The Legendre transform of ##f(x) = \exp(\lvert x\rvert )##

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Homework Statement


Let the single variable real function [itex]f:\mathbb{R}\rightarrow\mathbb{R}[/itex] be given by [itex]f(x)=e^{|x|}[/itex].
Determine the Legendre transform of [itex]f[/itex].

Homework Equations


Let [itex]I\subseteq\mathbb{R}[/itex]be an interval, and [itex] f:I\rightarrow\mathbb{R}[/itex]a convex function. Then its Legendre transform is the function [itex]f^{\ast}:I^{\ast}\rightarrow\mathbb{R}[/itex]defined by : [itex]f^{\ast}(p) = \sup\lbrace xp - f(x)\hspace{1mm}\vert\hspace{1mm}x\in\mathbb{R}\rbrace[/itex].

The Attempt at a Solution


The function f is clearly a convex function and the supremum can easily by evaluated by finding the global maximum for [itex]xp-f(x)[/itex]. This yields

[itex]f^{\ast}(p) = \left\lbrace
\begin{aligned}
&p\big(\ln p - 1\big) \hspace{6pt},\hspace{12pt}p>0 \\
&-1 \hspace{6pt},\hspace{46pt}p=0\\
&f(-p) \hspace{6pt},\hspace{36pt}p<0
\end{aligned}
\right.[/itex]

My problem is that this function isn't convex nor isn't continuous on [itex]\mathbb{R}[/itex] as I would expect the Legendre transform to be... Have I misunderstood something or simply done the legendre transform wrong? If the latter what have I then done wrong?

Thanks in advance.
 

Answers and Replies

  • #2
Ray Vickson
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Homework Statement


Let the single variable real function [itex]f:\mathbb{R}\rightarrow\mathbb{R}[/itex] be given by [itex]f(x)=e^{|x|}[/itex].
Determine the Legendre transform of [itex]f[/itex].

Homework Equations


Let [itex]I\subseteq\mathbb{R}[/itex]be an interval, and [itex] f:I\rightarrow\mathbb{R}[/itex]a convex function. Then its Legendre transform is the function [itex]f^{\ast}:I^{\ast}\rightarrow\mathbb{R}[/itex]defined by : [itex]f^{\ast}(p) = \sup\lbrace xp - f(x)\hspace{1mm}\vert\hspace{1mm}x\in\mathbb{R}\rbrace[/itex].

The Attempt at a Solution


The function f is clearly a convex function and the supremum can easily by evaluated by finding the global maximum for [itex]xp-f(x)[/itex]. This yields

[itex]f^{\ast}(p) = \left\lbrace
\begin{aligned}
&p\big(\ln p - 1\big) \hspace{6pt},\hspace{12pt}p>0 \\
&-1 \hspace{6pt},\hspace{46pt}p=0\\
&f(-p) \hspace{6pt},\hspace{36pt}p<0
\end{aligned}
\right.[/itex]

My problem is that this function isn't convex nor isn't continuous on [itex]\mathbb{R}[/itex] as I would expect the Legendre transform to be... Have I misunderstood something or simply done the legendre transform wrong? If the latter what have I then done wrong?

Thanks in advance.
I suggest you plot y = exp(|x|) over some x-interval (-a,a) to clarify your thoughts about the function.
 

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