Relation of Polar and Fenchel transforms

Polar and Fenchel Transforms of a Gauge

Let $X, Y$ be a dual pair with bilinear pairing $\langle x, y \rangle$.
Let $C \subset X$ be a nonempty convex set containing $0$.

We define two main objects:

• The gauge function (Minkowski functional):
  $$\gamma_C(x) := \inf\{t > 0 : x \in tC\}.$$

• The indicator function:
  $$\delta_{C^\circ}(y) = \begin{cases} 0, & y \in C^\circ, \\ +\infty, & y \notin C^\circ. \end{cases}$$

We also define three fundamental transformations:

• The polar transform of a set:
  $$C^\circ := \{ y \in Y : \langle x, y \rangle \le 1 ; \forall x \in C\}.$$

• The polar transform of a gauge:
  $$k^\circ(y) := \inf\{ \mu \ge 0 : \langle x, y \rangle \le \mu k(x) ; \forall x \in X\}.$$

• The Fenchel transform (Fenchel conjugate) of a convex function:
  $$f^*(y) := \sup\{ \langle x, y \rangle - f(x) : x \in X \}.$$

In this note, we prove the following results:

1. Polar transform of a gauge: $(\gamma_C)^\circ = \gamma_{C^\circ}$.

2. Fenchel transform of a gauge: $(\gamma_C)^* = \delta_{C^\circ}$.

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NOTE. There is a more general definition of the polar transform on the class of nonnegative convex functions vanishing at $0$:

$$k^\circ(y) := \inf\{ \mu \ge 0 : \langle x, y \rangle \le 1 + \mu k(x) ; \forall x \in X \}.$$

On this class of functions, the following diagram commutes (see Convex Analysis - Rockafellar - Corollary 15.5.1 - P.138):

$$\gamma_C \dashrightarrow \gamma_{C^\circ} \\ \downarrow \qquad\quad \downarrow \\ \delta_{C^\circ} \dashrightarrow \delta_{C}$$

where $\dashrightarrow$ denotes the Polar transform and $\downarrow$ denotes the Fenchel transform.

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1. Polar of a Gauge

Fix $\mu > 0$. The condition

$$\langle x, y \rangle \le \mu, \gamma_C(x) \quad \forall x$$

is equivalent to checking it only on $C$.

If $x \in C$, then $\gamma_C(x) \le 1$, hence

$$\langle x, y \rangle \le \mu \quad \forall x \in C.$$

This is equivalent to

$$\langle x, \mu^{-1} y \rangle \le 1 \quad \forall x \in C,$$

i.e. $\mu^{-1} y \in C^\circ$, or equivalently $y \in \mu C^\circ$.

Therefore

$$(\gamma_C)^\circ(y) =\inf\{ \mu \ge 0 : y \in \mu C^\circ \} =\gamma_{C^\circ}(y).$$

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2. Fenchel Conjugate of a Gauge

Normalize any $x$ as

$$x = t c, \quad t := \gamma_C(x), \quad c \in C.$$

Then

$$\langle x, y \rangle - \gamma_C(x) t(\langle c, y \rangle - 1).$$

Hence

$$(\gamma_C)^*(y) =\sup_{t \ge 0,; c \in C} t(\langle c, y \rangle - 1).$$

Case analysis

Case 1: $y \notin C^\circ$

Then there exists $c \in C$ such that $\langle c, y \rangle > 1$.

Letting $t \to +\infty$ gives $+\infty$.

Case 2: $y \in C^\circ$

Then

$$\langle c, y \rangle \le 1 \quad \forall c \in C.$$

Hence $t(\langle c, y \rangle - 1) \le 0$ for all $t \ge 0$.

Taking $t = 0$ yields value $0$, so the supremum equals $0$.

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Conclusion

We obtain the compact formula

$$(\gamma_C)^*(y) = \begin{cases} 0, & y \in C^\circ, \\ +\infty, & y \notin C^\circ, \end{cases}$$

or equivalently

$$(\gamma_C)^* = \delta_{C^\circ}.$$