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let X be the space of bounded real sequences with sup norm.define a linear operator T:X$\to$X by

$$T(x) = (x_1,\frac{x_2}{2},\frac{x_3}{3},......) \forall x=(x_1,x_2,....)\in X$$

then,which one of the following is correct:

(1)T is bounded but not 1-1

(2)T is 1-1 but not bounded

(3)T is bounded and it's inverse exists and is not bounded

(4)T is bounded and it's inverse exists is bounded

my thought:

$$ ||T|| := \mathrm{sup}_{||x|| = 1} ||Tx|| < +\infty. $$

If $||x|| = 1$ then $|x_n| \leq 1$ for all $n \in \mathbb{N}$ and then $|(T(x_n))_m| = \left| \frac{x_n}{m^2} \right| \leq \frac{1}{m^2} \leq 1$ for all $m \in \mathbb{N}$ and thus $||T|| \leq 1$.so, T is bounded.

also,it is easily seen that T is 1-1

now,$T^{-1}$ is $T^{-1}(x_1,x_2,....)$=$(x_1,2x_2,3x_3,....)$

now, i am stuck in this step??how to check whether this operator is bounded or not??if possible,please explain....

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  • $\begingroup$ What do you mean by $T(x_n)$? It doesn't make sense because $T$ acts on sequences, but $x_n$ is only a term of a sequence (namely, the sequence $x$). $\endgroup$ Commented Jan 26, 2016 at 7:09
  • $\begingroup$ it is just a representation .. $\endgroup$ Commented Jan 26, 2016 at 7:31
  • $\begingroup$ What do you mean by that? Representation of what? $\endgroup$ Commented Jan 26, 2016 at 7:32
  • $\begingroup$ on each position of $x_n$ we have in range $\frac{x_n}{n}$,so,i just wrote it but technically it is wrong.. $\endgroup$ Commented Jan 26, 2016 at 7:38

1 Answer 1

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Take $e_n=(0,0,..,0,1,0,...)$ ($1$ in the $n$th-position). Then $\|T^{-1}e_n\|=n$.

Thus $T^{-1}$ is not bounded.

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  • $\begingroup$ but the answer given is (4) $\endgroup$ Commented Jan 26, 2016 at 7:08
  • $\begingroup$ Suppose it is bounded. Then, there is $M>0$ such that $\|T^{-1}x\|\le M$ for all $x$ with $\|x\|\le 1$. But $\|T^{-1}e_n\|=n$ and $\|e_n\|=1$ $\endgroup$
    – sinbadh
    Commented Jan 26, 2016 at 7:11
  • $\begingroup$ so,(4)is wrong,(3) is the correct option $\endgroup$ Commented Jan 26, 2016 at 7:13
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    $\begingroup$ Yes, (3) is the correct answer $\endgroup$
    – sinbadh
    Commented Jan 26, 2016 at 9:43

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