Understanding the Legendre transform
$begingroup$
In physics, I've seen the Legendre transform motivated by "changing the variable $x$ of a function $x mapsto f(x)$ to the variable $u = frac{df}{dx}$."
I don't quite see what that means and why the Legendre transform is the answer to this heuristic. I'd understand it as follows:
Let $f: mathbb{R} tomathbb{R}$ be strictly convex and differentiable. Then for every $x_0 in mathbb{R}$ the slope $frac{df(x_0)}{dx}$ is unique. We want to find a function $f^*: f'(mathbb{R}) to mathbb{R}$ such that $f(x_0)=f^*(frac{df(x_0)}{dx})$ for every $x_0 in mathbb{R}$. In fact we can view the function $f^*$ as a function on a subset of the dual space $mathbb{R^*}$ such that $f^*(df(x_0))=f(x_0)$.
However, this doesn't seem to capture the Legendre transform, for example by the above the Legendre transform of the exponential function should be the identity function. How can the physicists heuristic be made precise?
real-analysis functions
asked Jan 7 at 22:26
Jannik PittJannik Pitt
646517
$endgroup$
add a comment |
$begingroup$
In physics, I've seen the Legendre transform motivated by "changing the variable $x$ of a function $x mapsto f(x)$ to the variable $u = frac{df}{dx}$."
I don't quite see what that means and why the Legendre transform is the answer to this heuristic. I'd understand it as follows:
Let $f: mathbb{R} tomathbb{R}$ be strictly convex and differentiable. Then for every $x_0 in mathbb{R}$ the slope $frac{df(x_0)}{dx}$ is unique. We want to find a function $f^*: f'(mathbb{R}) to mathbb{R}$ such that $f(x_0)=f^*(frac{df(x_0)}{dx})$ for every $x_0 in mathbb{R}$. In fact we can view the function $f^*$ as a function on a subset of the dual space $mathbb{R^*}$ such that $f^*(df(x_0))=f(x_0)$.
However, this doesn't seem to capture the Legendre transform, for example by the above the Legendre transform of the exponential function should be the identity function. How can the physicists heuristic be made precise?
real-analysis functions
asked Jan 7 at 22:26
Jannik PittJannik Pitt
646517
$endgroup$
add a comment |
$begingroup$
In physics, I've seen the Legendre transform motivated by "changing the variable $x$ of a function $x mapsto f(x)$ to the variable $u = frac{df}{dx}$."
I don't quite see what that means and why the Legendre transform is the answer to this heuristic. I'd understand it as follows:
Let $f: mathbb{R} tomathbb{R}$ be strictly convex and differentiable. Then for every $x_0 in mathbb{R}$ the slope $frac{df(x_0)}{dx}$ is unique. We want to find a function $f^*: f'(mathbb{R}) to mathbb{R}$ such that $f(x_0)=f^*(frac{df(x_0)}{dx})$ for every $x_0 in mathbb{R}$. In fact we can view the function $f^*$ as a function on a subset of the dual space $mathbb{R^*}$ such that $f^*(df(x_0))=f(x_0)$.
However, this doesn't seem to capture the Legendre transform, for example by the above the Legendre transform of the exponential function should be the identity function. How can the physicists heuristic be made precise?
real-analysis functions
asked Jan 7 at 22:26
Jannik PittJannik Pitt
646517
$endgroup$
In physics, I've seen the Legendre transform motivated by "changing the variable $x$ of a function $x mapsto f(x)$ to the variable $u = frac{df}{dx}$."
I don't quite see what that means and why the Legendre transform is the answer to this heuristic. I'd understand it as follows:
Let $f: mathbb{R} tomathbb{R}$ be strictly convex and differentiable. Then for every $x_0 in mathbb{R}$ the slope $frac{df(x_0)}{dx}$ is unique. We want to find a function $f^*: f'(mathbb{R}) to mathbb{R}$ such that $f(x_0)=f^*(frac{df(x_0)}{dx})$ for every $x_0 in mathbb{R}$. In fact we can view the function $f^*$ as a function on a subset of the dual space $mathbb{R^*}$ such that $f^*(df(x_0))=f(x_0)$.
However, this doesn't seem to capture the Legendre transform, for example by the above the Legendre transform of the exponential function should be the identity function. How can the physicists heuristic be made precise?
real-analysis functions
real-analysis functions
asked Jan 7 at 22:26
Jannik PittJannik Pitt
646517
asked Jan 7 at 22:26
Jannik PittJannik Pitt
646517
edited Jan 8 at 12:29
Jannik Pitt
asked Jan 7 at 22:26
Jannik PittJannik Pitt
646517
asked Jan 7 at 22:26
Jannik PittJannik Pitt
646517
asked Jan 7 at 22:26
Jannik PittJannik Pitt
646517
646517
add a comment |
add a comment |
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