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I have a homework problem involving a Hilbert space and letting $B:H\times H \rightarrow \mathbb{C}$ be a functional which is linear with respect to the first argument and conjugate linear with respect to the second argument.

I am not sure what this means, and I am trying to research it but I'm not using good keywords. I am trying to learn as much as I can as quickly as I can, so I would appreciate some direction for rigorous understanding of these terms and concepts. (I have a textbook...) I just don't understand this question.

Thank you

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Linear in the first argument means that for all $x,y,z\in H$ and $\alpha\in\mathbb C$, $$B(\alpha x+y,z)=\alpha B(x,z)+B(y,z).$$ That is, for each $z\in H$, the map $x\mapsto B(x,z)$ is linear.

Conjugate linear in the second argument means that for all $x,y,z\in H$ and $\alpha\in\mathbb C$, $$B(z,\alpha x+y)=\overline\alpha B(z,x)+B(z,y).$$ That is, for each $z\in H$, the map $x\mapsto\overline{B(z,x)}$ is linear.

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  • $\begingroup$ Thank you Aweygan. Would you happen to know of any resource that goes more in depth about this concept? Where can I read more about it and see examples? (I think my book assumes I already know this)... $\endgroup$
    – PBJ
    Apr 18 '18 at 15:06
  • $\begingroup$ I'm not aware of any resource that gives a very concise treatment, but try googling "sesquilinear forms" and see what comes up. $\endgroup$
    – Aweygan
    Apr 18 '18 at 15:15
  • $\begingroup$ Oh, and you're welcome. Glad to help! $\endgroup$
    – Aweygan
    Apr 18 '18 at 15:16
  • $\begingroup$ Oh great, thank you! I have also been wondering about that "sesquilinear," ive been stuck in that too. Knowing that these 2 concepts are connected is more helpful than you can imagine :) $\endgroup$
    – PBJ
    Apr 18 '18 at 15:21
  • $\begingroup$ Great, I'm glad to hear it! $\endgroup$
    – Aweygan
    Apr 18 '18 at 15:22

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