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.