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

In summary: The graph has a kink at the origin and is not differentiable there. Note that the derivative is negative for x<0 and positive for x>0. That makes it non-convex.
  • #1
Wuberdall
34
0

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.
 
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  • #2
Wuberdall said:

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.
 

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

What is the Legendre transform of f(x) = exp(|x|)?

The Legendre transform of f(x) = exp(|x|) is a mathematical operation that transforms the original function into its conjugate function. In this case, the conjugate function would be f*(y) = ylog(y) - y. This transformation is useful in solving optimization problems and finding the convex hull of a set of points.

Why is the Legendre transform important?

The Legendre transform is important because it allows us to find the dual function of a given function, which can often be easier to solve or analyze. It also helps in understanding the geometry of a function and its properties, such as convexity and concavity.

How do you perform the Legendre transform?

To perform the Legendre transform, you first need to find the derivative of the original function f(x). Then, set the derivative equal to y and solve for x. This will give you the conjugate function f*(y). You can also use the Legendre-Fenchel transformation formula, which is f*(y) = sup(xy - f(x)).

What is the relationship between the Legendre transform and the Fourier transform?

The Legendre transform and the Fourier transform are both mathematical operations that involve transforming a function into a different representation. However, the Legendre transform is used for convex functions, while the Fourier transform is used for periodic functions. They both have applications in physics, engineering, and other fields of science.

What are some real-world applications of the Legendre transform?

The Legendre transform has many real-world applications, including in thermodynamics, statistical mechanics, and economics. It can be used to solve optimization problems and find the equilibrium state of a system. It is also used in data analysis and signal processing to analyze and manipulate data in a more efficient way.

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