लजान्द्र रूपान्तर

गणित में किसी वास्तविक मान वाले, तथा सभी बिन्दुओं पर अवकलनीय फलन f तथा g में निम्नलिखित सम्बन्ध हो तो g को f का लजान्द्र रूपान्तर (LegendreTransform) कहा जाता है। इस रूपान्तर का नाम फ्रांसीसी गणितज्ञ आद्रियें मारि लजान्द्र (Adrien-Marie Legendre) के नाम पर पड़ा है।

${\displaystyle Df=\left(Dg\right)^{-1}}$

जहाँ D , अवकलज (डिफरेंशियल) का प्रतीक है तथा दाहिनी ओर आने वाला -1 , प्रतिलोम फलन को सूचित कर रहा है। यह आसानी से दिखाया जा सकता है कि g , f का लजान्द्र रूपान्तर हो तो f, g का लजान्द्र रूपान्तर होगा

उदाहरण के लिये, फलन ${\displaystyle f(x)={\tfrac {x^{2}}{2}}}$ तथा फलन ${\displaystyle g(p)=-{\tfrac {p^{2}}{2}}}$ एक दूसरे के लजान्द्र रूपान्तर हैं।

एक विशेष स्थिति में, यदि फलन f एक उत्तल फलन (कान्वेक्स फंक्शन) हो तो इसका लजान्द्र रूपान्तर ƒ^* निम्नलिखित सम्बन्ध द्वारा अभिव्यक्त किया जा सकता है-

${\displaystyle f^{\star }(p)=\sup _{x}{\bigl (}px-f(x){\bigr )}.}$

• (१) ऊष्मागतिकी में - ऊष्मागतिक विभव प्राप्त करने के लिये,
• (२) क्लासिकल यांत्रिकी में -- to derive the Hamiltonian formalism out of the Lagrangian formalism
• (३) सांख्यिकीय यांत्रिकी में,
• (३) अनेक चरों वाले अवकल समीकरणों का हल प्राप्त करने के लिये।

Let f(x) = cx² defined on R, where c > 0 is a fixed constant.

For x* fixed, the function x*x – f(x) = x*x – cx² of x has the first derivative x* – 2cx and second derivative −2c; there is one stationary point at x = x*/2c, which is always a maximum. Thus, I* = R and

${\displaystyle f^{*}(x^{*})={\frac {{x^{*}}^{2}}{4c}}}$

Clearly,

${\displaystyle f^{**}(x)={\frac {1}{4(1/4c)}}x^{2}=cx^{2},}$

namely f ** = f.

Let f(x) = x² for x ∈ I = [2, 3].

For x* fixed, x*x − f(x) is continuous on I compact, hence it always takes a finite maximum on it; it follows that I* = R. The stationary point at x = x*/2 is in the domain [2, 3] if and only if 4 ≤ x* ≤ 6, otherwise the maximum is taken either at x = 2, or x = 3. It follows that

${\displaystyle f^{*}(x^{*})={\begin{cases}2x^{*}-4,\quad &x^{*}<4\\{\frac {{x^{*}}^{2}}{4}},&4\leqslant x^{*}\leqslant 6,\\3x^{*}-9,&x^{*}>6\end{cases}}}$ .

The function f(x) = cx is convex, for every x (strict convexity is not required for the Legendre transformation to be well defined). Clearly x*x − f(x) = (x* − c)x is never bounded from above as a function of x, unless x* − c = 0. Hence f* is defined on I* = {c} and f*(c) = 0.

One may check involutivity: of course x*x − f*(x*) is always bounded as a function of x* ∈ {c}, hence I ** = R. Then, for all x one has

${\displaystyle \sup _{x^{*}\in \{c\}}(xx^{*}-f^{*}(x^{*}))=xc,}$

and hence f **(x) = cx = f(x).

Let

${\displaystyle f(x)=\langle x,Ax\rangle +c}$

be defined on X = Rⁿ, where A is a real, positive definite matrix. Then f is convex, and

${\displaystyle \langle p,x\rangle -f(x)=\langle p,x\rangle -\langle x,Ax\rangle -c,}$

has gradient p − 2Ax and Hessian −2A, which is negative; hence the stationary point x = A⁻¹p/2 is a maximum. We have X* = Rⁿ, and

${\displaystyle f^{*}(p)={\frac {1}{4}}\langle p,A^{-1}p\rangle -c}$ .

${\displaystyle f(x)}$	${\displaystyle \operatorname {dom} f}$	${\displaystyle f^{\star }(x^{\star })}$	${\displaystyle \operatorname {dom} f^{\star }}$	शर्तें
${\displaystyle af(x)}$	${\displaystyle \operatorname {dom} f}$	${\displaystyle af^{\star }(x^{\star }/a)}$	${\displaystyle a\cdot \operatorname {dom} f^{\star }}$	${\displaystyle a>0}$
${\displaystyle f(ax)}$	${\displaystyle a^{-1}\cdot \operatorname {dom} f}$	${\displaystyle f^{\star }(x^{\star }/a)}$	${\displaystyle a\cdot \operatorname {dom} f^{\star }}$	${\displaystyle a>0}$
${\displaystyle f(x)+a}$	${\displaystyle \operatorname {dom} f}$	${\displaystyle f^{\star }(x^{\star })-a}$	${\displaystyle \operatorname {dom} f^{\star }}$	${\displaystyle a\in \mathbb {R} }$
${\displaystyle f(x-a)}$	${\displaystyle a+\operatorname {dom} f}$	${\displaystyle f^{\star }(x^{\star })+ax^{\star }}$	${\displaystyle \operatorname {dom} f^{\star }}$	${\displaystyle a\in \mathbb {R} }$
${\displaystyle f(x)+ax}$	${\displaystyle \operatorname {dom} f}$	${\displaystyle f^{\star }(x^{\star }-a)}$	${\displaystyle a+\operatorname {dom} f^{\star }}$	${\displaystyle a\in \mathbb {R} }$
${\displaystyle f(x)+g(x)}$	${\displaystyle \operatorname {dom} f\cap \operatorname {dom} g}$	${\displaystyle (f^{\star }\star _{\text{inf}}g^{\star })(x^{\star })}$	${\displaystyle \operatorname {dom} f^{\star }+\operatorname {dom} g^{\star }}$	${\displaystyle (f\star _{\text{inf}}g)(x)=\inf _{y}\{f(x-y)+g(y)\}}$
${\displaystyle (f\star _{\text{inf}}g)(x)}$	${\displaystyle \operatorname {dom} f+\operatorname {dom} g}$	${\displaystyle f^{\star }(x^{\star })+g^{\star }(x^{\star })}$	${\displaystyle \operatorname {dom} f^{\star }\cap \operatorname {dom} g^{\star }}$	${\displaystyle (f\star _{\text{inf}}g)(x)=\inf _{y}\{f(x-y)+g(y)\}}$
${\displaystyle ax+b}$	${\displaystyle \mathbb {R} }$	${\displaystyle -b}$	${\displaystyle \{a\}}$
${\displaystyle |x|^{p}/p}$	${\displaystyle \mathbb {R} }$	${\displaystyle |x^{\star }|^{p^{\star }}/p^{\star }}$	${\displaystyle \mathbb {R} }$	${\displaystyle 1/p+1/p^{\star }=1}$, ${\displaystyle p>1}$
${\displaystyle -x^{p}/p}$	${\displaystyle [0,\infty )}$	${\displaystyle -|x^{\star }|^{p^{\star }}/p^{\star }}$	${\displaystyle (-\infty ,0]}$	${\displaystyle 1/p+1/p^{\star }=1}$, ${\displaystyle p<1}$
${\displaystyle \exp(x)}$	${\displaystyle \mathbb {R} }$	${\displaystyle x^{\star }(\ln(x^{\star })-1)}$	${\displaystyle \mathbb {R} ^{+}}$
${\displaystyle x\ln(x)}$	${\displaystyle \mathbb {R} ^{+}}$	${\displaystyle \exp(x-1)}$	${\displaystyle \mathbb {R} }$
${\displaystyle -1/2-\ln x}$	${\displaystyle \mathbb {R} ^{+}}$	${\displaystyle -1/2-\ln |x^{\star }|}$	${\displaystyle \mathbb {R} ^{-}}$
${\displaystyle x\exp(x+1)}$	${\displaystyle \mathbb {R} }$	${\displaystyle x^{\star }(W(x^{\star })-1)^{2}/W(x^{\star })}$	${\displaystyle [-1/e,\infty )}$	लैम्बर्ट का W फलन

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