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The bad news is that many closed convex cones in infinite dimensions have empty interior, e.g., the natural cone of positive functions in $L^p(\Omega)$ for all $\Omega \subset \mathbb{R}^n$; or in $W^ …

The radial cone of $C$ is defined via $$ \mathcal R_C(x) := \bigcup_{\lambda > 0} \lambda ( C - x)$$ for all $x \in C$ and we can show $$ \mathcal R_C(x) = C + \operatorname{span}(x), $$ since $C$ is …

If you are willing to replace $a_i > 0$ by $a_i \ge 0$, then this becomes a quadratic program. Indeed, it can be formulated as \begin{align*} \text{Minimize}\quad & \frac12 a^\top Q a + q^\top a, \\ …

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 …

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) …

Let us assume that your $f$ is at least directionally differentiable at $x_0$ and that the directional derivative depends continuously on the direction (this may be satisfied under rather general assu …

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

If your problem has a solution $x^* \ne 0$, then $0$ is also a solution. Indeed, consider the function $$\varphi(t) = c(t \, x^*) - k\cdot (t \, x^*).$$ Since $x^*$ is a solution, we have $$\varphi(0) …

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 …

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 …

I was also struggling with Exercise 2.36... I think that I am now able to solve it, although it seems that it is more difficult than it appears... The key seems to be the following theorem. Theorem: …

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 …