Convex dual
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% .. sets
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\newcommand{\gttlog}{\slog\gtt}
\newcommand{\gtttlog}{\slog\gttt}$
Given a convex function $f$, its convex conjugate $f^*(a)$ gives the (negative) intercept of the tangent to $f$ of slope $a$:
\[
f^*(a) = \sup_t \p{at-f(t)}
\]
#figure
This gives an alternate way to describe $f$, not as a series of points $(t, f(t))$ drawing its curve, but as series of lines $\setco{at-f^*(a)}{t \in \R}$ hugging its curve.
Duality
Put another way, $f^*$ is minimized under the constraint that
\[
f(t) + f^*(a) \ge at,\tag{$*$}
\]
which shows by symmetry that $f$ is also $f^*$’s convex conjugate: indeed, there is also no way to make $f$ smaller without lowering one of its tangents, which would increase $f^*(a)$, so $f$ is also minimized under constraint $(*)$.
Derivatives
For a moment, suppose that $f$ is derivable and strictly convex, which means $f'(t)$ exists and is smoothly increasing.
Consider the tangent of slope $a$, and suppose it touches $f$ at $t$ (which means $\f{\d f}{\d t}(t) = a$). Then intuitively, if we slightly increase the slope $a$, gliding the tangent further along $f$, the intercept will decrease at a rate proportional to $t$ since it is a distance $t$ away, meaning that $\f{\d f^*}{\d a}(a) = t$. That is, the derivatives of $f$ and $f^*$ are inverses of each other, and in particular this shows that $f^*$ is also convex.
#figure
Slightly more formally, since $f$ is smooth, we have $f(t') \approx at' - f^*(a)$ for $t'$ close to $t$, so for a small slope increase $\eps$, if we take the point $t'$ such that $\f{\d f}{\d t}(t') = a+\eps$, we have
\[
\al{
f^*(a+\eps)
&= (a+\eps)t' - f(t')\\
&\approx f^*(a) + \eps t'\\
&\approx f^*(a) + \eps t,
}
\]
which shows that $\f{\d f^*}{\d a}(a) = t$ as desired.