Continuous Linear Images of Bounded Sets in a LCTVS

# A Continuous Linear Image of a Bounded Set is a Bounded Set in a LCTVS

Proposition 1: Let $E$ and $F$ be locally convex topological vector spaces. If $A \subseteq E$ is bounded and $t : E \to F$ is continuous then $t(A)$ is bounded. |

**Proof:**Let $A \subseteq E$ be bounded and let $t : E \to F$ be continuous.

- Let $\mathcal V$ be a base of absolutely convex neighbourhoods of $o_F$ and let $V \in \mathcal V$. Since $t$ is continuous, $t^{-1}(V)$ is an absolutely convex neighbourhood of $o_E$.

- Indeed, $t^{-1}(V)$ is absolutely convex, for if $x_1, x_2 \in t^{-1}(V)$ and $\lambda, \mu \in \mathbf{F}$ are such that $|\lambda| + |\mu| \leq 1$, then $t(x_1), t(x_2) \in V$, so that by the absolute convexity of $V$, we have that $\lambda t(x_1) + \mu t(x_2) \in V$. So by linearity of $t$, $t(\lambda x_1 + \mu x_2) \in V$, so that $\lambda x_1 + \mu x_2 \in t^{-1}(V)$.

- Since $A$ is bounded and $t^{-1}(V)$ is an absolutely convex neighbourhood of the origin, there exists a $\lambda > 0$ such that:

\begin{align} \quad A \subseteq \lambda t^{-1}(V) \end{align}

- Hence $\lambda^{-1} A \subseteq t^{-1}(V)$, so that $t(\lambda^{-1}A) \subseteq V$, or equivalently, $t(A) \subseteq \lambda V$. So $t(A)$ is bounded. $\blacksquare$