Let $X$ be a real normed linear space of all real sequences which are eventually zero with the 'sup' norm and $T:X \to X$ be a bijective linear operator defined by $$T(x_1,x_2,x_3,....)=\left(x_1,\frac{x_2}{2^2},\frac{x_3}{3^2},....\right)$$ How to check whether $T$ and $T^{-1}$ is bounded or not ?

$$\left\lVert Tx\right\rVert=\sup \Big\{\vert x_1 \vert,\frac{\vert x_2 \vert}{2^2},...\Big\}=\sup_n\Big\{\frac{\vert x_n \vert}{n^2}\Big\} \leq \sup_n\Big\{\frac{\vert x_n \vert}{n}\Big\}$$

How to make the RHS of above in the form $K \vert\vert x \vert \vert$ if possible ? Any hint ?

On the otherhand, $T^{-1}:X \to X$ is a map by $$T^{-1}(x_1.x_2,...)=(x_1,2^2x_2,3^2x_3,...)$$

$$\left\lVert T^{-1}x\right\rVert=\sup_n\Big\{n^2 \vert x_n \vert\Big\} \geq n$$ so $T^{-1}$ is not bounded. Am I right? Any help ?


1 Answer 1

  1. $\sup_n \{\frac{|x_n|}{n}\} \le \sup_n\{|x_n|\} =||x||.$
  2. Your considerations concerning $T^{-1}$ are correct.
  • $\begingroup$ Thanks! Is there any text book which deals with lot of this type of problems ? $\endgroup$ Commented Jan 22, 2019 at 6:43
  • $\begingroup$ Why no response? Is I'm asking anything wrong? $\endgroup$ Commented Jan 22, 2019 at 7:06

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