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1 files changed, 21 insertions, 21 deletions

diff --git a/source/know/concept/legendre-transform/index.md b/source/know/concept/legendre-transform/index.md
index 354b7ae..51c003e 100644
--- a/source/know/concept/legendre-transform/index.md
+++ b/source/know/concept/legendre-transform/index.md
@@ -8,32 +8,32 @@ categories:
 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,
+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.
+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 $-C$:
+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 $-C$) for all $x_0 \in [a, b]$,
-where $C$ corresponds to the tangent line at $x = x_0$. This yields:
+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:
 
 $$\begin{aligned}
     L(f'(x)) = f'(x) \: x - f(x)
 \end{aligned}$$
 
-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
+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:
 
 $$\begin{aligned}
@@ -43,12 +43,12 @@ $$\begin{aligned}
 \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, 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
+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:
 
@@ -59,7 +59,7 @@ $$\begin{aligned}
 \end{aligned}$$
 
 Furthermore, Legendre transformation is an *involution*, meaning it is
-its own inverse. Let $g(L')$ be the Legendre transform of $L(f')$:
+its own inverse. Let $$g(L')$$ be the Legendre transform of $$L(f')$$:
 
 $$\begin{aligned}
     g(L') = L' \: f'(L') - L(f'(L'))
@@ -68,7 +68,7 @@ $$\begin{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
+of $$f(x)$$ means that $$f''(x) > 0$$ for all $$x \in [a, b]$$, which yields
 the following proof:
 
 $$\begin{aligned}