why is this bounded linear map not invertible

76 Views Asked by At

let $T:L^2 \to L^2$ be a bounded linear map st for all $x=x_1,x_2,x_3...x_n$ which are elements of $L^2$,

$$Tx=(x_1,x_2/\sqrt 2,x_3/\sqrt 3,....x_n/\sqrt n)$$

show that this is not invertible

I dont understand why this isn't invertible , i assumed

$$T(x)=(x_1,\sqrt 2 x_2, \sqrt 3 x_3.. . etc)$$

would be the inverse.

I know that i need to show its not surjective to prove its not invertible, ie for some $T:u \to v$, there exists a $v \in V$ st $T(u)=v$ but how to I apply this when $L^2 \to L^2$?

1

There are 1 best solutions below

0
On

Let $a_n=\frac 1 n$. Then $(a_n)\in \ell^{2}$. Suppose $(a_n)=T(x_n)$. Then $\frac 1 n =\frac {x_n} {\sqrt n}$ so $x_n=\frac 1 {\sqrt n}$. But then $\sum x_n^{2}=\infty$, a contradiction. So $T$ is not surjective.