Smax = 2
n = 100
np = 3
f = function(s) { return s*s + 1 }
// Legendre transform of f
g = function(s) { return s*s/4 - 1 }
points = {
var points = new Array()
var idx = 0
for( var i = -n ; i <= n; i++) {
var s = Smax*i/n
points[idx++] = { point: s, value: f(s) }
}
return points
}
legendre = {
var points = new Array()
var idx = 0
for( var i = -np; i <= np; i++) {
var p = Smax*i/np
var gp = g(p)
points[idx++] = { x: -Smax, y: -Smax*p - gp, index: i }
points[idx++] = { x: Smax, y: Smax*p - gp, index: i }
}
return points
}46 Duality
46.1 Basic Intuition
We start by stating a basic fact
Characterization of convex functions
Any convex function can be represented as a pointwise supremum of convex functions. More formally, for any convex \(f \colon \reals^n \to \reals\), there exists a countable index set \(I\) and a family of \(\{a_i \in \reals^n\}_{i \in I}\) and \(\{b_i \in \reals\}_{i \in I}\) such that \[ f(x) = \sup_{i \in I} \{ a_i^\TRANS x + b_i \} \]
An illustration of this fact is shown in Figure 46.1.
Why countable index set?
From the figure you can convince yourself of the result for an uncountable index set \(\bar I\). Now consider the set \(\bar S = \{a_i, b_i\}_{i \in \bar I} \subset \reals^{n+1}\) (for uncountable \(\bar I\)). Since \(\reals^{n+1}\) is separable, let \(S = \{a_i, b_i\}_{i \in I}\) be a countable dense subset of \(\bar S\) (where \(I\) is countable). Since the supremum over any set is the same as the supremum of its dense subset, we can replace \(\bar I\) by \(I\).
The Lengendre-Fenchel transform is a compact way of representing this basic fact. To fix ideas, consider a twice differentiable and strictly convex function \(f \colon \reals \to \reals\). Arbitrarily fix a point \(x^∘\) and consider the tangent \(τ(x)\) to the plot of \(f(⋅)\) at \(x^∘\). This tangent has a slope \(p = f'(x^∘)\). Let \(g\) denote the negative \(y\)-intercept of the tangent (using the negative \(y\)-intercept is just a matter of convention). Thus, the equation for the tangent is: \[ τ(x) = p x - g, \quad \forall x \in \reals \] at at the point \(x\), the tangent \(τ\) intercepts the function \(f\); thus, \(τ(x^∘) = f(x^∘)\), or equivalently: \[ f(x^∘) = p x^∘ - g \quad\hbox{or, equivalently}\quad g = p x^∘ - f(x^∘) \]
Note that the \(g\) above depends on \(x^∘\). Since \(f\) is strictly convex, \(f'\) is strictly increasing. Thus, there is a one-to-one relationship between the point \(x^∘\) and its slope \(p = f'(x^∘)\). The Lengendre-Fenchel transform is a method to define the intercept \(g\) as a function of \(p\). In particular, it is defined as \[ g(p) = \sup_{x \in \reals} \{ px - f(x) \}. \]
To understand this definition, suppose \(x^∘\) achieves the supremum in the above equation. Then, (i) \(f(x^∘) = p x^∘ - g(p)\); that is the line \(px - g(p)\) touches the function \(f\) at \(x^∘\); and (ii) at all other points \(x \neq x^∘\), \(g(p) > p x - f(x)\) or \(f(x) > px - g(p)\); that is the line \(px - g(p)\) lies below the function \(f\). Hence, \(px - g(p)\) is a :supporting line of \(f\) (and since \(f\) is differentiable, equal to its tangent).
An implication of this definition is that for differentiable and convex \(f\), the Legendre transform \(g\) of \(f\) solves \[ g(p) = p x^∘ - f(x^∘) \] where \(x^∘\) solves \(p = f'(x^∘)\), which is called the duality condition.
Example 46.1 Let \(f(x) = x^2\). For a fixed \(p\), the duality condition is \(p = 2x^∘\) or \(x^∘ = p/2\). Therefore, \[ g(p) = p x^∘ - f(x^∘) = \frac{p^2}{2} - \frac{p^2}{4} = \frac{p^2}{4}. \]
Example 46.2 Let \(f(x) = (x^α)/α\), where \(α > 1\). For a fixed \(p\), the duality condition is \(p = (x^∘)^{α-1}\) or \(x^∘ = p^{1/(p-1)}\). Therefore, \[ g(p) = p x^∘ - f(x^∘) = p^{α/(α-1)} - \frac{1}{α} p^{α/(α-1)} = \frac{α-1}{α} p^{α/(α-1)}. \] We can compactly write the above expression is \(g(p) = p^β/β\), where \(1/β = 1 - 1/α\).
Example 46.3 Let \(f(x) = x \log x + (1-x) \log (1-x)\), where the domain is \([0,1]\). This is the negative binary entropy. For a fixed \(p\), the duality condition is \[ p = f'(x^∘) = \log(x^∘) + 1 - \log(1-x^∘) - 1 % = \log(x^∘) - \log(1-x^∘) = \log \frac{x^∘}{1-x^∘} \] Thus, \(x^∘ = e^p/(1 + e^p)\). Therefore, \[ g(p) = p x^∘ - f(x^∘) = \log(1 + e^p), \] where the last step follows after some algebra.
Fenchel-Young inequality
Let \(g\) be the Legendre-Fenchel transform of \(f\). Then, by definition (changing the variable names for convenience), we have \[\begin{equation}\tag{Fenchel's inequality} xy \le f(x) + g(y). \end{equation}\] For the special case of Example 46.2, we have \[\begin{equation} \tag{Young's inequality} xy \le \frac{x^p}{p} + \frac{y^q}{q} \end{equation}\] where \(p, q > 1\) are such that \(\frac 1p + \frac 1q = 1\).
An interesting example of Fenchel-Young inequality is the following. Consider a real-valued, strictly increasing, continuous function \(h\) on \(\reals\) which satisfies \(h(0) = 0\), \(h(x) \to -∞\) as \(x \to -∞\), and \(h(x) \to ∞\) as \(x \to ∞\). Since \(h\) is continuous and strictly increasing, it has an inverse. Define \[\begin{equation}\label{eq:legendre-example} f(x) = \int_{0}^x h(s) ds \quad\hbox{and}\quad g(y) = \int_{0}^y h^{-1}(t) dt. \end{equation}\]
The graph of \(t = h(s)\) is shown in Figure 46.2, where the shaded portions represent \(f(x)\) and \(g(y)\). From the plots, we can infer Fenchel-Young’s inequality: \[ xy \le f(x) + g(y) \] with equality if and only if \(y = h(x) = \dot f(x)\). This immediately implies that \[ g(y) = \sup_{x \in \reals} \bigl\{ xy - f(x) \bigr\} \] and \[ f(x) = \sup_{y \in \reals} \bigl\{ xy - g(y) \bigr\}. \]
An alternative view
Suppose \(f\) is a convex function. Then it can be shown that its Legendre transform \(g\) is the unique function such that:
- \(f'\) and \(g'\) are inverses of each other, i.e., \(g'(f'(x)) = x\) and \(f'(g'(p)) = p\).
- It satisfies the Fenchel inequality with equality, i.e., \(f(x) + g(p(x)) = x p(x)\), for all \(x\)
The first property shows that \(g\) can be computed up to additive constants by solving: \[ g(p) = \int_0^p (f')^{-1}(q) dq + \text{constant} \] where we can use the second property to find the constant.
We illustrate this via the above examples.
In Example 46.1, \(f'(x) = 2x\). Therefore, \((f')^{-1}(x) = x/2\). Hence,
\[ g(p) = \int_{0}^p \frac{q}{2} dq + c = \frac{p^2}{4} + c. \]
In Example 46.2, \(f'(x) = x^{α-1}\). Therefore, \((f')^{-1} = x^{1/(α-1)}\). Hence \[ g(p) = \int_{0}^p q^{1/(α-1)} dq + c = \frac{α-1}{α} p^{α/(α-1)} + c \]
In Example 46.3, \(f'(x) = \log(x) + \log(1-x)\). Therefore, \((f')^{-1}(x) = e^x/(1 + e^x)\). Hence \[ g(p) = \int_{0}^p \frac{e^q}{1 + e^q} dq + c = \log(1 + e^p) + c'. \]
46.2 General definition
Definition 46.1 For any function \(f \colon \reals^n \to \bar {\reals}\), the function \(f^* \colon \reals^n \to \bar \reals\) defined by \[\begin{equation}\label{eq:conjugate} f^*(p) \coloneqq \sup_{x \in \reals^n} \bigl\{ \langle p, x \rangle - f(x) \bigr\} \end{equation}\] is conjugate to \(f\), while the function \(f^{**} = (f^*)^*\) defined by \[\begin{equation}\label{eq:biconjugate} f^{**}(x) \coloneqq \sup_{p \in \reals^n} \bigl\{ \langle p, x \rangle - f^*(p) \bigr\} \end{equation}\] is biconjugate to \(f\). The mapping \(f \mapsto f^*\) is called Legendre-Fenchel transform.
The significance of the conjugate can be understood in terms of epigraph relationships. Note that \eqref{eq:conjugate} implies that \[ (p,β) \in \text{epi } f^* \iff β \ge \langle p, x \rangle - α \text{ for all } (x,α) \in \text{epi } f. \] Let \(\ell_{p,β}(x) \coloneqq \langle p, x \rangle - β\), then we can write the above relationship as \[ (p,β) \in \text{epi } f^* \iff \ell_{p,β} \le f, \] that is, \(f^*\) describes a family of affine functions majorized by \(f\). Similarly, \[ β \ge f^*(p) \iff β \ge \ell_{x,α}(p) \text{ for all } (x,α) \in \text{epi } f. \]
Theorem 46.1 For any function \(f \colon \reals^n \to \bar \reals\) with \(\text{con } f\) proper, both \(f^*\) and \(f^{**}\) are proper, lsc, convex and \[ f^{**} = \text{cl con } f. \] Thus, \(f^{**} \le f\) and when \(f\) is itself proper, lsc, and convex, one has \(f^{**} = f\).
46.3 Properties of Legendre-Fenchel transform
Thus, Legendre-Fenchel transform sets up a one-to-one correspondence in the class of proper, lsc, and convex functions: if \(f\) is conjugate to \(g\), then \(g\) is a conjugate to \(f\): \[ f \xleftrightarrow{\hskip0.5em*\hskip0.5em} g \text { when } \begin{cases} g(p) = \sup_{x \in \reals^n} \bigl\{ \langle p, x \rangle - f(x) \bigr\} \\ f(x) = \sup_{x \in \reals^n} \bigl\{ \langle p, x \rangle - g(p) \bigr\} \end{cases} \]
Given \(f \xleftrightarrow{\hskip0.5em*\hskip0.5em} g\), we have the following properties:
Scaling properties. For any \(λ > 0\), \[\begin{align*} λ f(x) &\xleftrightarrow{\hskip0.5em*\hskip0.5em} λg(λ^{-1} p), \\ f(λx) &\xleftrightarrow{\hskip0.5em*\hskip0.5em} g(λ^{-1} p), \end{align*}\]
Translation properties. \[\begin{align*} f(x) - \langle a, x \rangle &\xleftrightarrow{\hskip0.5em*\hskip0.5em} g(p + a) , \\ f(x + b) &\xleftrightarrow{\hskip0.5em*\hskip0.5em} g(p) - \langle p, b \rangle, \\ f(x) + c &\xleftrightarrow{\hskip0.5em*\hskip0.5em} g(p) - c, \\ \end{align*}\]
Proposition 46.1 Let \(f\) be a finite, coercive, twice differentiable, and strongly convex function, then the conjugate \(g = f^*\) is also finite, coercive, twice differentiable, and strongly convex. Moreover,
the gradient mapping \(\GRAD f\) is one-to-one from \(\reals^n\) to \(\reals^n\), and its inverse is \(\GRAD g\); one has \[\begin{align*} g(p) &= \bigl\langle (\GRAD f)^{-1}(p), p \bigr\rangle - f\bigl((\GRAD f)^{-1}(p)\bigr), \\ f(x) &= \bigl\langle (\GRAD g)^{-1}(x), x \bigr\rangle - g\bigl((\GRAD g)^{-1}(x)\bigr), \end{align*}\]
The matrices \(\GRAD^2 f(x)\) and \(\GRAD^2 g(p)\) are inverse to one another when \(p = \GRAD f(x)\) or, equivalently, \(x = \GRAD g(p)\).
Strongly convex functions on a simplex have the following properties.
Proposition 46.2 Let \(Δ\) be the simplex in \(\reals^n\) and \(f \colon Δ \to \reals\) be twice differentiable and strongly convex. Let \(g \colon \reals^n \to \reals\) be the Legendre-Fenchel conjugate of \(f\). Then,
Unique maximizing argument: \(\GRAD g\) is Lipschitz and satisfies \[\GRAD g(p) = \arg\max_{x \in Δ}\bigl\{ \langle x, p \rangle - f(x) \bigr\}.\]
Boundedness: If there are constants \(L\) and \(U\) such that for all \(x \in Δ\), we have \(L \le f(x) \le U\), then \[ \| p \|_{∞} - U \le g(p) \le \| p \|_{∞} - L. \]
Distributivity: For any \(c \in \reals\), \[ g(p + c \ONES) = g(p) + c. \]
Monotonicity: If \(p_1 \le p_2\), then \(g(p_1) \le g(p_2)\).
Notes
The material on the intuition behind Legendre-Fenchel transform is adapted from Kennerly (2011). The example for Young-Fenchel inequality and Figure 46.2 is taken from Ellis (1985). The material on general definition and properties is adapted from Rockafellar and Wets (2009). Proposition 46.1 is stated as Example 11.9 in Rockafellar and Wets (2009). Proposition 46.2 is from Geist et al. (2019).