# A result on separable Fréchet spaces

In the proof of the following theorem:

if $X$ is a separable Fréchet space, then $\mathcal{E}(X)=\mathcal{B}(X)$.

Where $\mathcal{E}(X)$ is the $\sigma$-algebra generated by the cylindrical sets on $X$ and $\mathcal{B}(X)$ is the Borel $\sigma$-algebra.

They start by taking $(x_n)$ a dense sequence in $X$, and a family of seminorms $(p_k)$ which defines the topology of $X$. The part of the proof that I didn't get is the following:

By the Hahn-Banach theorem for every $n$ and $k$ there is $l_{n,k}\in X^*$ such that $p_k(x_n)=l_{n,k}(x_n)$ and $\sup\{l_{n,k}(x):\ p_k(x)\leq 1\}=1$. As a consequence, for every $x\in X$ and $k\in\mathbb{N}$ we have $p_k(x)=\sup_n\{l_{n,k}(x)\}$.

Questions:

*on which linear application did they apply the Hahn-Banach theorem ?

*what did they use to obtain $p_k(x)=\sup_n\{l_{n,k}(x)\}$ as a consequence ?

Define a linear functional $l_{n,k}(\alpha x_n) = \alpha p_k(x_n)$ on the line through $x_n$. Check that $l_{n,k}(\alpha x_n) \le p_k(\alpha x_n)$ for all $\alpha$.

Now extend $l_{n,k}$ to all of $X$ using Hahn Banach such that $l_{n,k}(x) \le p_k(x)$ for all $x$.

Now suppose $p_k(x) \le 1$, then we have $l_{n,k}(x) \le 1$. Furthermore, we have $l_{n,k}({1 \over p_k(x_n)} x_n) = {1 \over p_k(x_n)} p_k(x_n) = 1$, and so $\sup_{p_k(x) \le 1} l_{n,k}(x) = 1$.

For the last part, note that $l_{n,k}(x) \le p_k(x)$ for all $x$ and hence $\sup_n l_{n,k}(x) \le p_k(x)$ for all $x$. So, all that remains is to show equality.

Fix $x$ and note that $l_{n,k}(y) = l_{n,k}(x) + l_{n,k}(y-x) \le l_{n,k}(x) + p_k(y-x)$, or $l_{n,k}(y) -p_k(y-x) \le l_{n,k}(x)$.

Since $x_n$ is dense, there is some subsequence $x_{n_i} \to x$, so replacing $n$ by $n_i$ and $y$ by $x_{n_i}$ we get $l_{n_i,k}(x_{n_i}) -p_k(x_{n_i}-x) = p_k(x_{n_i})-p_k(x_{n_i}-x) \le l_{n_i,k}(x) \le \sup_n l_{n,k}(x)$

Since $p_k(x_{n_i}) \to p_k(x)$, $p_k(x_{n_i}-x) \to 0$, we get $p_k(x) \le \sup_n l_{n,k}(x)$.

• I get it now ! but can you please tell me why $\sup_{p_k(x) \le 1} l_{n,k}(x) = 1$ is considered for $x$ and not for $x_n$ . – houda Nov 17 '15 at 18:32
• I don't understand your question. $x_n$ is a specific point, the $\sup$ is taken over all $x$ satisfying $p_k(x) \le 1$. – copper.hat Nov 17 '15 at 18:33
• I thought that $x$ may be the limit of $x_n$ since it's a dense sequence. – houda Nov 17 '15 at 18:35
• Not in this case, the $\sup$ is over all $x$ such that $p_k(x) \le 1$ (and this includes the point ${1 \over p_k(x_n) } x_n$). – copper.hat Nov 17 '15 at 18:35
• but in the last consequence $p_k(x)=\sup_n\{l_{n,k}(x)\}$ they used the limit isn't it ? :/ – houda Nov 17 '15 at 18:41