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 $p_k(x)=\sup_n\{l_{n,k}(x)\}$.


*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 ?

Thank you for your time.


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) $.

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.