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For two functions $f$ and $g$, if $\nabla f(x) = \nabla g(x)$, $f = g + c$ for some constant $c$. Does the same hold if the gradient is replaced by the (convex) subdifferential, ie $\partial f(x) = \partial g(x)$ for all $x$ ?

And, as a stronger result, can we characterize pairs $(f, g)$ for which $\partial f(x) \cap \partial g(x) \neq \emptyset$ for all $x$ ?

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  • $\begingroup$ On which space are $f$ and $g$ defined? Are they finite everywhere? $\endgroup$
    – gerw
    Commented Feb 14 at 11:41
  • $\begingroup$ @gerw This is really an open question so I'd be interested in the simple case where the space is Euclidean and both functions have full domain, or a similar reasonable assumption. If some result holds without this, I'm glad to hear about it too! $\endgroup$ Commented Feb 14 at 12:18

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You need some extra assumptions on $f$ and $g$. If $f,g \colon X \to \bar{\mathbb R}$ are convex and lower semicontinuous and if $X$ is a Banach space, then $\partial f = \partial g$ imply that $f$ and $g$ differ by a constant. A proof can be found in the 1970 paper "On the maximal monotonicity of subdifferential mappings" by Rockafellar, see https://doi.org/10.2140/pjm.1970.33.209.

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  • $\begingroup$ This makes a lot of sense. This post characterizes $\partial f(x)$ in terms of the Fenchel conjugate $f^*\,.$ So, when we have two functions $f,g$ that are different but have the same Fenchel conjugate we have $\partial f(x)\not=\partial g(x)\,.$ $\endgroup$
    – Kurt G.
    Commented Feb 14 at 13:16
  • $\begingroup$ @KurtG. I do not understand your comment. If $f$ and $g$ have the same Fenchel conjugate, then we trivially have $f = f^{**} = g^{**} = g$. $\endgroup$
    – gerw
    Commented Feb 14 at 13:53
  • $\begingroup$ $f^*$ is convex and therefore $f^{**}$ is convex. When the function $f$ we started with is not convex we can't have $f^{**}=f\,.$ $\endgroup$
    – Kurt G.
    Commented Feb 14 at 13:55
  • $\begingroup$ When $f$ is not convex, $\partial f$ makes (almost) no sense anyway... $\endgroup$
    – gerw
    Commented Feb 14 at 13:58
  • $\begingroup$ ... except to provide another answer to this post. :) $\endgroup$
    – Kurt G.
    Commented Feb 14 at 14:05

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