= Fenchel–Young inequality
{c}
{title2=$f(x)+f^*(p)\geq p\cdot x$}
The <Fenchel–Young inequality> says that every input and every slope obey
$$
f(x)+f^*(p)\geq p\cdot x,\qquad x,p\in\mathbb R^n,
$$
where $f$ is any <proper extended-real function> and $f^*$ is its <convex conjugate>. No <convexity> assumption on $f$ is needed for this inequality: the defining <supremum> is at least the particular value $p\cdot x-f(x)$. If either term on the left is $+\infty$, the inequality still holds, but cannot be an equality because the <dot product> on the right is finite.
For a <proper convex function>, equality is equivalent to the <subgradient> condition
$$
f(x)+f^*(p)=p\cdot x\quad\Longleftrightarrow\quad p\in\partial f(x).
$$
To see this, with $f(x)$ finite, equality says that $x$ attains the <supremum> defining $f^*(p)$, so $p\cdot z-f(z)\leq p\cdot x-f(x)$ for every $z$. Rearranging gives the <subgradient inequality> $f(z)\geq f(x)+p\cdot(z-x)$. Geometrically, the <affine minorant> $\ell_p(z)=p\cdot z-f^*(p)$ touches $f$ at $x$.
If the <proper convex function> is also closed, meaning its <epigraph> is a <closed set>, or equivalently it is <lower semicontinuous>, then
$$
p\in\partial f(x)\quad\Longleftrightarrow\quad x\in\partial f^*(p).
$$
The forward implication holds without closedness; the reverse one requires the additional hypothesis in general. Applying the equality criterion to $f^*$ relates $x\in\partial f^*(p)$ to $f^{**}(x)+f^*(p)=p\cdot x$, and the <Fenchel-Moreau theorem> supplies $f^{**}=f$ when $f$ is closed. See https://ocw.mit.edu/courses/6-253-convex-analysis-and-optimization-spring-2012/db7bde2cd43e8dbaba11f6911311b87c_MIT6_253S12_summary.pdf#page=46[Bertsekas, Convex Optimization Theory: A Summary, §5.4.1, Proposition 5.4.3] for precisely this distinction.
A concrete failure of the reverse implication occurs for the <indicator functional of a constraint set> $f$ of the open interval $(-1,1)$. Its <convex conjugate> is $f^*(p)=|p|$: the endpoint gives the <supremum>, even though it is excluded from the interval. At $(x,p)=(1,1)$, $x\in\partial f^*(p)=\{1\}$, but $f(1)=+\infty$, so equality fails and $\partial f(1)$ is empty. Closing the interval changes the endpoint value and restores equality there.
For the <convex quadratic function> $f(x)=x^2/2$, the <Fenchel–Young inequality> is $x^2/2+p^2/2\geq px$, and equality holds exactly when $x=p$. For the <absolute value> $f(x)=|x|$, finite equality requires $p=\operatorname{sgn}(x)$ at $x\ne0$, where $\operatorname{sgn}$ is the <sign function>; at $x=0$ every $p\in[-1,1]$ gives equality. These are exactly the <subgradients> of the <absolute value>, including its entire interval of slopes at the corner.
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