Trying to solve this problem:
Problem 8: a) Suppose $\mathcal{P}$ is a separating family of seminorms on a vector space $X$. Let $\mathcal{Q}$ be the smallest family of seminorms on $X$ that contains $\mathcal{P}$ and is closed under max. [This means: if $p1,p2 \in\mathcal{Q}$ and $p=\max(p1,p2)$ then $p \in Q$]. If the construction of Theorem 1.37 is applied to $\mathcal{P}$ and $\mathcal{Q}$ show that the two resulting topologies coincide. The main difference is that $\mathcal{Q}$ leads directly to a base, rather than a subbase. b) Suppose $\mathcal{Q}$ as in part (a) and $\Lambda$ is a linear functional on $X$. Show that $\Lambda$ is continuous if and only if there exists a $p\in Q$ such that $|\Lambda x|≤Mp(x)$ for all $x\in X$ and some constant $M < \infty$.
This is my attempt given my recent review of basis and subbasis of a topology.
a) About the topology $\mathcal{Q}$ we have the following
$$ \begin{array}{l} p \in \mathcal{P} \Rightarrow q \in \mathcal{Q} \\ p_1, p_2 \in \mathcal{Q} \Rightarrow \max \left\{p_1,p_2\right\} \in \mathcal{Q} \end{array} $$
So the seminorms on $\mathcal{Q}$ are defined recursively. We can easily prove that for each $q \in \mathcal{Q}$ we have $p_1,\ldots, p_n \in \mathcal{P}$ such that
$$ q = \max \left\{p_1, \ldots, p_n \right\} $$
To prove that $\mathcal{P}$ and $\mathcal{Q}$ generate the same topology it suffice to show they generate the same basis.
This is easy to show since if $V(q,m)$ is essentially a basis element
$$ V(q,m) = \left\{x : q(x) < \frac{1}{m} \right\} = \left\{ x : p_1(x),\ldots, p_n(x) < \frac{1}{m}, p_i \in \mathcal{Q} \right\} = \\\bigcap_{i=1}^n \left\{x : p_i(x) < \frac{1}{m} \right\} $$
So each $V(q,m)$ is actually a basis element generated by the subbasis defined by the seminorms of $\mathcal{P}$, hence they generate the same topology.
b) I'm stuck at the moment for this one. Suppose there are $M > 0$ and $p \in \mathcal{Q}$ such that for all $x \in X$
$$ \left| \Lambda x \right| \leq M p(x) \Leftrightarrow \left| \Lambda \left( \frac{x}{p(x)} \right) \right| \leq M $$
Since $p \left( \frac{x}{p(x)} \right)= 1$ and for $0 < \epsilon < 1$ we have $$ p \left(\frac{\epsilon x}{p(x)} \right) < 1 $$ also the $V(p,n)$ are absorbing we can easily pick a basis element (which is open) such that $\Lambda$ is bounded.
On the other way suppose $\Lambda$ is continuous then it is bounded in some neighborhood of $0$, which implies there's an element of the local base where $\Lambda$ is bounded. This means we have a $p \in \mathcal{Q}$ and a natural $n$ such that
$$ \left| \Lambda x \right| \leq M \; \text{for $x \in V(p,n)$} $$
I define now a map from $X$ to $V(p,n)$ as $x \to \frac{x}{2np(x)}$. This is well defined since
$$ p\left( \frac{x}{2np(x)} \right) = \frac{1}{2n} \frac{p(x)}{p(x)} = \frac{1}{2n} < \frac{1}{n}. $$
Therefore for each $x \in X$
$$ \left|\Lambda \left( \frac{x}{2np(x)} \right) \right| \leq M \iff \left|\Lambda \left( x \right) \right| \leq 2nM p(x) $$
defining $M' = 2nM$ I have
$$ \left|\Lambda \left( x \right) \right| \leq M' p(x) $$
which is the result.