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Show that if \(X\) is a finite-dimensional Banach space, then every linear functional \(f\) on \(X\) is continuous on \(X\).

Short Answer

Expert verified
Every linear functional on a finite-dimensional Banach space is continuous because linear functionals on \( \text{\bf R}^n \) are continuous and every finite-dimensional Banach space is isomorphic to \( \text{\bf R}^n \).

Step by step solution

01

- Understand the Given Problem

We need to show that any linear functional on a finite-dimensional Banach space is continuous. Let's begin by understanding the terms: A Banach space is a complete normed vector space, and a linear functional is a linear map from the space to the real or complex numbers.
02

- Finite-dimensional Spaces are Isomorphic to \(\text{\bf R}^n\)

Recall that any finite-dimensional Banach space is isomorphic to \(\text{\bf R}^n\) (or \(\text{\bf C}^n\) if working with complex numbers). An isomorphism here means there exists a linear bijection between the Banach space and \(\text{\bf R}^n\).
03

- Linear Maps on \( \text{\bf R}^n \) are Continuous

In finite-dimensional spaces such as \( \text{\bf R}^n \), any linear map is continuous. This is a known result from linear algebra and is based on the fact that all norms on finite-dimensional vector spaces are equivalent.
04

- Use the Isomorphism to Finalize the Proof

Given the isomorphism between our Banach space \(X\) and \( \text{\bf R}^n \), any linear functional on \(X\) can be translated into a linear functional on \(\text{\bf R}^n\). Since linear functionals on \( \text{\bf R}^n \) are continuous, so are the linear functionals on \(X\) due to the isomorphism.

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Key Concepts

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

linear functional
A linear functional is a very special kind of function in mathematics. It is a linear map that takes a vector from a vector space and returns a real or complex number. Importantly, it preserves the operations of addition and scalar multiplication. This means that for a linear functional \(f \) and any two vectors \(x, y \) in the vector space, and any scalar \(a, b \) from the respective field: \ f(ax + by) = af(x) + bf(y) \. This property makes linear functionals particularly useful in various fields including functional analysis, quantum mechanics, and optimization.
In simpler terms, a linear functional stretches or scales vectors and sums their results in a linear way, turning them into simple numbers. Understanding this will make further steps, such as studying continuity and isomorphism, clearer and more logical.
continuity
Continuity, in the context of functions, means that small changes in the input lead to small changes in the output. For linear functionals on finite-dimensional Banach spaces, continuity is guaranteed.
To see why, let's start by recalling that a Banach space is a vector space with a norm, which is complete (meaning it contains all its limit points). Now, in any real-world application, we love Banach spaces as they allow us to handle infinite sequences nicely under the operation of taking limits.
We can show that a linear functional \(f \) is continuous by noting that in a finite-dimensional Banach space, any linear map is continuous. This comes from the equivalence of all norms in finite-dimensional spaces. In other words, no matter which norm we use, distances and sizes of vectors remain proportional. Therefore, any small change in the vector results in a proportionally small change in the output number given by the linear functional, ensuring continuity.
isomorphism
Isomorphism is a concept that gives a really powerful idea: two mathematical structures can be considered the same if there is a way to map one structure onto the other perfectly, respecting their operations.
When we say a finite-dimensional Banach space is isomorphic to \( \text{\bf R}^n \), we mean there's a bijective (one-to-one and onto) linear map that preserves vector addition and scalar multiplication. Essentially, it means these spaces are structurally identical for practical purposes.
This is a critical concept because it tells us that what happens in the relatively simple and intuitive \( \text{\bf R}^n \) also happens in any finite-dimensional Banach space. Thus, properties like continuity of linear maps that are well-known in \( \text{\bf R}^n \) transfer directly. Therefore, if every linear functional is continuous on \( \text{\bf R}^n \), they are also continuous on any finite-dimensional Banach space through the isomorphism. Understanding isomorphisms helps in transferring and applying results from simpler spaces to more complex ones seamlessly.

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Most popular questions from this chapter

Let \(Y\) be a subspace of a Banach space \(X\) and \(\|\cdot\|\) be an equivalent norm on \(Y\). Show that \(\|\cdot\|\) can be extended to an equivalent norm on \(X\). Hint: Let \(B_{2}\) be the unit ball of the original norm of \(X .\) Assume without loss of generality that the unit ball \(B_{1}\) of \(\|\cdot\|\) on \(Y\) contains \(B_{2} \cap Y\). The Minkowski functional of the set \(\operatorname{conv}\left(B_{1} \cup B_{2}\right)\) yields the desired norm.

Let \(C\) be a convex subset of a real Banach space \(X\) that contains a neighborhood of 0 (then \(\mu_{C}\) is a positive homogeneous sublinear functional on \(X\) ). Prove the following: (i) If \(C\) is also open, then \(C=\left\\{x ; \mu_{C}(x)<1\right\\}\). If \(C\) is also closed, then \(C=\left\\{x ; \mu_{C}(x) \leq 1\right\\}\) (ii) There is \(c>0\) such that \(\mu_{C}(x) \leq c\|x\|\). (iii) If \(C\) is moreover symmetric, then \(\mu_{C}\) is a seminorm, that is, it is a homogeneous sublinear functional. (iv) If \(C\) is moreover symmetric and bounded, then \(\mu_{C}\) is a norm that is equivalent to \(\|\cdot\|_{X} .\) In particular, it is complete, that is, \(\left(X, \mu_{C}\right)\) is a Banach space. Note that the symmetry condition is good only for the real case. In a complex normed space \(X\), we have to replace it by \(C\) being balanced; that is, \(\lambda x \in C\) for all \(x \in C\) and \(|\lambda|=1\). Hint: (i): Assume that \(C\) is open. If \(x \in C\), then also \(\delta x \in C\) for some small \(\delta>1\), hence \(x \in \frac{1}{\delta} C\) and \(\mu_{C}(x)<1\). Assume that \(C\) is closed. If \(\mu_{C}(x)=1\) then there are \(\lambda_{n}>1\) such that \(x \in \lambda_{n} C\) and \(\lambda_{n} \rightarrow 1\). Then \(\frac{1}{\lambda_{n}} x \rightarrow x\), and by convexity and closedness of \(C, x=\lim \left(\frac{1}{\lambda_{n}} x+\frac{1-\lambda_{n}}{\lambda_{n}} 0\right) \in C\). (ii): Find \(c>0\) such that \(\frac{1}{c} B_{X} \subset C\), then use previous exercises. (iii): Observing that \(\mu_{C}(-x)=\mu_{C}(x)\) and positive homogeneity are enough to prove \(\mu_{C}(\lambda x)=|\lambda| \mu_{C}(x)\) for all \(\lambda \in \mathbf{R}, x \in X\). (iv): From (iii) we already have the homogeneity and the triangle inequality. We must show that \(\mu_{C}(x)=0\) implies \(x=0\) (the other direction is obvious). Indeed, \(\mu_{C}(x)=0\) implies that \(x \in \lambda C\) for all \(\lambda>0\), which by the boundedness of \(C\) only allows for \(x=0\). In (ii) we proved \(\mu_{C}(x) \leq c\|x\| ;\) an upper estimate follows from \(C \subset\) \(d B_{X}\). The equivalence then implies completeness of the new norm.

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 \(\operatorname{dim}(X) \leq 4 n+1\)

Let \(\Gamma\) be a set and let \(p \in[1, \infty), q \in(1, \infty]\) be such that \(\frac{1}{p}+\frac{1}{q}=1\). Show that \(c_{0}(\Gamma)^{*}=\ell_{1}(\Gamma)\) and \(\ell_{p}(\Gamma)^{*}=\ell_{q}(\Gamma)\). Hint: See the proofs of Propositions \(2.14,2.15\), and \(2.16\).

Show that \(c^{*}\) is isometric to \(\ell_{1}\). Hint: We observe that \(c=c_{0} \oplus \operatorname{span}\\{e\\}\), where \(e=(1,1, \ldots)\) (express \(x=\left(\xi_{i}\right) \in c\) in the form \(x=\xi_{0} e+x_{0}\) with \(\xi_{0}=\lim _{i \rightarrow \infty}\left(\xi_{i}\right)\) and \(x_{0} \in\) \(\left.c_{0}\right)\). If \(u \in c^{*}\), put \(v_{0}^{\prime}=u(e)\) and \(v_{i}=u\left(e_{i}\right)\) for \(i \geq 1 .\) Then we have \(u(x)=u\left(\xi_{0} e\right)+u\left(x_{0}\right)=\xi_{0} v_{0}^{\prime}+\sum_{i=1}^{\infty} v_{i}\left(\xi_{i}-\xi_{0}\right)\) and \(\left(v_{1}, v_{2}, \ldots\right) \in \ell_{1}\) as in Proposition 2.14. Put \(\tilde{u}=\left(v_{0}, v_{1}, \ldots\right)\), where \(v_{0}=v_{0}^{\prime}-\sum_{i=1}^{\infty} v_{i}\), and write \(\tilde{x}=\left(\xi_{0}, \xi_{1}, \ldots\right) .\) We have \(u(x)=\xi_{0} v_{0}+\sum_{i=1}^{\infty} v_{i} \xi_{i}=\tilde{u}(\tilde{x})\) Conversely, if \(\tilde{u} \in \ell_{1}\), then the above rule gives a continuous linear functional \(u\) on \(c\) with \(\|u\| \leq\|\tilde{u}\|\), because \(|\tilde{u}(\tilde{x})| \leq\left(\sum_{i=0}^{\infty}\left|v_{i}\right|\right) \sup _{i \geq 0}\left|\xi_{i}\right|=\) \(\left|\tilde{u}\left\|\sup _{i \geq 0}\left|\xi_{i}\right|=\right\| \tilde{u}\left\|_{1}\right\| x \|_{\infty} .\right.\) The inequality \(\|\tilde{u}\| \leq\|u\|\) follows like this: Let \(\xi_{i}\) be such that \(\left|v_{i}\right|=\xi_{i} v_{i}\) if \(v_{i} \neq 0\) and \(\xi_{i}=1\) otherwise, \(i=0,1, \ldots\) Set \(x^{n}=\left(\xi_{1}, \ldots, \xi_{n}, \xi_{0}, \xi_{0}, \ldots\right) .\) Then \(\left\|x^{n}\right\|_{\infty}=1\) and \(\left|u\left(x^{n}\right)\right|=\left|\tilde{u}\left(\tilde{x}^{n}\right)\right| \geq\left|v_{0}\right|+\sum_{i=1}^{n}\left|v_{i}\right|-\sum_{i=n+1}^{\infty}\left|v_{i}\right| .\) Since \(\left|u\left(x^{n}\right)\right| \leq\|u\|\), we have \(\|u\| \geq\left|v_{0}\right|+\sum_{i=1}^{n}\left|v_{i}\right|-\sum_{i=n+1}^{\infty}\left|v_{i}\right| .\) By letting \(n \rightarrow \infty\), we get \(\|\tilde{u}\| \leq\|u\|\).

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