# 勒壤得轉換

（重定向自勒让德变换

## 概述

${\displaystyle f^{\star }(p)=pu-f(u)|_{{\mathrm {d} [pu-f(u)] \over \mathrm {d} u}=0}\,\!}$

## 定義

${\displaystyle f^{*}(x^{*})=\sup _{x\in I}(x^{*}x-f(x)),\quad x^{*}\in I^{*}}$

${\displaystyle I^{*}=\left\{x^{*}\in \mathbb {R} :\sup _{x\in I}(x^{*}x-f(x))<\infty \right\}~.}$

f(x)凸函数时，这个函数有良好的定义。

${\displaystyle X^{*}=\left\{x^{*}\in \mathbb {R} ^{n}:\sup _{x\in X}(\langle x^{*},x\rangle -f(x))<\infty \right\}}$

${\displaystyle f^{*}(x^{*})=\sup _{x\in X}(\langle x^{*},x\rangle -f(x)),\quad x^{*}\in X^{*}~,}$

### 最大值式定義

${\displaystyle {\frac {\partial }{\partial x}}\left(px-f(x)\right)=p-{\mathrm {d} f(x) \over \mathrm {d} x}=0\,\!}$

${\displaystyle p={\mathrm {d} f(x) \over \mathrm {d} x}\,\!}$ (1)

${\displaystyle {\frac {\partial ^{2}}{\partial x^{2}}}(xp-f(x))=-{\mathrm {d} ^{2}f(x) \over \mathrm {d} x^{2}}<0\,\!}$

${\displaystyle f^{\star }(p)=g(p)\ p-f(g(p))\,\!}$

1. 找出导函數 ${\displaystyle p={\frac {\mathrm {d} f}{\mathrm {d} x}}\,\!}$
2. 計算导函數 ${\displaystyle p={\frac {\mathrm {d} f}{\mathrm {d} x}}\,\!}$  的反函數 ${\displaystyle x=g(p)\,\!}$
3. 代入 ${\displaystyle F(x)\,\!}$  方程式來求得新函數 ${\displaystyle f^{\star }(p)=g(p)\ p-f(g(p))\,\!}$

### 反函數式定義

${\displaystyle {\frac {\mathrm {d} }{\mathrm {d} x}}f(x)=\left({\frac {\mathrm {d} }{\mathrm {d} p}}f^{*}\right)^{-1}(x)\,\!}$

${\displaystyle {\frac {\mathrm {d} }{\mathrm {d} p}}f^{*}(p)=\left({\frac {\mathrm {d} }{\mathrm {d} x}}f\right)^{-1}(p)\,\!}$

${\displaystyle f\,\!}$ ${\displaystyle f^{\star }\,\!}$  互相為彼此的勒壤得轉換。

${\displaystyle {\mathrm {d} f(x) \over \mathrm {d} x}=p\,\!}$
${\displaystyle {\mathrm {d} f^{\star }(p) \over \mathrm {d} p}=x\,\!}$

${\displaystyle {\frac {\mathrm {d} }{\mathrm {d} p}}(xp-f(x))=x+p{\frac {\mathrm {d} x}{\mathrm {d} p}}-{\frac {\mathrm {d} f}{\mathrm {d} x}}{\frac {\mathrm {d} x}{\mathrm {d} p}}=x={\mathrm {d} f^{\star }(p) \over \mathrm {d} p}=x\,\!}$

${\displaystyle f^{\star }(p)=xp-f(x)=g(p)\ p-f(g(p))\,\!}$

${\displaystyle f^{\star }(p)=f(x)-xp\,\!}$

## 數學性質

### 標度性質

${\displaystyle f(x)=a\cdot g(x)\rightarrow f^{\star }(p)=a\cdot g^{\star }\left({\frac {p}{a}}\right)\,\!}$
${\displaystyle f(x)=g(a\cdot x)\rightarrow f^{\star }(p)=g^{\star }\left({\frac {p}{a}}\right)\,\!}$

${\displaystyle {\frac {1}{r}}+{\frac {1}{s}}=1\,\!}$

### 平移性質

${\displaystyle f(x)=g(x)+b\rightarrow f^{\star }(p)=g^{\star }(p)-b\,\!}$
${\displaystyle f(x)=g(x+y)\rightarrow f^{\star }(p)=g^{\star }(p)-p\cdot y\,\!}$

### 反演性質

${\displaystyle f(x)=g^{-1}(x)\rightarrow f^{\star }(p)=-p\cdot g^{\star }\left({\frac {1}{p}}\right)\,\!}$

### 線形變換性質

${\displaystyle A\,\!}$  成為一個從 ${\displaystyle R^{n}\,\!}$ ${\displaystyle R^{m}\,\!}$  的線形變換。對於任何定義域為 ${\displaystyle R^{n}\,\!}$  的凸函數 ${\displaystyle f\,\!}$  ，必有

${\displaystyle \left(Af\right)^{\star }=f^{\star }A^{\star }\,\!}$

${\displaystyle \left\langle Ax,y^{\star }\right\rangle =\left\langle x,A^{\star }y^{\star }\right\rangle \,\!}$

## 例子

### 例一

${\displaystyle f(x)=e^{x}}$

${\displaystyle f^{*}(p)=p(\ln p-1)}$

## 應用

### 熱力學

${\displaystyle U=U(S,\ V,\ \{N_{i}\})\,\!}$

${\displaystyle H(S,\ P,\ \{N_{i}\})=U+PV\,\!}$
${\displaystyle P=-\left({\frac {\partial U}{\partial V}}\right)_{S,\ \{N_{i}\}}\,\!}$

${\displaystyle G(T,\ P,\ \{N_{i}\})=H-TS\,\!}$
${\displaystyle T=\left({\frac {\partial H}{\partial S}}\right)_{P,\ \{N_{i}\}}\,\!}$

${\displaystyle A(T,\ V,\ \{N_{i}\})=U-TS\,\!}$
${\displaystyle T=\left({\frac {\partial U}{\partial S}}\right)_{V,\ \{N_{i}\}}\,\!}$

### 古典力學(哈密頓力學)

${\displaystyle \mathbf {p} ={\frac {\partial {\mathcal {L}}}{\partial {\dot {\mathbf {q} }}}}\,\!}$
${\displaystyle {\mathcal {H}}(\mathbf {q} ,\ \mathbf {p} ,\ t)=\mathbf {p} \cdot {\dot {\mathbf {q} }}-{\mathcal {L}}(\mathbf {q} ,\ {\dot {\mathbf {q} }}(\mathbf {q} ,\ \mathbf {p} ,\ t),\ t)\,\!}$

### 正則變換

${\displaystyle {\frac {\partial {\mathcal {K}}}{\partial \mathbf {P} }}={\dot {\mathbf {Q} }}\,\!\,\!}$
${\displaystyle {\frac {\partial {\mathcal {K}}}{\partial \mathbf {Q} }}=-{\dot {\mathbf {P} }}\,\!\,\!}$
${\displaystyle \mathbf {p} \cdot {\dot {\mathbf {q} }}-{\mathcal {H}}=\mathbf {P} \cdot {\dot {\mathbf {Q} }}-{\mathcal {K}}+{\frac {dG}{dt}}\,\!}$

## 參考文獻

• Arnold, Vladimir. Mathematical Methods of Classical Mechanics (second edition). Springer. 1989. ISBN 0-387-96890-3.
• Rockafellar, Ralph Tyrell. Convex Analysis. Princeton University Press. 1996. ISBN 0-691-01586-4.