Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I'm working on to prove the following: Show that the convergence in the space $\omega$ (space of all sequences with respect to the Fréchet metric) is the coordinate convergence. Any hint is appreciated, thanks.

share|improve this question
4  
Are you talking about the metric $d(x,y) = \sum_{n = 1}^{\infty} 2^{-n} \frac{|x_n - y_n|}{1 + |x_{n} - y_{n}|}$? –  t.b. Feb 11 '11 at 16:58
    
Isn't the "natural" metric on the space $\mathbb{R}^\mathbb{N}$ this one: $d(x,x) = 0$ and $d(x,y) = 1/N_{xy}$, where $N_{xy}$ is the first index where $x$ and $y$ differ –  kahen Feb 11 '11 at 21:05
    
@kahen: Something's wrong with your metric. It'll give the discrete topology on the subspace of sequences with only finitely many non-zero entries, so it can't be a topological vector space b/c scalar multiplication is not continuous. I don't know for sure what you have in mind, but your suggestion only works for pro-discrete things, I guess. –  t.b. Feb 11 '11 at 22:22
    
@matheww: Out of curiosity: who writes $w$ for the space of all sequences and why, I've never seen this before. –  t.b. Feb 11 '11 at 22:30
1  
Anyway, the original question seems to be pretty well answered by Wikipedia's article on Fréchet spaces. In particular check out the section on constructing Fréchet spaces and the bit in the Examples section on the family of seminorms on $\mathbb{R}^\omega$ –  kahen Feb 11 '11 at 22:47

1 Answer 1

In the following $\mathbb{K}$ always denotes a topological field. In most applications this will be the real or complex numbers.

Definition. A vector space with a Hausdorff topology over $\mathbb{K}$ such that the vector space operations are continuous is said to be a topological vector space.

Note in particular that any normed vector space or metric linear space (i.e. a metric space $(E,d)$ where $E$ is a vector space over the real or complex numbers such that $d$ is translation invariant and the vector space operations are continuous) is a TVS.

Definition. A family of seminorms $(\lVert \cdot \rVert_\alpha)_{\alpha \in A}$ on an $\mathbb{K}$-vector space $E$ is said to be a fundamental system of seminorms (or a sufficient family of seminorms) if $x \neq 0 \implies \exists \alpha \in A : \lVert x \rVert_\alpha \neq 0$.

Proposition. If $(\lVert \cdot \rVert_\alpha)$ is a fundamental system of seminorms on a $\mathbb{K}$-vector space $E$, then the topology generated by the open balls of the seminorms gives $E$ the structure of a TVS.

This is fairly easy to prove. Sufficiency is (of course) needed to prove that the topology is Hausdorff.

Proposition. If $(\lVert \cdot \rVert_n)$ is a countable fundamental system of seminorms on a real or complex vector space $E$, then $$d(x,y) = \sum_{j=1}^\infty \frac{\lVert x-y \rVert_j}{1 + \lVert x-y \rVert_j} 2^{-j} $$ defines a metric on $E$ that gives it the structure of a metric linear space.

Proposition. If $(\lVert \cdot \rVert_n)$ is a countable fundamental system of seminorms on a real or complex vector space $E$ such that for all sequences $(x_k)$ that are Cauchy w.r.t. each seminorm $\lVert\cdot\rVert_j$ there exists an $x \in E$ such that $\lVert x-x_k\rVert_j \to 0$ as $k \to \infty$ for all $j$, then the metric defined in the proposition above is complete, i.e. it gives $E$ the structure of a Fréchet space.

Of course checking that $\lVert (x_n) \rVert_j := \lvert x_j \rvert$ defines a countable fundamental system of seminorms on $\mathbb{C}^\mathbb{N}$ that satisfies the conditions of the above proposition is easy to do. Now proving that proposition on the other hand is probably going to be rather tedious but not terribly difficult since you have a lot to work with.

Sources: Wikipedia's article on Fréchet spaces & other (linked) articles and Meise & Vogt - Introduction to Functional Analysis.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.