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If we are working in a complex inner product space with elements f,g then is $\langle f,g \rangle = \overline{\langle g,f \rangle} = \langle \overline{g},\overline{f} \rangle$ ? Obviously the first equality is true but I am wondering if the complex conjugate can be brought into the inner product.

We can assume that $\overline{f}$ is defined. In the specific problem I am working on we are in a reproducing kernel Hilbert space.

TY!

kreitz
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    In an abstract inner product space, there is no such thing as the "conjugate" of a vector. If you pick coordinates, then you can conjugate the coordinates, but this operation depends on what coordinates you picked. At any rate, in that context the answer is yes. – coiso May 04 '23 at 17:49
  • I was wondering the same thing, but as @coiso remarked you cannot make this statement generally across all inner products, but this property can be showed for certain inner products. For example, on the complex $L^2$ space of complex-valued random variables the associated inner product $\langle X, Y \rangle := \mathbb{E}[X \bar{Y}]$ satisfies the equality, using this fact – Dylan Dijk Jul 20 '23 at 15:02

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