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Is the Legendre transform connected to identity in any sense? Is there a "good" way to interpolate continuously between a convex function and its Legendre transform in the sense that

$\gamma: [0,1] \rightarrow \{ \text{mappings of convex functions into convex functions} \}$

such that

$\gamma(0) = id \quad \text{and} \quad \gamma(1) = \text{Legendre transform}$

dan-ros
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2 Answers2

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Yes, this is possible using the proximal average. The link to the official paper is https://epubs.siam.org/doi/10.1137/060664513

max_zorn
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I doubt there's anything like what you're seeking. Let $X$ be a Banach space, for a proper function $f:X\to \Bbb R\cup \{\infty\}$, its Legendre-Fenchel transform $f^*:X^*\to\Bbb R\cup\{\infty\}$ is defined on a totally different space.

Given that you want $\gamma$ such that $\gamma(0)=f$ and $\gamma(1)=f^*$, the domain of the function $\gamma(t)$ for $t\in (0,1)$ is already a problem. Do you want it to be $X,X^*$ or something else entirely? You'd need to think of a continuous way to transform $X$ into $X^*$ (and I'm not even sure how to make that precise).

BigbearZzz
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  • doesn't need to be general; what about special cases for $X=[\text{min}(f),\infty]$? – dan-ros Dec 13 '18 at 17:14
  • It just happens to be the case that $(\Bbb R^n)^\simeq\Bbb R^n$, normally there's no connection between a point $x$ in our original space and a linear functional. My main point is there's no "middle ground" between $X$ and $X^$ so you can't expect anything "in between". – BigbearZzz Dec 13 '18 at 17:18