[proofplan] We prove the equivalences in the cycle (i) $\Rightarrow$ (ii) $\Rightarrow$ (iii) $\Rightarrow$ (ii) $\Rightarrow$ (i). The implication (i) $\Rightarrow$ (ii) is immediate from the definition of continuity. The implication (ii) $\Rightarrow$ (i) uses linearity of $T$ together with translation-invariance of the lcs topologies (continuity of vector addition). The substantive direction (ii) $\Rightarrow$ (iii) extracts a quantitative seminorm bound from the topological condition by using that the open seminorm-balls form a neighbourhood basis at $0$ in an lcs, and exploits homogeneity of $T$ to scale arbitrary $x$ into the basic neighbourhood. The reverse (iii) $\Rightarrow$ (ii) packages the seminorm bounds into a basic open neighbourhood that maps into a prescribed one. [/proofplan] [step:Establish (i) $\Leftrightarrow$ (ii) using translation invariance] The implication (i) $\Rightarrow$ (ii) is immediate: continuity at every point includes continuity at $0$. For (ii) $\Rightarrow$ (i), fix any $x_0 \in X$ and a neighbourhood $V$ of $T(x_0)$ in $Y$. The lcs topology on $Y$ is translation-invariant (vector addition $Y \times Y \to Y$ is jointly continuous in the lcs topology), so $V - T(x_0)$ is a neighbourhood of $0$ in $Y$. By continuity of $T$ at $0$, there exists a neighbourhood $U_0$ of $0$ in $X$ with $T(U_0) \subseteq V - T(x_0)$. By translation invariance of the lcs topology on $X$, $U := x_0 + U_0$ is a neighbourhood of $x_0$. For every $x \in U$, write $x = x_0 + u$ with $u \in U_0$; by linearity of $T$, \begin{align*} T(x) = T(x_0) + T(u) \in T(x_0) + (V - T(x_0)) = V. \end{align*} Hence $T(U) \subseteq V$, showing $T$ is continuous at $x_0$. Since $x_0$ was arbitrary, $T$ is continuous on $X$. [guided] The reduction "continuity at $0$ $\Rightarrow$ continuity everywhere" is standard for linear maps between topological vector spaces, but it has two ingredients that are worth naming explicitly: 1. **Linearity of $T$** lets us decompose $T(x_0 + u) = T(x_0) + T(u)$, so the increment $T(x) - T(x_0)$ depends only on the increment $x - x_0 = u$, not on $x_0$ itself. 2. **Translation invariance** of the topologies on $X$ and $Y$ — i.e. the maps $x \mapsto x + a$ are homeomorphisms — is what lets us translate neighbourhoods of $0$ into neighbourhoods of $x_0$ in $X$ and into neighbourhoods of $T(x_0)$ in $Y$ without losing the topological information. Both are non-trivial: a non-linear map need not have continuity at $0$ implying continuity everywhere; and a non-translation-invariant topology (e.g. one that distinguishes the origin) need not satisfy this either. Locally convex spaces always have translation-invariant topologies because the seminorm-balls $\{x : p(x - x_0) < \varepsilon\}$ form a basis at $x_0$ and are clearly translates of those at $0$. [/guided] [/step] [step:Establish (ii) $\Rightarrow$ (iii) by extracting a seminorm bound from a basic neighbourhood] Assume $T$ is continuous at $0$ and fix $q \in \mathcal{Q}$. Define \begin{align*} V := \{y \in Y : q(y) < 1\}, \end{align*} which is open in $Y$ in the lcs topology by definition of the topology generated by $\mathcal{Q}$, and contains $0$ since $q(0) = 0$. By continuity of $T$ at $0$, the preimage $T^{-1}(V)$ is open in $X$ and contains $0$, so it contains a basic open neighbourhood of $0$. By the definition of the lcs topology on $X$, there exist $\varepsilon > 0$, $n \in \mathbb{N}$, and $p_1, \ldots, p_n \in \mathcal{P}$ with \begin{align*} U := \{x \in X : p_k(x) < \varepsilon \text{ for all } 1 \le k \le n\} \subseteq T^{-1}(V). \end{align*} Define $p: X \to [0, \infty)$ by $p(x) = \max_{1 \le k \le n} p_k(x)$ (a seminorm, as the maximum of finitely many seminorms). Then $U = \{x : p(x) < \varepsilon\}$, and the inclusion $U \subseteq T^{-1}(V)$ states \begin{align*} p(x) < \varepsilon \implies q(T(x)) < 1. \tag{$\star$} \end{align*} We use $(\star)$ together with positive homogeneity of $T$, $p$, and $q$ to derive the global bound. Two cases. **Case 1: $p(x) > 0$.** For any $\delta > 0$, the element $\frac{\varepsilon}{p(x) + \delta} x$ has \begin{align*} p\!\left(\frac{\varepsilon}{p(x) + \delta} x\right) = \frac{\varepsilon}{p(x) + \delta} p(x) < \varepsilon \end{align*} (the equality uses $\frac{\varepsilon}{p(x)+\delta} > 0$ and positive homogeneity of $p$; the inequality uses $p(x) < p(x) + \delta$). Applying $(\star)$ and using positive homogeneity of $q$ and linearity of $T$: \begin{align*} q(T(x)) = \frac{p(x) + \delta}{\varepsilon} \cdot q\!\left( T\!\left( \frac{\varepsilon}{p(x) + \delta} x \right)\right) < \frac{p(x) + \delta}{\varepsilon}. \end{align*} Letting $\delta \downarrow 0$, $q(T(x)) \le p(x)/\varepsilon$. **Case 2: $p(x) = 0$.** For every $t > 0$, the scaled element $tx$ has $p(tx) = t p(x) = 0 < \varepsilon$, so $(\star)$ gives $q(T(tx)) < 1$. By linearity of $T$ and positive homogeneity of $q$, $q(T(tx)) = t q(T(x))$, so $t q(T(x)) < 1$ for every $t > 0$. Letting $t \to \infty$ forces $q(T(x)) = 0$. The bound $q(T(x)) \le p(x)/\varepsilon$ then holds with both sides equal to $0$. In either case, with $C := 1/\varepsilon$, \begin{align*} q(T(x)) \le C \cdot \max_{1 \le k \le n} p_k(x) \qquad \text{for all } x \in X, \end{align*} which is condition (iii). [guided] We have a topological condition — $T$ is continuous at $0$ — and we want to extract a quantitative inequality. The bridge is that in an lcs, the topology has a **basis of open sets of a specific form**: finite intersections of seminorm balls. Concretely, every open neighbourhood of $0$ in $X$ contains a set of the form $\{x : p_1(x) < \varepsilon, \ldots, p_n(x) < \varepsilon\}$ for some seminorms $p_1, \ldots, p_n \in \mathcal{P}$ and some $\varepsilon > 0$. (One could allow different $\varepsilon_k$ for each $p_k$, but taking the smallest yields a single $\varepsilon$.) So the topological condition $T(U_0) \subseteq V$ for some neighbourhood $U_0$ of $0$ in $X$ reduces to: for some basic neighbourhood $U = \{x : \max_k p_k(x) < \varepsilon\}$, we have $T(U) \subseteq V = \{y : q(y) < 1\}$. This is the implication $(\star)$: a uniform bound on $\max_k p_k$ implies a uniform bound on $q \circ T$. The remaining work is to globalise: $(\star)$ only controls $T$ on the small set $\{p < \varepsilon\}$, but we want a bound for every $x \in X$. The trick is **scaling**. For any $x$ with $p(x) > 0$, we can shrink it to $\lambda x$ with $\lambda$ small enough that $\lambda x \in U$, namely $\lambda < \varepsilon / p(x)$. Apply $(\star)$ to $\lambda x$ to bound $q(T(\lambda x))$, and then **unscale** using linearity of $T$ and positive homogeneity of $q$: \begin{align*} q(T(\lambda x)) = q(\lambda T(x)) = \lambda q(T(x)). \end{align*} This rearranges to $q(T(x)) = \lambda^{-1} q(T(\lambda x)) < \lambda^{-1}$, and choosing $\lambda$ close to $\varepsilon / p(x)$ gives the bound $q(T(x)) \le p(x)/\varepsilon$. The case $p(x) = 0$ is the corner where $x$ lies in the kernel of every $p_k$. Then we can scale $x$ by **any** positive $t$ without leaving $U$ — the entire ray $\{tx : t > 0\}$ is in $U$. This forces $q(T(x))$ to be smaller than every positive number, hence zero. So the bound $q(T(x)) \le p(x)/\varepsilon$ holds with both sides zero. [/guided] [/step] [step:Establish (iii) $\Rightarrow$ (ii) by lifting seminorm bounds to neighbourhood inclusions] Assume (iii). To verify continuity of $T$ at $0$, let $V \subseteq Y$ be an arbitrary neighbourhood of $0$. By definition of the lcs topology on $Y$, $V$ contains a basic neighbourhood: there exist $m \in \mathbb{N}$, $q_1, \ldots, q_m \in \mathcal{Q}$, and $\delta > 0$ with \begin{align*} \{y \in Y : q_j(y) < \delta \text{ for all } 1 \le j \le m\} \subseteq V. \end{align*} Apply (iii) to each $q_j$: there exist $C_j \ge 0$, $n_j \in \mathbb{N}$, and seminorms $p_{j,1}, \ldots, p_{j,n_j} \in \mathcal{P}$ such that \begin{align*} q_j(T(x)) \le C_j \cdot \max_{1 \le k \le n_j} p_{j,k}(x) \qquad \forall x \in X. \end{align*} Set $\varepsilon := \min_{1 \le j \le m} \delta / (C_j + 1) > 0$ (the $+1$ avoids division by zero in case some $C_j = 0$, and we use $C_j + 1 \ge C_j$ throughout). Define \begin{align*} U := \{x \in X : p_{j,k}(x) < \varepsilon \text{ for all } 1 \le j \le m, 1 \le k \le n_j\}, \end{align*} which is a basic open neighbourhood of $0$ in $X$ (a finite intersection of seminorm balls). For $x \in U$ and each $j$: \begin{align*} q_j(T(x)) \le C_j \cdot \max_{1 \le k \le n_j} p_{j,k}(x) < C_j \varepsilon \le C_j \cdot \frac{\delta}{C_j + 1} \le \delta. \end{align*} Hence $T(U) \subseteq \{y : q_j(y) < \delta \text{ for all } j\} \subseteq V$, showing $T$ is continuous at $0$. [guided] Going from (iii) back to (ii) is the easier direction: a quantitative bound is generally stronger than a topological neighbourhood condition, so we just have to package the bounds correctly. The strategy is to take an arbitrary neighbourhood $V$ of $0$ in $Y$, refine it to a basic neighbourhood specified by finitely many seminorms $q_1, \ldots, q_m$ and a single threshold $\delta$, and then use (iii) **on each $q_j$ separately** to manufacture seminorms on $X$ that control $q_j(T(\cdot))$. The final neighbourhood $U$ is the joint intersection of all these seminorm balls. Why is the choice $\varepsilon = \min_j \delta/(C_j + 1)$? Because we need \begin{align*} C_j \varepsilon \le \delta \qquad \text{for every } j, \end{align*} which forces $\varepsilon \le \delta/C_j$ for every $j$. Taking the minimum over $j$ ensures all bounds hold simultaneously. The $C_j + 1$ in the denominator (rather than $C_j$) is a defensive choice to handle the degenerate case $C_j = 0$ — there, the bound $q_j(T(x)) \le 0$ would already be trivial, but the $+1$ keeps the formula uniform. The strict inequality $q_j(T(x)) < C_j \varepsilon$ comes from the strict inequality $p_{j,k}(x) < \varepsilon$ used in defining $U$ — combined with the bound $q_j(T(x)) \le C_j \max_k p_{j,k}(x)$, which is strict on the strict inequality side. [/guided] [/step] [step:Conclude the equivalence] Step 1 established (i) $\Leftrightarrow$ (ii). Step 2 established (ii) $\Rightarrow$ (iii). Step 3 established (iii) $\Rightarrow$ (ii). Concatenating, the three statements (i), (ii), (iii) are equivalent. [/step]