Add "Legendre transform"

3 files changed, 92 insertions, 11 deletions

diff --git a/latex/know/concept/fourier-transform/source.md b/latex/know/concept/fourier-transform/source.md
index 3e25980..10cb021 100644
--- a/latex/know/concept/fourier-transform/source.md
+++ b/latex/know/concept/fourier-transform/source.md
@@ -51,8 +51,8 @@ $$\begin{aligned}
 \end{aligned}$$
 
 But that still gives a lot of freedom. The exact choices of $A$ and $B$
-are generally motivated by the convolution theorem and Parseval's
-theorem.
+are generally motivated by the [convolution theorem](/know/concept/convolution-theorem/)
+and [Parseval's theorem](/know/concept/parsevals-theorem/).
 
 The choice of $|s|$ depends on whether the frequency variable $k$
 represents the angular ($|s| = 1$) or the physical ($|s| = 2\pi$)
diff --git a/latex/know/concept/legendre-transform/source.md b/latex/know/concept/legendre-transform/source.md
new file mode 100644
index 0000000..954b6fc
--- /dev/null
+++ b/latex/know/concept/legendre-transform/source.md
@@ -0,0 +1,79 @@
+% Legendre transform
+
+
+# Legendre transform
+
+The **Legendre transform** of a function $f(x)$ is a new function $L(f')$,
+which depends only on the derivative $f'(x)$ of $f(x)$, and from which
+the original function $f(x)$ can be reconstructed. The point is, just
+like other transforms (e.g. Fourier), that $L(f')$ contains the same
+information as $f(x)$, just in a different form.
+
+Let us choose an arbitrary point $x_0 \in [a, b]$ in the domain of
+$f(x)$. Consider a line $y(x)$ tangent to $f(x)$ at $x = x_0$, which has
+a slope $f'(x_0)$ and intersects the $y$-axis at $-C$:
+
+$$\begin{aligned}
+    y(x) = f'(x_0) (x - x_0) + f(x_0) = f'(x_0) x - C
+\end{aligned}$$
+
+The Legendre transform $L(f')$ is defined such that $L(f'(x_0)) = C$ (or
+sometimes $-C$ instead) for all $x_0 \in [a, b]$, where $C$ is the
+constant corresponding to the tangent line at $x = x_0$. This yields:
+
+$$\begin{aligned}
+    L(f'(x)) = f'(x) \: x - f(x)
+\end{aligned}$$
+
+We want this function to depend only on the derivative $f'$, but
+currently $x$ still appears here as a variable. We fix that problem in
+the easiest possible way: by assuming that $f'(x)$ is invertible for all
+$x \in [a, b]$. If $x(f')$ is the inverse of $f'(x)$, then $L(f')$ is
+given by:
+
+$$\begin{aligned}
+    \boxed{
+        L(f') = f' \: x(f') - f(x(f'))
+    }
+\end{aligned}$$
+
+The only requirement for the existence of the Legendre transform is thus
+the invertibility of $f'(x)$ in the target interval $[a,b]$, which can
+only be true if $f(x)$ is either convex or concave, i.e. its derivative
+$f'(x)$ is monotonic.
+
+Crucially, the derivative of $L(f')$ with respect to $f'$ is simply
+$x(f')$. In other words, the roles of $f'$ and $x$ are switched by the
+transformation: the coordinate becomes the derivative and vice versa.
+This is demonstrated here:
+
+$$\begin{aligned}
+    \boxed{
+        \dv{L}{f'} = \dv{x}{f'} \: f' + x(f') - \dv{f}{x} \dv{x}{f'} = x(f')
+    }
+\end{aligned}$$
+
+Furthermore, Legendre transformation is an *involution*, meaning it is
+its own inverse. Let $g(L')$ be the Legendre transform of $L(f')$:
+
+$$\begin{aligned}
+    g(L') = L' \: f'(L') - L(f'(L'))
+    = x(f') \: f' - f' \: x(f') + f(x(f')) = f(x)
+\end{aligned}$$
+
+Moreover, the inverse of a (forward) transform always exists, because
+the Legendre transform of a convex function is itself convex. Convexity
+of $f(x)$ means that $f''(x) > 0$ for all $x \in [a, b]$, which yields
+the following proof:
+
+$$\begin{aligned}
+    L''(f')
+    = \dv{x(f')}{f'}
+    = \dv{x}{f'(x)}
+    = \frac{1}{f''(x)}
+    > 0
+\end{aligned}$$
+
+Legendre transformation is important in physics,
+since it connects Lagrangian and Hamiltonian mechanics to each other.
+It is also used to convert between thermodynamic potentials.
diff --git a/latex/know/concept/parsevals-theorem/source.md b/latex/know/concept/parsevals-theorem/source.md
index 41af734..9f406da 100644
--- a/latex/know/concept/parsevals-theorem/source.md
+++ b/latex/know/concept/parsevals-theorem/source.md
@@ -3,23 +3,24 @@
 
 # Parseval's theorem
 
-**Parseval's theorem** relates the inner product of two functions to the
-inner product of their [Fourier transforms](/know/concept/fourier-transform/).
+**Parseval's theorem** relates the inner product of two functions $f(x)$ and $g(x)$ to the
+inner product of their [Fourier transforms](/know/concept/fourier-transform/)
+$\tilde{f}(k)$ and $\tilde{g}(k)$.
 There are two equivalent ways of stating it,
 where $A$, $B$, and $s$ are constants from the Fourier transform's definition:
 
 $$\begin{aligned}
     \boxed{
-        \braket{f(x)}{g(x)} = \frac{2 \pi B^2}{|s|} \braket{\tilde{f}(k)}{\tilde{g}(k)}
+        \braket{f(x)}{g(x)} = \frac{2 \pi B^2}{|s|} \braket*{\tilde{f}(k)}{\tilde{g}(k)}
     }
     \\
     \boxed{
-        \braket{\tilde{f}(k)}{\tilde{g}(k)} = \frac{2 \pi A^2}{|s|} \braket{f(x)}{g(x)}
+        \braket*{\tilde{f}(k)}{\tilde{g}(k)} = \frac{2 \pi A^2}{|s|} \braket{f(x)}{g(x)}
     }
 \end{aligned}$$
 
-For this reason many physicists like to define their Fourier transform
-with $|s| = 1$ and $A = B = 1 / \sqrt{2\pi}$, because then the FT nicely
+For this reason, physicists like to define their Fourier transform
+with $A = B = 1 / \sqrt{2\pi}$ and $|s| = 1$, because then the FT nicely
 conserves the total probability (quantum mechanics) or the total energy
 (optics).
 
@@ -40,14 +41,15 @@ $$\begin{aligned}
     &= 2 \pi B^2 \iint \tilde{f}^*(k_1) \: \tilde{g}(k) \: \delta(s (k_1 - k)) \dd{k_1} \dd{k}
     \\
     &= \frac{2 \pi B^2}{|s|} \int_{-\infty}^\infty \tilde{f}^*(k) \: \tilde{g}(k) \dd{k}
-    = \frac{2 \pi B^2}{|s|} \braket{\tilde{f}}{\tilde{g}}
+    = \frac{2 \pi B^2}{|s|} \braket*{\tilde{f}}{\tilde{g}}
 \end{aligned}$$
 
 Where $\delta(k)$ is the [Dirac delta function](/know/concept/dirac-delta-function/).
-We can just as well do it in the opposite direction, with equivalent results:
+Note that we can just as well do it in the opposite direction,
+which yields an equivalent result:
 
 $$\begin{aligned}
-    \braket{\tilde{f}}{\tilde{g}}
+    \braket*{\tilde{f}}{\tilde{g}}
     &= \int_{-\infty}^\infty \big( \hat{\mathcal{F}}\{f(x)\}\big)^* \: \hat{\mathcal{F}}\{g(x)\} \dd{k}
     \\
     &= A^2 \int