- #76

member 587159

By linearity of [itex]T[/itex], it suffices to show [itex]T[/itex] is continuous at [itex]0[/itex]. Take [itex](x_n) \subseteq H_1[/itex] such that [itex]x_n \xrightarrow[n\to\infty]{}0\in H_1[/itex]. Let [itex]z\in H_1[/itex], then

[tex]

\langle z,Tx_n \rangle = \langle Sz, x_n \rangle \xrightarrow[n\to\infty]{}0 \in \mathbb K.

[/tex]

As [itex]z[/itex] is arbitrary it implies [itex]Tx_n\xrightarrow[n\to\infty]{}0 \in H_2[/itex].

I think you mean ##z\in H_2##.

Please justify the step

$$\forall z \in H_2: \langle z, Tx_n\rangle\to 0\implies T x_n\to 0$$

PS: Glad to see you back here!