# Every continuous linear functional in $(\mathbb{R}^\infty)^*$ is of the form $\sum_n a_n x(n)$

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?

• 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). 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 $$(c_{00})^{\ast}\simeq V$$.
• @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. May 20 '19 at 21:09