Proving that a Banach space is of finite dimension

Let X be a Banach space with dimX≤∞. such that X∗ (the dual of X ) is finite dimensional. Then, show that X is of finite dimensional too and dimX=dimX∗. Note If wherenever we know that X is of finite dimensision, it is easy using linear algebra tools to show that dimX=dimX∗. Therefore the assumption … Read more

Inner product in dual space

I don’t understand this from a textbook. “Dual space could have an inner product that is induced from the vector space.” Suppose there is a vector space $V$. The inner product is determined $<v,v>=v^i v^i$ or in matrix form $$\begin{bmatrix} v^1 & v^2 & v^3\\ \end{bmatrix} \begin{bmatrix} v^1 \\ v^2 \\ v^3 \\ \end{bmatrix}$$ How … Read more

Is it true that ‖\|f\|^2 = \sup_{x\in E} \left [ 2 \langle f, x \rangle – |x|^2 \right ]?

Let (E, |\cdot|) be a normed linear space and (E’, \| \cdot \|) its dual. Then \|f\| := \sup_{x\in E} \frac{\langle f, x \rangle}{ |x|}, \quad \forall f \in E’. We have \left [ \frac{\langle f, x \rangle}{ |x|} \right ]^2 \ge 2 \langle f, x \rangle – |x|^2, \quad \forall f \in E’, x\in … Read more

What is the dual vector space of C\mathbb{C}\,?

Given that the complex numbers C are considered to be canonically isomorphic to R2, I am wondering if the dual space of C is the same es that of R, i.e. the vector space of linear functionals. I never heard about dual space of C until today, so I couldnt find much information in the … Read more

A lemma in functional analysis about distance

I want to prove the following result, I know it is correct: If $X$ is a Banach Space, $G \subset X$ is a closed linear subspace, $f \in X^*$ is a linear functional, then we have: \begin{equation*} \text{dist}(f,G^\perp) = \sup_{\|x\| \leq 1, x \in G}\left<f,x \right> \end{equation*} How should we do this? Is $G$ necessarily … Read more

Viewing tensors as multilinear maps, what is the difference between a vector and a covector?

I am currently taking a module on Differential Geometry, and as such I have been mostly looking at tensors as objects following the usual transformation rule. However I would like to understand how that is equivalent to viewing tensors as multilinear maps from direct products of Vector and their Dual spaces to the corresponding field … Read more

Evaluation under tensor products

Could anyone explain to me the highlighted step in this calculation? Can I just exaluate the tenor product component wise? The first step is basically (evV⊗idV)∘(idV(v)⊗coevV(v))=(evV⊗idV)(viei⊗(ϵj⊗ej)) and then the second equality would look like (evV⊗idV)(viei⊗(ϵj⊗ej))=(evV((viei⊗(ϵj⊗ej)))⊗idV((viei⊗(ϵj⊗ej)))). But I think I’m wrong there. Answer \DeclareMathOperator{\ev}{ev} \DeclareMathOperator{\coev}{coev} \DeclareMathOperator{\id}{id} \renewcommand{\eps}{\epsilon}Yes, you are indeed wrong. This is really confusing, so … Read more

Why is $V^{\vee}\otimes W^{\vee}\longrightarrow (V\otimes W)^{\vee}$ always injective?

Let $R$ be a commutative ring with $1$. For all $R$-modules $V,W$ we have a canonical $R$-linear map $V^{\vee}\otimes W^{\vee}\longrightarrow (V\otimes W)^{\vee}$ from tensor product of dual modules into the dual of tensor product. My question is, why is this map always injective? Answer This is false. For instance, let $k$ be a field and … Read more

Inclusion of dual space: X⊂Y⟹Y∗⊂X∗X\subset Y \implies Y^*\subset X^*

I’m confused at dual space of Hilbert space. Perhaps it is easy, but I don’t know, please help me. Let X, Y be Hilbert spaces and X∗, Y∗ are the dual spaces of X and Y respectively. Suppose that X⊂Y. Then, Y∗⊂X∗. This is correct according to books about functional analysis. But, I have no … Read more