Let $(V, (\cdot, \cdot))$ be a complex inner product space, say a space of complex-valued functions, with $(\cdot, \cdot)$ linear in the second position and sesquilinear in the first. Assume that $V$ is closed under conjugation. I want to prove/disprove that the function $f\colon V\times V\to \mathbb{C}$ defined by $f(x, y)=(\bar{x}, y)$ is symmetric.

If $V$ is finite dimensional, then one can fix a basis $\mathfrak{B}$ for it and then obtain $$(x, y)=[x]_{\mathfrak{B}}^*P[y]_{\mathfrak{B}}$$ where $P$ is the positive definite matrix of the inner product with respect to the chosen basis. From here the truth of the above statement about $f$ is clear if $\mathcal{B}$ consists only of real valued functions.

However, when $V$ is infinite dimensional to prove this statement I take the completion of $V$ and use the fact that there is an isometric isomorphism from the completion of $V$ to some $\ell^2(S)$ space with the standard inner product of functions where symmetry is clear.

I would like to see how other people solve this problem both in the finite and (especially) infinite dimensional case. Thank you!

Edit: This statement is now disproved. See the comments below.

  • $\begingroup$ What do you mean by $\overline y$? $\endgroup$ – azarel Oct 18 '14 at 2:38
  • $\begingroup$ I am assuming that $V$ is a space of complex valued functions. $\endgroup$ – EPS Oct 18 '14 at 3:09
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    $\begingroup$ I tripped up below and am still looking for a fix/assumption. We have $\|x\| = \|\overline{x} \|$ for all $x$ iff $\overline{\langle x , y \rangle} = \langle \overline{x}, \overline{y} \rangle $ for all $x,y$. $\endgroup$ – copper.hat Oct 18 '14 at 20:42
  • $\begingroup$ This I agree; by a polarization. $\endgroup$ – EPS Oct 18 '14 at 21:17
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    $\begingroup$ By the way, it is not clear to me that the result is true in finite dimensions. You can always restrict to the subspace $\operatorname{sp} \{ x,y, \overline{x}, \overline{y} \}$ to prove the result for any pair $x,y$. $\endgroup$ – copper.hat Oct 19 '14 at 6:22

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