2
$\begingroup$

Let $X$ be a locally convex space whose topology is defined by a family of seminorms $\mathcal{P}$. Let $f$ be a linear functional on $X$. Then, I am trying to prove that the following statements are equivalent.

  1. $f$ is continuous.
  2. There are seminorms $p_1,...,p_n\in \mathcal{P}$ and positive scalars $\alpha_1,...,\alpha_n$ such that $$|f(x)|\leq \sum_{k=1}^n\alpha_kp_k(x).$$

I am stuck in proving $1\Rightarrow 2$. Applying the definition, and that $f$ is continuous at $0$, I get that there exist seminorms $p_1,...,p_n$ and $\epsilon_1,...,\epsilon_n>0$, such that whenever $$x\in \bigcap_{k=1}^n\{y:p_k(y)<\epsilon_k\}\Rightarrow |f(x)|<1.$$ But I am not able to conjure up positive $\alpha_1,..,\alpha_n$ such that above inequality holds. A hint will be very appreciated. Thanks.

$\endgroup$
3
$\begingroup$

Since $f$ is continuous at $0$, the preimage $U=f^{-1}[(-1,1)]$ is open. By linearity, $0\in U$, so $U$ contains an open neighborhood of $0$. Since sets of the form $$V=\{x\in X: p_1(x),\ldots,p_n(x)<\epsilon\}$$ form a local basis for $0$, we have some such set $V\subseteq U$. Thus we have some $\epsilon>0$ and $p_1,\ldots,p_n\in \mathcal P$ such that $$\sum_{i=1}^n p_i(x)<\epsilon\implies p_1(x),\ldots,p_n(x)<\epsilon\implies |f(x)|<1$$ and for any $x\in X$ we have $$\sum_{i=1}^n p_i\left(\frac{\epsilon}{2\sum_{i=1}^n p_i(x)}x\right)=\frac{\epsilon}{2\sum_{i=1}^n p_i(x)}\sum_{i=1}^n p_i\left(x\right)=\epsilon/2$$ thus $$\frac{\epsilon}{2\sum_{i=1}^n p_i(x)}|f(x)|=\left|f\left(\frac{\epsilon}{2\sum_{i=1}^n p_i(x)}x\right)\right|<1$$ which gives us $$f(x)\leq \frac{2\sum_{i=1}^n p_i(x)}{\epsilon}=\sum_{i=1}^n\frac{2}{\epsilon}p_i(x).$$

Edit: If $\sum\limits_{i=1}^n p_i(x)=0$ for some $x\in V$, then for any $\lambda\in\mathbb R$ we have $$\sum\limits_{i=1}^n p_i(\lambda x)=|\lambda| \sum\limits_{i=1}^n p_i(x)=0$$ thus $\lambda x\in V$, so $|\lambda||f(x)|=|f(\lambda x)|<1$. Since $\lambda$ is arbitrary, this means $|f(x)|=0$.

$\endgroup$
  • $\begingroup$ Ohk, so the key step was to choose $p_i$'s in such a manner that $\sum_ip_i(x)<\epsilon$. Thanks for the help. $\endgroup$ – Nirakar Neo Jul 3 '12 at 4:10
  • $\begingroup$ just a small doubt - what if $\sum_{i=1}^np_i(x)=0$? how do we manage that? $\endgroup$ – Nirakar Neo Jul 3 '12 at 14:39
  • $\begingroup$ @NirakarNeo Good point. Editing to fix. $\endgroup$ – Alex Becker Jul 4 '12 at 0:44

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.