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Show that if \(X\) is an infinite-dimensional Banach space, then \(X\) admits a discontinuous linear functional.

Short Answer

Expert verified
By the Hahn-Banach theorem, extend a bounded functional from a finite-dimensional subspace to the whole space. Apply this theorem to show the existence of a discontinuous functional.

Step by step solution

01

- Understand the Hahn-Banach Theorem

To show that an infinite-dimensional Banach space admits a discontinuous linear functional, first recall the Hahn-Banach theorem. The Hahn-Banach theorem states that if you have a linear subspace of a Banach space and a bounded linear functional defined on this subspace, you can extend this functional to the entire space without increasing its norm.
02

- Consider a Finite-Dimensional Subspace

Since any infinite-dimensional space contains finite-dimensional subspaces, consider a finite-dimensional subspace of our infinite-dimensional Banach space, and assume it is denoted by \([y_1, y_2, ..., y_n]\).
03

- Define a Bounded Linear Functional

On this finite-dimensional subspace, define a bounded linear functional \(f_0\). Such a bounded linear functional always exists in finite dimensions because you can always assign values to basis elements.
04

- Extend Using Hahn-Banach Theorem

Using the Hahn-Banach theorem, extend this bounded linear functional \(f_0\) defined on the finite-dimensional subspace to a bounded linear functional \(f\) on the entire Banach space \(X\). This extended functional \(f\) is necessarily bounded.
05

- Construct a Discontinuous Functional

To find a discontinuous linear functional, consider the quotient space \(X/Y\), where \(Y\) is a closed subspace of \(X\). This new space \(X/Y\) is itself an infinite-dimensional Banach space. By the Hahn-Banach theorem, one can show that there exists a bounded linear functional on this quotient space that is not continuous on the original space.

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Key Concepts

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

Hahn-Banach Theorem
The Hahn-Banach Theorem is a fundamental result in functional analysis. It allows us to extend a bounded linear functional defined on a subspace of a Banach space to the entire space. This extension can be done without increasing the functional’s norm. To understand this, suppose you have:
  • A linear subspace of a Banach space
  • A bounded linear functional defined on this subspace
According to the Hahn-Banach theorem, we can extend this bounded linear functional to the whole Banach space while keeping the original norm. This property is crucial for proving the existence of certain functionals, like discontinuous ones in an infinite-dimensional space.
Banach Space
A Banach space is a complete normed vector space. This means that every Cauchy sequence in the space converges within that space. For example, the set of all continuous functions on [0,1] with the supremum norm is a Banach space. These spaces are important in analysis and are generalizations of Euclidean spaces. They provide a framework for discussing convergence, boundedness, and continuity in infinite dimensions.
Bounded Linear Functional
A bounded linear functional is a linear map from a Banach space to the real numbers (or complex numbers) that satisfies a particular property: it does not increase too rapidly. Mathematically, a linear functional f on a Banach space X is bounded if there exists a constant C such that |f(x)| ≤ C ||x|| for all x in X . Such a constant C is called the bound of f .
Quotient Space
In functional analysis, quotient spaces help us deal with subspaces. Given a Banach space X and a closed subspace Y , the quotient space X/Y consists of cosets of Y in X . Formally, an element of X/Y is an equivalence class of the form x + Y for x in X . This new space X/Y has a natural norm and is itself a Banach space.
Finite-Dimensional Subspace
Even infinite-dimensional spaces have finite-dimensional subspaces. Consider an infinite-dimensional Banach space, which means it has infinite basis elements. In practice, we can often restrict attention to some finite-dimensional portion of this space. For instance, a subspace spanned by a finite set of vectors, such as {y1, y2, ..., yn}, is finite-dimensional. The properties and behaviors of such subspaces are often simpler to study and help us to extend our understanding to the entire space, especially when using results like the Hahn-Banach theorem.

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Most popular questions from this chapter

Let \(X\) be a closed subspace of \(C[0,1]\) such that every element of \(X\) is a continuously differentiable function on \([0,1]\). Show that \(X\) is finitedimensional. Hint: Let \(T: X \rightarrow C[0,1]\) be defined for \(f \in X\) by \(T(f)=f^{\prime} .\) Show that the graph of \(T\) is closed: If \(f_{n} \rightarrow f\) uniformly and \(f_{n}^{\prime} \rightarrow g\) uniformly, then \(f^{\prime}=g .\) Therefore \(T\) is continuous by the closed graph theorem. Thus for some \(n \in \mathbf{N}\) we have \(\left\|f^{\prime}\right\|_{\infty} \leq n\) whenever \(f \in X\) satisfies \(\|f\|_{\infty} \leq 1\). Let \(x_{i}=\frac{i}{4 n}\) for \(i=0,1, \ldots, 4 n\). Define an operator \(S: X \rightarrow\) \(\mathbf{R}^{4 n+1}\) by \(S(f)=\left\\{f\left(x_{i}\right)\right\\} .\) We claim that \(S\) is one-to-one. It is enough to show that if \(\|f\|_{\infty}=1\), then for some \(i, S(f)\left(x_{i}\right) \neq 0 .\) Assume that this is not true. If \(f(x)=1\) and \(x \in\left(\frac{i}{4 n}, \frac{i+1}{4 n}\right)\), then by the Lagrange mean value theorem we have \(\left|f(x)-f\left(\frac{i}{4 n}\right)\right|=\left|f^{\prime}(\xi)\right|\left|x-\frac{i}{4 n}\right| \leq n \cdot \frac{1}{4 n}\), a contradiction. Therefore \(\operatorname{dim}(X) \leq 4 n+1\)

(i) Prove directly that if \(X\) is a Banach space and \(f\) is a nonzero linear functional on \(X\), then \(f\) is an open map from \(X\) onto the scalars. (ii) Let the operator \(T\) from \(c_{0}\) into \(c_{0}\) be defined by \(T\left(\left(x_{i}\right)\right)=\left(\frac{1}{i} x_{i}\right) .\) Is \(T\) a bounded linear operator? Is \(T\) an open map? Does \(T\) map \(c_{0}\) onto a dense subset in \(c_{0} ?\) Hint: (i): If \(f(x)=\delta>0\) for some \(x \in B_{X}^{O}\), then \((-\delta, \delta) \subset f\left(B_{X}^{O}\right)\). (ii): Yes. No. Yes (use finitely supported vectors).

Let \(X, Y\) be Banach spaces and \(T \in \mathcal{B}(X, Y)\). If \(T\) is one-to- one, is \(T^{*}\) necessarily onto? Hint: No, consider the identity map from \(\ell_{1}\) into \(\ell_{2}\).

Let \((X,\|\cdot\|)\) be a Banach space. Show that \(\mu_{B_{X}}(x)=\|x\|\). Hint: Use continuity of the norm.

Let \(X=\mathbf{R}^{2}\) with the norm \(\|x\|=\left(\left|x_{1}\right|^{4}+\left|x_{2}\right|^{4}\right)^{\frac{1}{4}} .\) Calculate directly the dual norm on \(X^{*}\) using the Lagrange multipliers. Hint: The dual norm of \((a, b) \in X^{*}\) is \(\sup \left\\{a x_{1}+b x_{2} ; x_{1}^{4}+x_{2}^{4}=1\right\\} .\) Define \(F\left(x_{1}, x_{2}, \lambda\right)=a x_{1}+b x_{2}-\lambda\left(x_{1}^{4}+x_{2}^{4}-1\right)\) and multiply by \(x_{1}\) and \(x_{2}\), respectively, the equations you get from \(\frac{\partial F}{\partial x_{1}}=0\) and \(\frac{\partial F}{\partial x_{2}}=0\)

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