Legendre transformation of a piecewise function
$begingroup$
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
asked Jan 10 at 15:48
UnknownWUnknownW
1,045922
$endgroup$
add a comment |
$begingroup$
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
asked Jan 10 at 15:48
UnknownWUnknownW
1,045922
$endgroup$
add a comment |
$begingroup$
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
asked Jan 10 at 15:48
UnknownWUnknownW
1,045922
$endgroup$
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
asked Jan 10 at 15:48
UnknownWUnknownW
1,045922
asked Jan 10 at 15:48
UnknownWUnknownW
1,045922
edited Jan 12 at 19:58
UnknownW
asked Jan 10 at 15:48
UnknownWUnknownW
1,045922
asked Jan 10 at 15:48
UnknownWUnknownW
1,045922
asked Jan 10 at 15:48
UnknownWUnknownW
1,045922
1,045922
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
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.
answered Jan 14 at 7:29
max_zornmax_zorn
3,44061429
$endgroup$
$begingroup$
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
add a comment |
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1 Answer
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1 Answer
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active
oldest
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oldest
votes
$begingroup$
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.
answered Jan 14 at 7:29
max_zornmax_zorn
3,44061429
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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.
answered Jan 14 at 7:29
max_zornmax_zorn
3,44061429
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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.
answered Jan 14 at 7:29
max_zornmax_zorn
3,44061429
$endgroup$
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.
answered Jan 14 at 7:29
max_zornmax_zorn
3,44061429
answered Jan 14 at 7:29
max_zornmax_zorn
3,44061429
answered Jan 14 at 7:29
max_zornmax_zorn
3,44061429
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
add a comment |
$begingroup$
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
add a comment |
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