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A norm \(\|\cdot\|\) of a Banach space \(X\) is said to have the \((2 R)\) -property if \(\left\\{x_{n}\right\\}\) is a convergent sequence whenever \(\left\|x_{n}+x_{m}\right\| \rightarrow 2\). Show that every space whose norm has the \((2 R)\) -property is reflexive. Note that every separable reflexive space has an equivalent norm with the \((2 R)\) -property \(([\mathrm{OdSc}])\)

Short Answer

Expert verified
A space with the (2R)-property ensures weak convergence, leading to reflexivity.

Step by step solution

01

Understand the (2R)-property

The (2R)-property states that for any sequence \(\(x_n\)\) in a Banach space, if \(\(\frac{1}{2}abla_n\right\rightarrow 1\) as \(\rightarrow\infty\)\) then the sequence is convergent.
02

Understand that reflexive spaces

Recall that a Banach space \(\X\) is reflexive if every bounded sequence in \(x_n\X\) has a weakly convergent subsequence.
03

Assume a space X with the (2R)-property

Given \(\X\) with the (2R)-property, let's show \(\X\) is reflexive.
04

Show boundedness from weak convergence

Every bounded sequence in \(\X\) will have a subsequence converges to \(\yyy\) by a weak convergence argument. By the (2R)-property, this subsequence is equivalent to the norm convergence.
05

Conclude reflexivity

Given the equivalent subsequences with the weak and norm convergence, X must reflexively be weakly which translates to norm convergent by the (2R)-property.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

(2R)-property
The (2R)-property is a condition related to the norm in a Banach space. It states that if a sequence \(\left\{x_{n}\right\}\) in a Banach space satisfies \(\left\|x_{n} + x_{m}\right\| \rightarrow 2\) as \(\ n, m \rightarrow \infty\), then the sequence \(\left\{x_{n}\right\}\) is convergent. This property ensures that the behavior of sums of sequence elements can provide crucial information about the convergence of the sequence itself.
  • Consider a sequence \(\left\{x_{n}\right\}\).
  • If \(\left\|x_{n}+x_{m}\right\| \rightarrow 2\), the sequence converges.
This property is quite useful for studying Banach spaces, as it connects norm behavior with sequence convergence. This connection is important for analyzing the structure and characteristics of Banach spaces.
Reflexive Spaces
A Banach space is called reflexive if every bounded sequence within it has a weakly convergent subsequence. Reflexive spaces exhibit nice properties that facilitate the analysis of bounded sequences.
  • Every bounded sequence in a reflexive space has a weakly convergent subsequence.
  • Weak convergence implies convergence in terms of the weak topology.
In reflexive spaces, the dual space also shares reflexive properties, making these spaces easier to work with in functional analysis. Reflexivity ensures that certain limits and conditions are met, simplifying many problems involving bounded sequences.
Convergent Sequences
A sequence \(\left\{x_{n}\right\}\) in a Banach space is said to be convergent if there exists an element \(x \) in the space such that \(\|x_{n} - x\| \rightarrow 0\) as \(n \rightarrow \infty\). Convergence is a fundamental concept in analysis.
  • A sequence converges if its elements get arbitrarily close to a certain point \(x\).
  • Convergent sequences have unique limits in normed spaces.
Understanding whether a sequence converges can help solve problems related to limits, continuity, and other functional properties in Banach spaces.
Weak Convergence
Weak convergence is a type of convergence that is less stringent than norm convergence. A sequence \(\left\{x_{n}\right\}\) in a Banach space weakly converges to an element \(x\) if every continuous linear functional \(f \) on the space satisfies \(f(x_{n}) \rightarrow f(x)\).
  • Weak convergence: \(x_{n} \rightharpoonup x\)
  • It implies that the behavior under any bounded linear functional converges to the same behavior of the limit element.
Weak convergence is particularly useful in reflexive spaces, as it provides a connection between bounded sequences and their limits without requiring strong (norm) convergence.
Equivalent Norm
Two norms \(\|\cdot\|_1\) and \(\|\cdot\|_2\) on a vector space are said to be equivalent if there exist positive constants \(c_1\) and \(c_2\) such that \(c_1 \|x\|_1 \leq \|x\|_2 \leq c_2 \|x\|_1\) for all \(x\) in the space.
  • Equivalent norms: Norms that differ only by a constant factor.
  • They induce the same topology, meaning they dictate the same concept of convergence and continuity.
Equivalent norms are useful because they allow flexibility in analysis. If one norm is easier to work with, you can use it without changing the underlying properties of the space.

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Most popular questions from this chapter

Let \(A\) be a subset of \(\mathbf{R}\) with \(\lambda(A)=0\) (Lebesgue measure). Follow the hint to show that there is a Lipschitz function \(f\) on \(\mathbf{R}\) not differentiable at points of \(A\) ([Zah]). Since there is a residual set of measure 0 in \(\mathbf{R}\), this shows that the set of differentiability points of a Lipschitz function need not be residual. Hint: \(([\mathrm{BeLi}])\) Take open sets \(G_{1} \supset G_{2} \supset \supset \ldots \supset A\) such that \(\lambda\left(G_{n}\right) \leq 2^{-n}\) and \(\lambda\left((a, b) \cap G_{n+1}\right) \leq \frac{b-a}{3}\) for every component \((a, b)\) of \(G_{n} .\) Put \(f_{n}(x)=\) \(\lambda\left((-\infty, x) \cap G_{n}\right)\) and \(f=\sum(-1)^{n+1} f .\) Then \(f\) is Lipschitz. If \(x \in \bigcap G_{n}\), then \(f\) is not differentiable at \(x .\) Indeed, let \(\left(\alpha_{n}, \beta_{n}\right)\) be components of \(G_{n}\) containing \(x\). Fix \(n\), define \(\gamma_{j}=\frac{f\left(\beta_{n}\right)-f\left(\alpha_{n}\right)}{\beta_{n}-\alpha_{n}}\), and observe that \(\left\\{\gamma_{j}\right\\}\) is decreasing, \(\gamma_{1}=\ldots=\gamma_{n}=1\), and \(\gamma_{n+1} \leq \frac{1}{3} .\) Thus if \(n\) is even, then \(\frac{f\left(\beta_{n}\right)-f\left(\alpha_{n}\right)}{\beta_{n}-\alpha_{n}}=1-1+\ldots-1+\gamma_{n+1}-\gamma_{n+2}+\ldots \leq \frac{1}{3}\), while if \(n\) is odd, \(\frac{f\left(\beta_{n}\right)-f\left(\alpha_{n}\right)}{\beta_{n}-\alpha_{n}}=1-1+\ldots+1-\gamma_{n+1}+\gamma_{n+2}-\ldots \geq \frac{2}{3}\)

Let \(X\) be an infinite-dimensional separable Banach space. Show that there is a continuous convex function \(f\) on \(X\) such that \(f\) is unbounded on \(B_{X}\)

Let \(X\) be a reflexive strictly convex Banach space, and \(C\) a closed convex set in \(X\). Show that there is a unique nearest point to \(x\) in \(C\). Hint: \(x\) has a nearest point because \(X\) is reflexive (Exercise 3.104). Assume that \(x=0\) and \(\operatorname{dist}(x, C)=1\). Let \(c_{1} \neq c_{2} \in C\) satisfy \(\left\|c_{1}\right\|=\left\|c_{2}\right\|=1\). Then \(\frac{1}{2}\left(c_{1}+c_{2}\right) \in C\), yet \(\left\|\frac{1}{2}\left(c_{1}+c_{2}\right)\right\|<1\) by the strict convexity, a contradiction.

Let \(C \neq \emptyset\) be a convex compact set in a Hilbert space \(H\). Show that \(C\) has at least one exposed point.

Let \(X\) be a Banach space and let \(f\) be a continuous convex function on \(X^{*}\) that is \(w^{*}\) -lower semicontinuous. Show that if \(f\) is Fréchet differentiable at \(x^{*} \in X^{*}\), then \(f^{\prime}\left(x^{*}\right) \in X\). Hint: The derivative, as a uniform limit of quotients in \(B_{X^{*}}\), is also \(w^{*}\) -lower semicontinuous. Then use its linearity to see that \(f^{\prime}\left(x^{*}\right)\) is a functional that is \(w^{*}\) -continuous on \(B_{X *}\) and apply Theorem \(4.44\).

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