Search Results

If you switch from your perspective of functions to their epigraph, I think that you end up with a "closure operator", see Wikipedia. I am not sure about other examples, but if $\mathcal{F}$ consists …

You can take $X = \ell^\infty$ (actually $\ell^p$ with $p > 2$ should work) and define $$ f(x) = \frac12 \| x \|_{\ell^2}^2. $$ Note that $f$ equals $\infty$ on $\ell^\infty \setminus \ell^2$. One can …

Here are just some thoughts. I think it is a matter of curvature, so let us assume that $\varphi$ and $\psi$ are smooth. Then, $y(x)$ solves the optimality condition $$ \psi'(y(x)) = \varphi'(x - y(x) …

I don't think that this is true. Let us take $p = 2$, $s = 1$ and $f(x) = \frac12 \|x - (1,1)\|^2$. Then, $\theta_0 = (1,0)$ is a minimizer, but with $\theta = (0,1)$ we get $$ \nabla f(\theta_0)^\top …

Such an extension is not always possible, a counterexample can be found in Section 2 of https://arxiv.org/abs/1812.02419v3.

Have a look at the example from https://mathoverflow.net/a/37683/32507. This gives two convex, closed sets which cannot be separated by linear functionals (even not by discontinuous ones). If I unders …

I think we can argue as in https://mathoverflow.net/a/423284/32507 to answer the question in the affirmative. Let $\mathcal R_K(x)$ be the radial cone of $K$ at $x$ (as defined in the other answer). F …

The set $K_A$ is essentially a polar of $A$. Indeed, we have $$ A = \{ x \in \mathbb R^n \mid l(x) \ge t \; \forall (l,t) \in K_A\} =: B.$$ The inclusion "$\subset$" is clear and in order to check "$\ …

Yes, the converse is also true. This follows from the answer in https://math.stackexchange.com/questions/4227159/characterization-of-lipschitz-derivative. In fact, your condition yields $$ | (\nabla f …

Finally, I was able to cook up a counterexample. We choose $X = c_0$ (zero sequences equipped with supremum norm). Thus, the dual spaces are (isometric to) $X^* = \ell^1$ and $X^{**} = \ell^\infty$. W …

I don't think that this is true. Let us take $$ C := \{ x \in \mathbb R^2 \mid x_1^2 \le x_2 \le 1\}$$ and $$ f(x) = \frac{x_1^2}{x_2} $$ for $x \in C \setminus \{0\}$, $f(0,0) = 0$. This function is …

Here is a simple proof that $$\frac1\lambda \mapsto \frac1\lambda \, \log \int \exp(\lambda \, \phi(x)) \, \mathrm d \gamma(x)$$ is convex. This does not need any assumptions on $\phi$ or $\gamma$. Ma …

For convex optimization problems, you do not need second-order conditions, because already the optimality conditions of first order characterize global optimality.

This has been shown by Dudley, see https://www.jstor.org/stable/24490947.

Let me sketch a proof that the KKT conditions imply global optimality in the case that the objective $f$ and $S$ is convex. No constraint qualification is needed. Let us assume that the KKT condition …