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This is a cute problem. I would look at it like this: if we are going to show $U$ is a ball, we have to identify a good candidate for its radius $R$ and its center $c$. For $R$, we should think it should be the largest possible radius of a ball contained in $U$. So let $A \subset \mathbb{R}$ be the set of all numbers $r$ such that $U$ contains a ball of ...
Note that, by definition $\tau(B^*)$ is dense in $\def\c{\mathop{\rm cl}}\c(\tau B^*)$. Given now $x,y \in V$ and $\epsilon > 0$, the set $$W := \{h \in \c(\tau B^*) : |h(x) - f(x)|, |h(y) - f(y)|, |h(x+y) - f(x+y)| < \epsilon/3 \}$$ is an open - by the definition of the product topology - set containing $f$. As $f \in \c(\tau B^*)$, there is an $g ... 2 Yes, the problem has a unique solution (by strong convexity of the objective), but you can't compute it closed-form... Let's concentrate on your "real" problem: computing the subdifferential of that composite term. To this end, define$g = \|.\|_\infty$and$f := g \circ A$. By basic properties of subdifferentials, it's clear that \partial ... 2 Counterexample: $$C:\mathbb{R}_2^2\to\mathbb{R}_2^2:(x,y)\mapsto(2x+y,-3x-2y)$$ Then$\Vert C((1,1))\Vert>\Vert(1,1)\Vert$2 Consider the map$T : X \to \mathbb{R}^k$defined by$Tx = (x_1^*(x), \dots, x_k^*(x))$. I claim$T$is surjective. If not, let$(a_1, \dots, a_k)$be a nonzero element of the orthogonal complement of the image$TX$. Then$\sum_{i=1}^k a_i x_i^*(x) = 0$for all$x$. This means$\sum_{i=1}^k a_i x_i^* = 0$, contradicting linear independence of the ... 1 Yes, we can. Let$\bar{x}$be the unique fixed point of$f^p$. Then, we note that:$f^p(f^1(\bar{x}))=f^1(f^p(\bar{x}))=f^1(\bar{x})$. This implies that$f^1(\bar{x})$is a fixed point of$f^p$. However, we already know that$\bar{x}$is the unique fixed point, implying that$f^1(\bar{x})=\bar{x}$. This, finally, implies that$\bar{x}$is a fixed point of ... 1 In general, this is not true. In fact, it is possible to construct an example with$x \ne y$, see the related post http://math.stackexchange.com/a/426499/58577. However, in your situation, life is a little bit easier as pointed out by @saz. Every sequence converging in$L^*$with$*=p$or$*=q$has a subsequence, which converges pointwise a.e. This can be ... 1 Consider$A^C$. Let,$p$be a limit point of$A^C$and let$\{x^{(n)}\}$be a sequence of points in$A^C$converging to$p$. Then,$\lim_{n\to \infty}x^{(n)}_k=p_k,\ k=1,2,\cdots$. Since$x^{(n)}\in A^C$,$\$x^{(n)}_1\ge \sum_{k\ge 1}x^{(n)}_k2^{-k}\implies \lim_{n\to \infty}x^{(n)}_1\ge \sum_{k\ge 1}\lim_{n\to \infty}x^{(n)}_k 2^{-k}\implies p_1\ge ...