Let $E$ and $F$ be complex Banach spaces. We denote by $\overline{E}$ the compex conjugate of $E$, that is, the vector space $E$ with the same norm but with the conjugate multiplication by a complex scalar. The identity map defines a natural antilinear isomorphism $ x\to \overline{x}$ from $E$ onto $\overline{E}$. Recall that if $S\colon E\to F$ is a linear map we can define $\overline{S}\colon \overline{E} \to \overline{F}$ by $\overline{S}(\overline{x})=\overline{S(x)}$.

Let $H$ be a complex Hilbert space. Let $T\colon H \to E$ be a bounded linear map. Do we have $$ \vert\vert T \vert\vert_{H \rightarrow E}^2=\vert\vert T\overline{T^*}\colon \overline{E^*} \rightarrow E\vert\vert\ ? $$ where $T^*\colon E^* \to H^*$ is the adjoint map.

  • $\begingroup$ What is $\overline{T^\ast}$? Or do you mean, $\overline{T\circ T^\ast}$ rather than $T\circ\overline{T^\ast}$? $\endgroup$ Jul 16 '12 at 10:18
  • $\begingroup$ The map $\overline{T^*}\colon \overline{E^*} \to \overline{H^*}=H$ is the conjugate of the adjoint of $T$. $\endgroup$
    – Zouba
    Jul 16 '12 at 10:35

This is true even in operator space setting. See proposition 7.2 in Introduction to operator space theory. G. Pisier


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