Equivalence of continuity of a functional on a locally convex space 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. 


*

*$f$ is continuous.

*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.
 A: 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 $|f(x)|=0$.
