why $q\left(T\left(x-x_j\right)\right) \rightarrow 0$ implies $T\left(x_j\right) \rightarrow T(x)$?

155 Views Asked by At

Let $\mathscr{X}$ and $\mathscr{Y}$ be locally convex vector spaces, let $\mathscr{P}$ and $\mathscr{Q}$ be inducing collections of seminorms for $\mathscr{X}$, respectively $\mathscr{Y}$, and let $T: \mathscr{X} \rightarrow \mathscr{Y}$ be a linear map. lways continuous).

Now let $\left\{x_j\right\}_{j \in J}$ be a net in $\mathscr{X}$ and $x \in \mathscr{X}$ such that $x_j \rightarrow x$ in $\mathscr{X}$ in this thesis An Introduction to FUNCTIONAL SPACES by Marcel de Reus at page 158 in Lemma A.1.2 we have

To prove that $T$ is continuous, we should prove that $T\left(x_j\right) \rightarrow T(x)$ in $\mathscr{Y}$, which is equivalent to the statement that $q\left(T(x)-T\left(x_j\right)\right)=q\left(T\left(x-x_j\right)\right) \rightarrow 0$ in $\mathbb{R}$ for every $q \in \mathscr{Q}$, which in

Why $q\left(T\left(x-x_j\right)\right) \rightarrow 0$ implies $T\left(x_j\right) \rightarrow T(x)$ ?

This is what I tried.

$q\left(T(x)-T\left(x_j\right)\right)\rightarrow 0$ means that given $\epsilon >0$ there exist an $j_{\epsilon}$ such that $q\left(T(x)-T\left(x_j\right)\right)<\epsilon$ whenever $j>j_{\epsilon}$.

Now a neighborhood of $U$ of $T(x)$ is of the form $$U=\{ y:q_1(y-T(x))<\epsilon \wedge... q_k(y-T(x))<\epsilon \}$$

From the expression above we have that $T\left(x_j\right) \in U$ for $j>j_\epsilon$ wich is the definition of convergence to $T(x)$

I am not sure of proof because I am not sure if all my definition are correct.