[proofplan] The forward implication is immediate: any continuous map to $\mathbb{F}$ has closed level sets, in particular $\ker(f) = f^{-1}(\{0\})$ is closed. For the reverse, we use the lcs structure: when $\ker(f)$ is closed and proper, basic open neighbourhoods of the form $U = \{x : p_k(x) \le \varepsilon, 1 \le k \le n\}$ separate a chosen point $x_0 \notin \ker(f)$ from $\ker(f)$. We then exploit the fact that $f(U)$ is a convex balanced subset of $\mathbb{F}$, hence either bounded or all of $\mathbb{F}$. The "all of $\mathbb{F}$" alternative would contradict the separation, so $f(U)$ is bounded. Boundedness on a basic neighbourhood of $0$ is exactly the seminorm bound that the [Continuity of Linear Maps Between Locally Convex Spaces](/theorems/2636) requires for continuity. [/proofplan] [step:Reduce to the case $\ker(f) \ne X$ and dispose of the easy direction] If $f$ is continuous, then $\{0\} \subset \mathbb{F}$ is closed and $f$ is continuous, so $\ker(f) = f^{-1}(\{0\})$ is closed. For the converse, suppose $\ker(f)$ is closed. If $\ker(f) = X$, then $f \equiv 0 \in X^*$. Henceforth assume $\ker(f) \ne X$ and pick a point $x_0 \in X \setminus \ker(f)$, so $f(x_0) \ne 0$. [/step] [step:Choose a basic open neighbourhood $U$ of $0$ separating $x_0$ from $\ker(f)$] Since $\ker(f)$ is closed and $x_0 \notin \ker(f)$, the point $x_0$ has an open neighbourhood disjoint from $\ker(f)$. By the definition of the lcs topology generated by the family $\mathcal{P}$, every neighbourhood of $x_0$ contains a translate of a basic open neighbourhood of $0$. Therefore there exist $\varepsilon > 0$, $n \in \mathbb{N}$, and $p_1, \ldots, p_n \in \mathcal{P}$ such that with \begin{align*} U := \{x \in X : p_k(x) \le \varepsilon, \; 1 \le k \le n\}, \end{align*} we have $(x_0 + U) \cap \ker(f) = \varnothing$. [guided] Why does separation of $x_0$ from $\ker(f)$ follow from closedness? The complement $X \setminus \ker(f)$ is open and contains $x_0$. By the definition of the lcs topology, the sets $\{x : p_k(x - y) \le \varepsilon, \; 1 \le k \le n\}$ for $y \in X$, $\varepsilon > 0$, finite collections $p_1, \ldots, p_n \in \mathcal{P}$ form a neighbourhood base. So there is a basic neighbourhood \begin{align*} x_0 + U = \{x : p_k(x - x_0) \le \varepsilon, \; 1 \le k \le n\} \subseteq X \setminus \ker(f), \end{align*} which is exactly the statement $(x_0 + U) \cap \ker(f) = \varnothing$. We will use this set $U$ as our test neighbourhood: bounding $f$ on $U$ is exactly what the seminorm characterisation of continuity from the [Continuity of Linear Maps Between Locally Convex Spaces](/theorems/2636) requires. [/guided] [/step] [step:Show that $f(U) \subset \mathbb{F}$ is convex and balanced] The seminorm conditions $p_k(x) \le \varepsilon$ are convex and symmetric under $x \mapsto \lambda x$ for $|\lambda| \le 1$ (since $p_k(\lambda x) = |\lambda| p_k(x) \le p_k(x) \le \varepsilon$). Therefore $U$ is convex and balanced (i.e. $\lambda U \subseteq U$ for $|\lambda| \le 1$). The image $f(U)$ inherits these properties: linearity of $f$ preserves convex combinations, so $f(U)$ is convex; and $f(\lambda U) = \lambda f(U) \subseteq f(U)$ for $|\lambda| \le 1$, so $f(U)$ is balanced. [/step] [step:Conclude that $f(U)$ is bounded by ruling out $f(U) = \mathbb{F}$] A convex balanced subset of $\mathbb{F}$ is either bounded or equal to $\mathbb{F}$: indeed, in $\mathbb{R}$ it is an interval containing $0$ symmetric about $0$, so it is either $(-r, r)$, $[-r, r]$, or $\mathbb{R}$ for some $r \in [0, \infty]$; in $\mathbb{C}$ a convex balanced set containing some $z$ contains the whole disc of radius $|z|$, so if it is unbounded it contains every disc, hence equals $\mathbb{C}$. Suppose for contradiction that $f(U) = \mathbb{F}$. Then in particular $-f(x_0) \in f(U)$, so there exists $u \in U$ with $f(u) = -f(x_0)$, giving $f(x_0 + u) = 0$, i.e. $x_0 + u \in \ker(f)$. But $x_0 + u \in x_0 + U$, so $x_0 + u \in (x_0 + U) \cap \ker(f) = \varnothing$, a contradiction. Hence $f(U)$ is bounded: there exists $C > 0$ such that $|f(x)| \le C$ for all $x \in U$. [guided] The dichotomy "bounded or everything" for convex balanced subsets of the scalar field is the geometric heart of the argument. Why is it true? In $\mathbb{R}$, a convex balanced set is symmetric about $0$ and convex, so it is a (possibly unbounded) interval $(-r, r)$, $[-r, r]$, $(-\infty, \infty)$, or similar — its sup of $|t|$ is either finite or $\infty$. In $\mathbb{C}$, balancedness means closed under multiplication by the closed unit disc, so containing any non-zero $z$ forces the entire closed disc of radius $|z|$ to lie in the set; if no upper bound on $|z|$ exists, the set is all of $\mathbb{C}$. Now why does $f(U) = \mathbb{F}$ contradict separation? If $f(U) = \mathbb{F}$, then in particular the value $-f(x_0)$ (which is non-zero by our choice of $x_0$) is hit by $f$ somewhere on $U$: pick $u \in U$ with $f(u) = -f(x_0)$. Then $f(x_0 + u) = f(x_0) - f(x_0) = 0$, so $x_0 + u \in \ker(f)$. But the choice of $U$ was precisely so that $x_0 + U$ avoids $\ker(f)$. This contradiction forces $f(U)$ to be bounded — say $|f(x)| \le C$ for $x \in U$. [/guided] [/step] [step:Convert the bound on $U$ to a seminorm estimate via scaling] Set $p(x) := \max_{1 \le k \le n} p_k(x)$. We claim $|f(x)| \le (C/\varepsilon) \, p(x)$ for all $x \in X$. Fix $x \in X$. If $p(x) > 0$, then for $y := \varepsilon x / p(x)$ we have $p_k(y) = \varepsilon p_k(x)/p(x) \le \varepsilon$ for each $k$, so $y \in U$ and $|f(y)| \le C$, giving $|f(x)| = (p(x)/\varepsilon)|f(y)| \le (C/\varepsilon) \, p(x)$. If $p(x) = 0$, then for every scalar $\lambda$ we have $p_k(\lambda x) = |\lambda| p_k(x) = 0 \le \varepsilon$, so $\lambda x \in U$, hence $|\lambda||f(x)| = |f(\lambda x)| \le C$. Letting $|\lambda| \to \infty$ forces $f(x) = 0 = (C/\varepsilon) \cdot 0 = (C/\varepsilon) p(x)$. Both cases give $|f(x)| \le (C/\varepsilon) p(x)$. [/step] [step:Invoke the seminorm characterisation of continuity] We have shown \begin{align*} |f(x)| \le (C/\varepsilon) \cdot \max_{1 \le k \le n} p_k(x) \quad \text{for all } x \in X, \end{align*} with the constant $C/\varepsilon \ge 0$, the integer $n \in \mathbb{N}$, and seminorms $p_1, \ldots, p_n \in \mathcal{P}$ from above. This is condition (iii) of the [Continuity of Linear Maps Between Locally Convex Spaces](/theorems/2636) (taking $Y = \mathbb{F}$ with the single seminorm $q(t) = |t|$, which generates the standard topology on $\mathbb{F}$). That theorem then yields $f \in X^*$. [guided] At this point we have repackaged the geometric separation into the clean inequality $|f(x)| \le (C/\varepsilon) \max_k p_k(x)$. The [Continuity of Linear Maps Between Locally Convex Spaces](/theorems/2636) characterises continuity of a linear map $T: (X, \mathcal{P}) \to (Y, \mathcal{Q})$ by exactly such estimates: for each $q \in \mathcal{Q}$ there must exist a constant and finitely many $p_1, \ldots, p_n \in \mathcal{P}$ controlling $q \circ T$. Here $Y = \mathbb{F}$, equipped with the single seminorm $q(t) := |t|$ (which is the standard absolute value on $\mathbb{F}$ and generates the usual topology — its open balls $\{|t| < r\}$ are exactly the standard open discs/intervals around $0$). So the hypothesis of the theorem reduces to producing a single estimate of the form $|f(x)| \le C' \max_k p_k(x)$ for some constant $C' \ge 0$. We have produced such an estimate with $C' = C/\varepsilon$. Hence $f$ is continuous, i.e. $f \in X^*$. [/guided] [/step]