So, when we are finding $f^*$, its domain must be $[0,3]$, and so the result is $max{y-1,y-2}=y-1$ for all $yin [0,3]$. Do you agree?
            $endgroup$
            – UnknownW
            Jan 15 at 4:16


















          1












          $begingroup$

          Hint:



          $$f^*(p) = max Big{ sup_{x<1} big(y x -1big), sup_{xgeq 1} big( y x - 3x+1big)Big}$$



          Now let's evaluate the first supremum in the max.



          If $y<0$, then that supremum is $+infty$ (by letting $xto-infty$).

          If $y geq 0$, then that supremum is $lim_{xto 1^-} (yx-1)=y-1$.



          Now argue similarly for the second supremum and put it all together.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Wow, that's easier than I thought. I've not thought the way you have just shown. Thank you for that. I have another similar question about finding the supremum of ${xy-frac{x^2}{2}mid xin [-1,1]}$. Do you have time to help me with this one? Or can you confirm this: What I found out is that the supremum is $y^2/2$ if $yin[-1,1]$, and $y-1/2$ if $y>1$ and $-y-1/2$ if $y<-1$.
            $endgroup$
            – UnknownW
            Jan 14 at 19:25












          • $begingroup$
            Please post this as a separate question, the comments are not the right medium for that. The procedure is similar.
            $endgroup$
            – max_zorn
            Jan 15 at 0:38










          • $begingroup$
            I have evaluated the second supremum, which I got as follows: If $y>3$ then the supremum is infinite, while if $yleq 3$ then the supremum is $y-2$. So, when we are finding $f^*$, its domain must be $[0,3]$, and so the result is $max{y-1,y-2}=y-1$ for all $yin [0,3]$. Do you agree?
            $endgroup$
            – UnknownW
            Jan 15 at 4:16
















          1












          1








          1





          $begingroup$

          Hint:



          $$f^*(p) = max Big{ sup_{x<1} big(y x -1big), sup_{xgeq 1} big( y x - 3x+1big)Big}$$



          Now let's evaluate the first supremum in the max.



          If $y<0$, then that supremum is $+infty$ (by letting $xto-infty$).

          If $y geq 0$, then that supremum is $lim_{xto 1^-} (yx-1)=y-1$.



          Now argue similarly for the second supremum and put it all together.






          share|cite|improve this answer









          $endgroup$



          Hint:



          $$f^*(p) = max Big{ sup_{x<1} big(y x -1big), sup_{xgeq 1} big( y x - 3x+1big)Big}$$



          Now let's evaluate the first supremum in the max.



          If $y<0$, then that supremum is $+infty$ (by letting $xto-infty$).

          If $y geq 0$, then that supremum is $lim_{xto 1^-} (yx-1)=y-1$.



          Now argue similarly for the second supremum and put it all together.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Jan 14 at 7:29









          max_zornmax_zorn

          3,44061429




          3,44061429












          • $begingroup$
            Wow, that's easier than I thought. I've not thought the way you have just shown. Thank you for that. I have another similar question about finding the supremum of ${xy-frac{x^2}{2}mid xin [-1,1]}$. Do you have time to help me with this one? Or can you confirm this: What I found out is that the supremum is $y^2/2$ if $yin[-1,1]$, and $y-1/2$ if $y>1$ and $-y-1/2$ if $y<-1$.
            $endgroup$
            – UnknownW
            Jan 14 at 19:25












          • $begingroup$
            Please post this as a separate question, the comments are not the right medium for that. The procedure is similar.
            $endgroup$
            – max_zorn
            Jan 15 at 0:38










          • $begingroup$
            I have evaluated the second supremum, which I got as follows: If $y>3$ then the supremum is infinite, while if $yleq 3$ then the supremum is $y-2$. So, when we are finding $f^*$, its domain must be $[0,3]$, and so the result is $max{y-1,y-2}=y-1$ for all $yin [0,3]$. Do you agree?
            $endgroup$
            – UnknownW
            Jan 15 at 4:16




















          • $begingroup$
            Wow, that's easier than I thought. I've not thought the way you have just shown. Thank you for that. I have another similar question about finding the supremum of ${xy-frac{x^2}{2}mid xin [-1,1]}$. Do you have time to help me with this one? Or can you confirm this: What I found out is that the supremum is $y^2/2$ if $yin[-1,1]$, and $y-1/2$ if $y>1$ and $-y-1/2$ if $y<-1$.
            $endgroup$
            – UnknownW
            Jan 14 at 19:25












          • $begingroup$
            Please post this as a separate question, the comments are not the right medium for that. The procedure is similar.
            $endgroup$
            – max_zorn
            Jan 15 at 0:38










          • $begingroup$
            I have evaluated the second supremum, which I got as follows: If $y>3$ then the supremum is infinite, while if $yleq 3$ then the supremum is $y-2$. So, when we are finding $f^*$, its domain must be $[0,3]$, and so the result is $max{y-1,y-2}=y-1$ for all $yin [0,3]$. Do you agree?
            $endgroup$
            – UnknownW
            Jan 15 at 4:16


















          $begingroup$
          Wow, that's easier than I thought. I've not thought the way you have just shown. Thank you for that. I have another similar question about finding the supremum of ${xy-frac{x^2}{2}mid xin [-1,1]}$. Do you have time to help me with this one? Or can you confirm this: What I found out is that the supremum is $y^2/2$ if $yin[-1,1]$, and $y-1/2$ if $y>1$ and $-y-1/2$ if $y<-1$.
          $endgroup$
          – UnknownW
          Jan 14 at 19:25






          $begingroup$
          Wow, that's easier than I thought. I've not thought the way you have just shown. Thank you for that. I have another similar question about finding the supremum of ${xy-frac{x^2}{2}mid xin [-1,1]}$. Do you have time to help me with this one? Or can you confirm this: What I found out is that the supremum is $y^2/2$ if $yin[-1,1]$, and $y-1/2$ if $y>1$ and $-y-1/2$ if $y<-1$.
          $endgroup$
          – UnknownW
          Jan 14 at 19:25














          $begingroup$
          Please post this as a separate question, the comments are not the right medium for that. The procedure is similar.
          $endgroup$
          – max_zorn
          Jan 15 at 0:38




          $begingroup$
          Please post this as a separate question, the comments are not the right medium for that. The procedure is similar.
          $endgroup$
          – max_zorn
          Jan 15 at 0:38












          $begingroup$
          I have evaluated the second supremum, which I got as follows: If $y>3$ then the supremum is infinite, while if $yleq 3$ then the supremum is $y-2$. So, when we are finding $f^*$, its domain must be $[0,3]$, and so the result is $max{y-1,y-2}=y-1$ for all $yin [0,3]$. Do you agree?
          $endgroup$
          – UnknownW
          Jan 15 at 4:16






          $begingroup$
          I have evaluated the second supremum, which I got as follows: If $y>3$ then the supremum is infinite, while if $yleq 3$ then the supremum is $y-2$. So, when we are finding $f^*$, its domain must be $[0,3]$, and so the result is $max{y-1,y-2}=y-1$ for all $yin [0,3]$. Do you agree?
          $endgroup$
          – UnknownW
          Jan 15 at 4:16




















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