Question

Let X and Y be normed spaces, T∈B(X,Y) and (x_n) a sequence in X. If x_n→x₀, show that Tx_n→Tx₀.

Answer 1

ANSWER

Since $B(X,Y)$ is a space of linear bounded operators in which norma can be given $\left\| T \right\| =sup\left\{ { \left\| Tx \right\| }_{ Y } \right| x\in X,\ { \left\| x \right\| }_{ X }=1\} \quad ,$ then for $x$ non equal to zero ${ \left\| T\left( \frac { x }{ \left\| x \right\| } \right) \right\| }_{ Y }\le \ { \left\| T \right\| }_{ B(\{ ,Y) }$ . The mapping $T$ is a linear operator , so $T\left( \frac { x }{ \left\| x \right\| } \right) =\frac { 1 }{ \left\| x \right\| } Tx$ . Therefore, for all $x\in X$ the inequality

. . ${ \left\| Tx \right\| }_{ Y }\le \left\| T \right\| \cdot { \left\| x \right\| }_{ X }$ (1)

is true.

$T\left( { x }_{ n }-{x}_0\ \right) =T{ x }_{ n }-T{x}_0$ , ${ \left\| T{ x }_{ n }-T{x}_0 \right\| }_{ Y }={ \left\| T\left( { x }_{ n }-{x}_0\ \right) \right\| }_{ Y }$ . From (1) it follows

${ \left\| T\left( { x }_{ n }-{ x }_{ 0 }\ \right) \right\| }_{ Y }\le \left\| T \right\| \cdot { \left\| { x }_{ n }-{ { x }_{ 0 } } \right\| }_{ X }$ or

$0\le { \left\| T{ x }_{ n }-T{ x }_{ 0 } \right\| }_{ Y }\le \left\| T \right\| \cdot { \left\| { x }_{ n }-{ { x }_{ 0 } } \right\| }_{ X }$ (2)

By definition: ${ x }_{ n }\rightarrow { { x }_{ 0 } }\ \Leftrightarrow \ { \left\| { x }_{ n }-{ { x }_{ 0 } } \right\| }_{ X }\rightarrow 0$ .

By the Sandwich Theorem from (2) it follows ${ \left\| T{ x }_{ n }-T{ x }_{ 0 } \right\| }_{ Y }\rightarrow 0\quad$or $T{ x }_{ n }\rightarrow T{ { x }_{ 0 } }$ .