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I don't know what $e$ is, but if you need a subgradient of a maximum of convex functions, you can take a subgradient of one of the functions where the maximum is achieved, e.g. if $$ p(x) = \max_i f_i …

Not sure if this points in the right direction since I don't know enough context. It seems like you could in principle set $y= Ux$ and either work with the rotated variable all over the place or use " …

This is not an answer but it may be helpful to know that there exists the notion of modulus of convexity of a convex function $f:X\to ]-\infty,\infty]$ defined on a Banach space $X$ which quantifies h …

If $A$ is invertible and $BA^{-1}$ is, by any chance, symmetric positive definite, then you could write $$ \langle B(y-x),\nabla f(Ax)\rangle = \langle BA^{-1}A(y-x),\nabla f(Ax)\rangle $$ and use tha …

I do not know a definitive "weakest" condition and I doubt there is one. Many results in the realm of Hahn-Banach do the trick, i.e. there is the general result for $A,B$ convex and $A$ open (both ope …

Note that the conclusion is that the gradient of $g_r$ converges to zero, not $g_r$ itself! The first inequality and the fact that $r>0$ imply $$\ g_r(M(y^k)) - g_r(y^k) > r|\nabla g_r(y^k)|^2 \geq 0 …

First of all, the function $f_a$ is not strictly convex, and hence, you should expect multiple minimizers. As such, non-expansiveness (even in some generalized sense) does not seem likely. Consider th …

If I recall correctly, every proper, convex, weak*-lower semi-continuous function is a conjugate. In fact, it is the conjugate of its pre-conjugate $$ f(x) = \sup_{x^*\in X^*} \langle x^*,x\rangle - g …

That's a standard result in convex optimization. For example Theorem 2.1.5 in Nesterov's "Introductory Lectures on Convex Optimization" states that the following are equivalent: $f$ is $C^1$, convex …

I am not aware of results on the linear rate of this variant of the proximal point method. Let me note that convergence is usually shown by the following observation: Since $C$ is a bijection, you may …

One thing that is straightforward would be to extend $F$ to the full space $\mathcal{M}$ of signed measures by setting $F(\mu) = -\infty$ if $\mu$ is not a probability measure. This extension would pr …

The function $\ell$ is concave (it's a minimum of linear functions). Its conjugate $\ell^*(\theta)$ is the supremum over a linear function minus $\ell$. If I see correctly, $\ell^*(\theta) = \infty$ f …

I haven't seen this notion. Also note that Bachir did not use the whole space of continuous bounded functions for $\phi$ but certain subsets (doesn't the biconjugate get weird if you use the full spac …

A simple example is the convex function $$ f(x) = \begin{cases} \infty, & \text{if}\ x<0\\ x & \text{if}\ x\geq 0. \end{cases} $$ It holds that $\partial f(0) = ]-\infty,1]$. There are no examples wi …

I think Carlo Beenakker digged up the right reference for the notation of the set, but I think more can be said. First, there is some meaning for the subscript $0$ which can be found in the same pape …