Comments on The Fenchel–Moreau theorem; or, how to generalize convex conjugate intuition to the N-dimensional case?
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The Fenchel–Moreau theorem; or, how to generalize convex conjugate intuition to the N-dimensional case?
The Legendre transform or convex conjugate $f^*$ of a one-dimensional convex function $f$ is defined as $f^*(\lambda) = \text{max}_x (\lambda x - f(x))$. In other words, $f^*$ takes a slope $\lambda$ and returns the maximum difference between $f$ and the line $\lambda x$. As an illuminating example, in economics the convex conjugate of cost (a function of quantity) is profit (a function of the price, which is dual to quantity): for the profit at a price $\lambda$ is the maximum difference between the gross revenue at that price $\lambda x$ and the cost of production $c(x)$.1
Interestingly, though, the Legendre transformation is involutive: $(f^*)^* = f$ always! E.g., not only is profit the convex conjugate of cost but cost is the convex conjugate of profit.2
In the one-dimensional case I find this property quite intuitive: roughly speaking, convex conjugates measure the area between a curve and the x- or y- axis, as the image below illustrates. (In the above example, the green function would represent total cost, the yellow function total profit.)
But I am not quite sure how this generalizes to the convex conjugate in more dimensions3. If I know that involutiveness holds in 1 dimension, can I straightforwardly deduce that it holds in arbitrary vector spaces? (I assume not, because the Wikipedia page for the Fenchel–Moreau theorem doesn't have that proof.) If not, is there some other argument for the 1-dimensional case that generalizes relatively well, or any rate is there some other relatively simple proof of involutivity?
1(This assumes, of course, that the producer is acting rationally and thus keeps producing goods up to the point where the marginal cost and marginal revenue are equal.)
2 As an example, in an efficient market, given a price, the producer will continue to produce goods until the marginal cost equals that price, which maximizes the difference between total revenue and total cost; dually, the producer will only be able to sell a given quantity of goods at the price that equals the marginal cost of production, which maximizes the difference between total revenue and total profit.
3 This is defined by saying that, for all differentiable $f: X → \mathbb{R}$, $f^*: X^* → \mathbb{R}$ maps $\lambda \in X^*$ to $\text{sup } \lambda(x) - f(x)$.
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The following users marked this post as Works for me:
| User | Comment | Date |
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| clemens |
Thread: Works for me Nice exposition! In hindsight, I can't believe I missed that $$f^* (\lambda) \ge \lambda x - f(x) \Leftrightarrow f^*(x) \ge \lambda - f(x)$$ gener... |
Apr 11, 2026 at 15:11 |
It has been a bunch of years since I have thought about this stuff, but let me see if I still remember something.
In convex analysis, it is helpful to think of the function's epigraph, which (roughly speaking) is defined as the graph and all the points above it. The function has to be (extended) real-valued for this to make sense, and should be defined on a topological vector space so as to make sense of convexity and continuity and their related concepts. There might be more conditions on the space.
One can then define the convex conjugate as you do, though typically with supremum, rather than maximum. Also, in a more general space, one takes an element from the dual space instead of lambda, and then dual pairing between lambda and $x$. Then $f = f^{**}$ for closed, convex functions. Otherwise it recovers the function whose epigraph is the closed convex hull of the original function's epigraph.
This way of thinking about Legendre-Fenchel transform/convex conjugate is that it tries to set up supporting hyperplanes (lines in one dimension) to the epigraph of the function. Convex conjugate stores the information about the best supporting hyperplanes, but nothing more, which is equivalent to the closed convex hull of the epigraph, much like any closed, convex set can be defined as the intersection of (typically infinitely many) closed half-spaces, and for a non-closed or non-convex set the intersection is the closed convex hull.
To see this interpretation, consider that since $$ f^*(\lambda) = \sup_x (\lambda x - f(x)), $$ we also have, for every $x$, $$ f^*(\lambda) \ge \lambda x - f(x), $$ and thus $$ f(x) \ge \lambda x - f^*(\lambda). $$ Here $f$ is bounded below by an affine linear function, or, equivalently, the epigraph of $f$ is a subset of the closed half-space defined by that affine linear function. Each lambda defines one such half-space. Since the set we are interested in is an epigraph, there will never be any half-space cutting it off from "above", only "below", so they can be encoded by elements of the dual space like this.
To get intuition for this, I would calculate a few cases like $\frac{1}{p}|x|^p$, the absolute value, the indicator/characteristic function of $[-1, 1]$, the exponential function, etc.
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