I'm reading lecture notes about analysis on infinite dimensional spaces and I ran into this exercise:

Every continuous linear functional $f\in (\mathbb{R}^\infty)^*$ is of the form

$$f(x)=\sum_n^N a_n x(n)$$

for some $(a_n)_{n=1}^N\in \mathbb{R}$. Thus the space can be identified with $c_{00}$, the space of real sequences that eventually are zero.

How do you prove such statements (All X are of the form Y) in general? if this implies the sets are bijective do I have to prove double sided inclusion or find an isomorphism?

  • $\begingroup$ Assume you have a linear functional not of that form. Show it is not continuous. There are infinitely many $n$ so that $f(e_n) \ne 0$ where $e_n$ is the function equal to $1$ in coordinate $n$ and $0$ elsewhere. The statement you quoted is not "bijective" it is only one direction (but the other direction is easy). $\endgroup$
    – GEdgar
    Apr 9 '19 at 14:14

Here $V=\mathbb{R}^{\infty}=\mathbb{R}^{\mathbb{N}}$ is equipped with the product topology. This means that as a topological vector space, $V$ is the locally convex space with topology defined by the seminorms $$ ||x||_n=|x_n| $$ with $n\in\mathbb{N}$. A linear form $f:V\rightarrow \mathbb{R}$ is continuous iff it is bounded by a finite positive linear combination of the above seminorms, i.e., iff there is exists a constant $K\ge 0$ and a set of indices $\{n_1,\ldots,n_p\}$ such that for all sequence $x\in V$, $$ |f(x)|\le K(||x||_{n_1}+\cdots||x||_{n_p})=K(|x_{n_1}|+\cdots+|x_{n_p}|)\ . $$ It is then immediate that $f(x)$ only depends on the entries of $x$ at the locations $n_1,\ldots,n_p$. One thus has $$ f(x)=a_1x_{n_1}+\cdots+ a_p x_{n_p} $$ for some constants $a_1,\ldots,a_p$.

BTW, $V$ is reflexive. The space $c_{00}=\oplus_{\mathbb{N}}\mathbb{R}$ seen as the strong dual of $V$ is such that $(c_{00})^{\ast}\simeq V$.

  • $\begingroup$ This is the family of semi norms we use when we talk about this space being Frechet? $\endgroup$
    – user515599
    May 20 '19 at 15:41
  • $\begingroup$ @badatmath: careful about the use of "the" for the family of seminorms. It is highly nonunique. Frechet means metrizable in a way that turns the space into a complete metric space. Metrizability means you can define the topology by a countable collection of seminorms. For example, the family I used $x\mapsto ||x||_n=|x_n|$ indexed by $n\in\mathbb{N}$ works. It is countable and defines the same topology as the metric $d(x,y)=\sum_{n}2^{-n}\min\{1,||x-y||_n\}$ for which $\mathbb{R}^{\mathbb{N}}$ is complete. $\endgroup$ May 20 '19 at 21:09

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