0
$\begingroup$

Let $G$ be a group and $L^2(G) = \{f: G \rightarrow \mathbb{C} \}$. Now define an inner product on $L^2(G)$ by $$\langle f, g \rangle = \sum_{x \in G}f(x)\overline{g(x)}$$ Where $\overline{g(x)}$ is the complex conjugate.

I know what it means for the inner product described above to be linear in $f$, i.e. $\langle f + g, h \rangle = \langle f, h \rangle + \langle g, h \rangle$ and $\langle \lambda f, g \rangle = \lambda \langle f, g \rangle$. What does it mean for the inner product to be "conjugate linear" in $g$?

$\endgroup$
1
  • $\begingroup$ It implies that $<f,f>$ is non-negative. In case $<f,f>=0$ implies $f=0$ and it defines a norm. $\endgroup$
    – Urgje
    Feb 5, 2016 at 9:06

1 Answer 1

1
$\begingroup$

It means that for any scalar $s\in C$ and for all vectors $f,g_1,g_2$ we have $<f,s g_1+ g_2>=\bar s\cdot <f,g_1>+<f,g_2>.$

$\endgroup$
1
  • 1
    $\begingroup$ $\mathbb{C}...$ $\endgroup$
    – kk lm
    Feb 5, 2016 at 4:29

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .