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non-reflexive
non-reˈflexive, a. [non- 3.] 1. Gram. Not reflexive (reflexive a. 5).1949 O. Jespersen Mod. Eng. Gram. VII. 170 As a reflexive use of the self-forms we must also reckon the predicative in ‘he is quite himself again’ = ‘in a normal condition’, though NED takes it to be the emphatic, i.e. non-reflexiv... Oxford English Dictionary
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Distinguished space
There exists a distinguished semi-reflexive non-reflexive -quasibarrelled Mackey space whose strong dual is a non-reflexive Banach space. wikipedia.org
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The image of non-reflexive Banach space in its bidual is at large distance from some unit vector in the bidual I am having some difficulties with part of a problem, I am working on. > Let $X$ be a non-reflexive Banac...
Simply use the Riesz lemma as $i(X)$ is norm-closed in $X^{**}$.
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Do non-reflexive relations exist? I may be talking complete and utter trash here, but does there exist any relation that isn't by definition reflexive? I know anything must always be equal to itself but is there any w...
The relation "$x$ and $y$ are not equal" on any set. The relation "$x>y$" on a set of real numbers. The relation "$x$ is the boss of $y$" on the set of people.
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Dequeísmo
(Prescriptively incorrect, non-reflexive , usually "It pleases me...") References Bentivoglio, Paola. 1975. "" In Frances M. Aid, Melvyn C. wikipedia.org
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The difference between $(\partial f)^{-1}$ and $\partial f^*$ on non-reflexive space. Let $X$ be a Banach space, $f:X\to \Bbb R\cup\\{\infty\\}$ be a proper function. The multifunction $(\partial f)^{-1}$ is defined b...
The following should work: Let $X=c_0$ and $f=\tfrac{1}{2}\|\cdot\|^2$ on $c_0$. Then $f^*$ looks likewise, but uses the norm on $c_0^*=\ell_1$. Now take a sequence in $\ell_1$ such as $x^* =(1/2^n)_{n\geq 1}$. Then $\partial f^*(x^*)=\\{(1,1,\ldots)_{n\geq 1}\\}$ but $(\partial f)^{-1}=\varnothing$...
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irreflexive
irreˈflexive, a. [ir-2.] Not reflexive. spec. in Logic and Math., of a relation which no postulate can have to itself. Hence irreˈflexiveness, irrefleˈxivity, the quality or property of being irreflexive. See also non-reflexive a., reflexive a.1890 in Cent. Dict. 1933 Mind XLII. 36, a —fxx. 1952 S. ... Oxford English Dictionary
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Is there relation that is symmetrical, transitive and non-reflexive? We must show that there exists some kind of $\alpha$ relation $\alpha ⊆ X \times X$ which has these conditions : if this relation is I and II ty...
If $X=\\{x\\}$ or $X=\emptyset$, then it's easy to show that $\alpha=\emptyset$. Suppose $X$ has more than one element. Take $x\ne y$ - elements of $X$. Suppose also that $(x,y)\in\alpha$, then $(y,x)\in\alpha$, hence $(x,x)\in\alpha$, which leads to contradiction. Thus, $\alpha=\emptyset$.
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Image of the canonical map is closed in a non-reflexive space Let $X$ be a Banach space. We have $\tau : X \rightarrow$ $X^{**}$ is the canonical map, isomorphic and isometric but suppose that X is not reflexive, so t...
$\tau$ is an isometry so its range is complete. This implies that it is closed in $X^{**}$. More simply argue as follows: Let $\tau(x_n) \to x^{**}$ in $X^{**}$. Then $\|x_n-x_m\| =\|\tau (x_n)-\tau (x_m)\|\to 0$. Since $X$ is complete there exists $x \in X$ such that $x_n \to x$. Since $\tau$ is co...
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Is it possible for a Strongly Convex function to be unbounded below? Let $X$ be a non-reflexive Banach space and $f:X\rightarrow\mathbb{R}$ a $C^1$ function that is Strongly Convex, i.e. $$f(u)-f(v)\geq\langle f'(v),u...
No. For every $a$, $b$ and $v$, the function $g:u\mapsto 2\langle a,u\rangle+b+\|u-v\|^2$ is bounded below. To wit, $$ g(u)=\|u-v+a\|^2+b+2\langle a,v\rangle-\|a\|^2\geqslant b+2\langle a,v\rangle-\|a\|^2. $$ Apply this to $a=f'(v)/(2c)$ and $b=(f(v)-\langle f'(v),v\rangle)/c$. **Edit:** In a Banach...
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Weak-strong continuous non-compact operator $c_0 \to \ell_1$ In our lecture on partial differential equations we proved the following statement > Let $V$ be reflexive and $A: V \to V^*$ a weak-strong continuous opera...
What about $V=l^1$, $V^*=l^\infty$, $A=id$ the continuous embedding $l^1\hookrightarrow l^\infty$? Due to the Schur-property, it is trivially weak-strong continuous, but not compact.
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Metrizability of a subset in the weak topology. Let $X$ be a Banach space (not reflexive). It is well-known that $(X,w)$, which is $X$ with its weak topology, is not metrizable if $X$ is infinite dimensional. > I wan...
Reflexivity isn't the right kind of condition for metrizability of the weak topology. Instead we have the following result. > **Lemma:** Let $X$ be a normed space. The relative weak topology on the unit ball $B_X$ of $X$ is metrizable if and only if $X^*$ is separable. In particular, if $X$ is refle...
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Weak sequential completeness It is obvious that reflexive spaces are weakly sequentially complete. Can we have a kind of a converse to this fact? Is there a non-reflexive Banach space $X$ such that both $X$ and $X^*$ ...
Schur spaces are weakly sequentially complete and non-reflexive if infinite dimensional (see here for a proof of these facts).
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Is every Pre-Hilbert space reflexive? Good day, It is known that every Hilbert space is reflexive. Does the same hold true for Pre-Hilbert spaces? I guess not since completeness is a pretty strong property that is no...
No, if a pre-Hilbert space is not complete it is not reflexive, dual spaces are always complete! On the other hand if a pre-Hilbert space is complete, it is a Hilbert space.
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reference for "compactness" coming from topology of convergence in measure I have found this sentence in a paper of F. Delbaen and W. Schachermayer with the title: A compactness principle for bounded sequences of mart...
So that this question has an answer: t.b.'s comment suggests that the quotes passage relates to the paper's Theorem 1.3, which states: > **Theorem.** Given a bounded sequence $(f_n)_{n \ge 1} \in L^1(\Omega, \mathcal{F}, \mathbb{P})$ then there are convex combinations $$g_n \in \operatorname{conv}(f...
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