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In many sources including Wikipedia, we see that there is an antilinear isomorphism between $H$ and its dual given by $\phi(y) = \langle \cdot, y \rangle$. Why not change the definition to put $y$ in the first slot instead, and thereby get a plain old linear isomorphism?

keej
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    Because then $\phi'(y)$ wouldn't be a linear functional, but an antilinear functional. (So you have a linear isomorphism between $H$ and its antidual.) – Daniel Fischer May 02 '17 at 18:32
  • Ah, I see. So there's no avoiding it. – keej May 02 '17 at 18:34
  • Not if you look at complex Hilbert spaces. Since the inner product is sesquilinear and not bilinear, the conjugation must appear somewhere if you work coordinate-free. If you fix a Hilbert basis, you can give a linear (unitary) isomorphism between $H$ and its dual, but it's not canonical, that depends on the choice of basis. – Daniel Fischer May 02 '17 at 19:00

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