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In general, bits and pieces of operator theory can be done in incomplete normed spaces, too, but it's usually a hassle and people don't care to check how much of the theory goes through without completeness. In practice, this is of limited interest. In this specific case, you can always view $A$ as an operator $\tilde A$ between the completions $\tilde E$ ...


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This is false. On $\ell^{2}$ let $\phi ((x_n))=(x_2,x_3,x_4,..)$. Then $\phi^{*}((x_n))=(0,x_1,x_2,...)$. If $x=(1,0,0,...)$ then $\|\phi (x)\|=0<1=\|\phi^{*}(x)\|$. But there is no $x$ such that $\|\phi^{*}(x)\| <\|\phi (x)\|$.


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