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diff --git a/source/know/concept/legendre-transform/index.md b/source/know/concept/legendre-transform/index.md
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+++ b/source/know/concept/legendre-transform/index.md
@@ -9,76 +9,90 @@ layout: "concept"
 ---
 
 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,
-analogously to other transforms (e.g. [Fourier](/know/concept/fourier-transform/)),
-that $$L(f')$$ contains the same information as $$f(x)$$, just in a different form.
+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 that $$L(f')$$ contains the same information as $$f(x)$$,
+just in a different form,
+analogously to e.g. the [Fourier transform](/know/concept/fourier-transform/).
 
-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 $$-C$$:
+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 must have slope $$f'(x_0)$$, and intersects the $$y$$-axis at $$-C$$:
 
 $$\begin{aligned}
-    y(x) = f'(x_0) (x - x_0) + f(x_0) = f'(x_0) x - C
+    y(x)
+    = f'(x_0) (x - x_0) + f(x_0)
+    = f'(x_0) \: x - C
 \end{aligned}$$
 
-The Legendre transform $$L(f')$$ is defined such that $$L(f'(x_0)) = C$$
-(or sometimes $$-C$$) for all $$x_0 \in [a, b]$$,
-where $$C$$ corresponds to the tangent line at $$x = x_0$$. This yields:
+Where $$C \equiv f'(x_0) \: x_0 - f(x_0)$$.
+We now define the Legendre transform $$L(f')$$ such that
+for all $$x_0 \in [a, b]$$ we have $$L(f'(x_0)) = C$$
+(some authors use $$-C$$ instead).
+Renaming $$x_0$$ to $$x$$:
 
 $$\begin{aligned}
-    L(f'(x)) = f'(x) \: x - f(x)
+    L(f'(x))
+    = f'(x) \: x - f(x)
 \end{aligned}$$
 
-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:
+We want this function to depend only on the derivative $$f'$$,
+but currently $$x$$ still appears here as a variable.
+We fix this 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:
 
 $$\begin{aligned}
     \boxed{
-        L(f') = f' \: x(f') - f(x(f'))
+        L(f')
+        = f' \: x(f') - f(x(f'))
     }
 \end{aligned}$$
 
 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.
+the invertibility of $$f'(x)$$ in the target interval $$[a,b]$$,
+which can only be true if $$f(x)$$ is either convex or concave,
+meaning 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:
+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:
 
 $$\begin{aligned}
     \boxed{
-        \dv{L}{f'} = \dv{x}{f'} \: f' + x(f') - \dv{f}{x} \dv{x}{f'} = x(f')
+        \dv{L}{f'}
+        = f' \dv{x}{f'} + x(f') - f' \dv{x}{f'}
+        = x(f')
     }
 \end{aligned}$$
 
-Furthermore, Legendre transformation is an *involution*, meaning it is
-its own inverse. Let $$g(L')$$ be the Legendre transform of $$L(f')$$:
+Furthermore, Legendre transformation is an *involution*,
+meaning it is its own inverse.
+To show this, let $$g(L')$$ be the Legendre transform of $$L(f')$$:
 
 $$\begin{aligned}
-    g(L') = L' \: f'(L') - L(f'(L'))
-    = x(f') \: f' - f' \: x(f') + f(x(f')) = f(x)
+    g(L')
+    = L' \: f'(L') - L(f'(L'))
+    = x(f') \: f' - f' \: x(f') + f(x(f'))
+    = f(x)
 \end{aligned}$$
 
-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:
+Moreover, a Legendre transform is always invertible,
+because the transform of a convex function is itself convex.
+Convexity of $$f(x)$$ means that $$f''(x) > 0$$ for all $$x \in [a, b]$$,
+so a proof is:
 
 $$\begin{aligned}
     L''(f')
+    = \dv{}{f'} \Big( \dv{L}{f'} \Big)
     = \dv{x(f')}{f'}
     = \dv{x}{f'(x)}
     = \frac{1}{f''(x)}
     > 0
 \end{aligned}$$
 
+And an analogous proof exists for concave functions where $$f''(x) < 0$$.
+
 Legendre transformation is important in physics,
 since it connects [Lagrangian](/know/concept/lagrangian-mechanics/)
 and [Hamiltonian](/know/concept/hamiltonian-mechanics/) mechanics to each other.