Search Results

(Turning my comment into an answer here): Regarding the third question: Yes, reflexive spaces with the Schur property need to be finite-dimensional. To see that, two ingredients suffice, once you not …

Let $X$ be an infinite-dimensional Banach space and $\beta$ denote Istrățescu's spreading measure of noncompactness, i.e. $$\beta(M) = \sup \{ \varepsilon > 0 \colon \exists_{(x_n)^{\mathbb N} \in X^{ …

I've since come across the article Amann, Herbert. Compact embeddings of vector-valued Sobolev and Besov spaces. Dedicated to the memory of Branko Najman. Glas. Mat. Ser. III 35(55) (2000), no. 1, …

The following is known: Let $s \in (0,1)$ and $p \in [1,\infty)$ be such that $sp < n$. Let $q \in [1, p^*_{n,s})$ with $p^*_{n,s} = np/(n-sp)$, $\Omega \subset \mathbb R^n$ be a bounded extension …

Yes, this is correct; in fact, you could replace the map $|\cdot|$ with an arbitrary (uniformly) Lipschitz map $f$ (with $f(0) = 0$ if $G$ is unbounded) [1]. [1] Marcus, Moshe; Mizel, Victor J. Every …

The following is stated in the paper Moshe Marcus and Victor J. Mizel, MR 531975 Complete characterization of functions which act, via superposition, on Sobolev spaces, Trans. Amer. Math. Soc. 251 (1 …

Consider a bounded operator $T$ from a Hilbert space $H$ into itself, i.e. $T \in B(H)$. You can then define the numerical radius $r(T)$ as the radius of the numerical range $W(T)$, i.e. $$ W(T) = \{ …

Let $\Omega \subset \mathbb R^n$ be a bounded domain with a Lipschitz boundary and $n > p \ge 1$. It is well known that up to the critical exponent $p^* = pn/(n − p)$, i.e. $q < p^*$, the Sobolev-to- …