 Fenchel–Young inequality

The Fenchel–Young inequality says that every input and every slope obeywhere $f$ is any proper extended-real function and $f^{\ast }$ 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 conditionTo see this, with $f(x)$ finite, equality says that $x$ attains the supremum defining $f^{\ast }(p)$, so $p\cdot z-f(z)\le p\cdot x-f(x)$ for every $z$. Rearranging gives the subgradient inequality $f(z)\ge f(x)+p\cdot (z-x)$. Geometrically, the affine minorant ${\ell }_p(z)=p\cdot z-f^{\ast }(p)$ 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, thenThe forward implication holds without closedness; the reverse one requires the additional hypothesis in general. Applying the equality criterion to $f^{\ast }$ relates $x\in \partial f^{\ast }(p)$ to $f^{\ast \ast }(x)+f^{\ast }(p)=p\cdot x$, and the Fenchel-Moreau theorem supplies $f^{\ast \ast }=f$ when $f$ is closed. See 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 $f^{\ast }(p)=|p|$: the endpoint gives the supremum, even though it is excluded from the interval. At $(x,p)=(1,1)$, $x\in \partial f^{\ast }(p)=\{1\}$, 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 $x^2/2+p^2/2\ge px$, and equality holds exactly when $x=p$. For the absolute value $f(x)=|x|$, finite equality requires $p=\mathrm{sgn}(x)$ at $x\ne 0$, where $\mathrm{sgn}$ 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.