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Suppose $X$ is normed complex space and $h:X\to \mathbb{R}$ is bounded linear functional (real). Prove that $f:X\to \mathbb{C}$ defined by $f(x)=h(x)-ih(ix)$ belongs to the dual space of $X$ and $\left \| f \right \|=\left \| h \right \|$.

I proved that $f$ belongs to the dual space of $X$, however I couldn't show the equality $\left \| f \right \|=\left \| h \right \|$.

I'd appreciate a hint on how to show this. Thanks.

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  • $\begingroup$ Please include your work showing that $f$ is in the dual. Did you already get $||f|| \leq ||h||?$ $\endgroup$ – Vishal Gupta Sep 4 '13 at 2:44
  • $\begingroup$ I can show only the opposite direction. Any hint? $\endgroup$ – Galkina Sep 4 '13 at 3:31
  • $\begingroup$ How did you show $f$ is in the dual? $\endgroup$ – Vishal Gupta Sep 4 '13 at 7:12
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Showing that $\|f\|\leq\|h\|$ is the only part you seem to be stuck on.

For $x\in X$, write $f(x)=re^{i\theta}$. Then $|f(x)|=f(e^{-i\theta}x)=r=h(e^{-i\theta}x)=|h(e^{-i\theta}x)|\leq\|h\|\|e^{-i\theta}x\|=\|h\|\|x\|$.

(I probably read this trick in Douglas's Banach algebra techniques in operator theory.)

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