I am reading a section from the book "A Course in Functional Analysis" by J. B. Conway, and I came across this exercise.
Let $X$ and $Y$ be locally convex spaces and $T: X \rightarrow Y$ be a linear transformation. Then, $T$ is continuous if and only if for every continuous seminorm $p$ on $Y$, the seminorm $p \circ T$ is continuous on $X$.
For reference, we will use the following definitions:
Definition (Seminorm): Let $X$ be a vector space. A map $p : X \rightarrow \left[ 0, \infty \right)$ is a seminorm if
(1) For all $\alpha \in \mathbb{F}$ and for all $x \in X$, we have $p \left( \alpha x \right) = \left| \alpha \right| p \left( x \right)$.
(2) For all $x, y \in X$, we have $p \left( x + y \right) \leq p \left( x \right) + p \left( y \right)$.
Definition (Locally Convex Space): Let $X$ be a vector space and $\mathscr{P} = \left\lbrace p_{\alpha} \right\rbrace_{\alpha \in \Delta}$ be a family of seminorms on $X$. Then, we define a set $U \subseteq X$ to be open if and only if for each $x \in U$, there are $p_1, p_2, \cdots, p_n \in \mathscr{P}$ and $\epsilon_1, \epsilon_2 \cdots, \epsilon_n > 0$ such that $\bigcap\limits_{j = 1}^{n} \left\lbrace y \in X \mid p_j \left( x - y \right) < \epsilon_j \right\rbrace \subseteq U$.
Now, to prove the result stated above, one way is easy. If $T$ is continuous, then $p \circ T$ is also continuous for every continuous seminorm $p$ on $Y$. The map $p \circ T: X \rightarrow \left[ 0, \infty \right)$ is a seminorm because $T$ is linear.
However, the converse seems difficult. To prove that $T$ is continuous, I go with inverse images of open sets. So, we consider $V \subseteq Y$, an open set and then look at $T^{-1} \left( V \right) \subseteq X$. To prove that $T^{-1} \left( V \right)$ is open, we have to prove that for every $x \in T^{-1} \left( V \right)$, there are seminorms $p_1, p_2, \cdots, p_n$ on $X$ (through which the topology on $X$ is defined) and $\epsilon_1, \epsilon_2, \cdots, \epsilon_n > 0$ such that $\bigcap\limits_{j = 1}^{n} \left\lbrace y \in X \mid p_j \left( x - y \right) < \epsilon_j \right\rbrace \subseteq T^{-1} \left( V \right)$.
So, if we start with a point $x \in T^{-1} \left( V \right)$, we would mean that $Tx \in V$. Hence, we obtain seminorms $q_1, q_2, \cdots, q_n$ on $Y$ (through which the topology on $Y$ is defined) and $\epsilon_1, \epsilon_2, \cdots, \epsilon_n > 0$ such that $\bigcap\limits_{j = 1}^{n} \left\lbrace y \in Y \mid q_j \left( y - Tx \right) < \epsilon_j \right\rbrace \subseteq V$.
However, now I do know how to proceed! In particular, I know that somewhere I have to use the continuity of $p \circ T$, but where and how exactly is unknown to me. Any help in this regard will be appreciated!
Hints: Let $V$ be a convex balanced neighborhood of $0$ in $Y$. If we show that there exist a neighborhood $U$ of $0$ in $X$ such that $T(U) \subseteq V$ we can conclude that $T$ is continuous.
Let $p_V$ be the Minkowski functional of $V$. Then $P_V$ is a semi-norm, so $p_V \circ T$ is continuous. There exists an open set $U$ in $X$ containing $0$ such that $p_V(T(y))<1$ whenever $y \in U$. Use the definition of Minkowski functional to conclude that $Ty \in V$ whenever $y \in U$.