Let be $ T: l_{1} -> l_{1} $, $T(x_1,x_2,x_3,...)=(0,0,0,...x_1,0,0,...)$ .

( Coordinate x_1 is on n-place ). I proved that T is linear and bounded. But, I don't know how can prove that T is conjugate operator and what is $ \vert \vert T^{*} \vert \vert $ ? Please, help me.

  • $\begingroup$ By "conjugate" you mean "adjoint," right? Are you trying to show that $T$ is self-adjoint, or to find the adjoint of $T$? $\endgroup$ – Math1000 Nov 18 at 18:55
  • $\begingroup$ Yes, I mean adjoint. I need adjoint operator of operator T and his norm. $\endgroup$ – Math123a Nov 18 at 19:00
  • $\begingroup$ What is your norm on $\ell^1$? $\|x\| = \sum_{n=1}^\infty |x_n|$, correct? $\endgroup$ – Math1000 Nov 18 at 19:15
  • $\begingroup$ Norm is $ \vert\vert x\vert\vert = \Big( \sum\limits_{i=1}^{\infty} \vert x_i \vert\Big)$ $\endgroup$ – Math123a Nov 18 at 19:21
  • $\begingroup$ Recall that for a linear operator $T:X\to Y$ where $X,Y$ are Banach spaces, the adjoint operator $T^*$ is a map from $Y^*$ to $X^*$ where $Y^*$ and $X^*$ are the dual spaces of $Y$ and $X$, respectively. Now, what is the dual space of $\ell^1$, and what is its norm? $\endgroup$ – Math1000 Nov 18 at 19:22

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