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Reading the book "Classical and multilinear harmonic analysis, Vol. 1" by Muscalu, Schlag, 2013; I have a problem understanding the first step of the proof of Lemma 11.3.
The relevant parts are:

Let $\mu$ be a finite measure on $\mathbb R^n$ with $n\geq 2$ and $g\in L^2(\mu) \cap \mathcal S(\mathbb R^n)$. Then $$\Vert g\Vert_{L^2(\mu)} = \sup_{f\in \mathcal S(\mathbb R^n), \Vert f\Vert_{L^2(\mu)} = 1} \left| \int_{\mathbb R^n} \hat g(\xi) f(\xi) d\mu(\xi) \right|$$

I understand that $(L^2(\mu))' = L^2(\mu)$, but why is it justified to say that $$\Vert f \Vert_{L^2(\mu)} = 1 \Leftrightarrow \Vert \overline{\hat f} \Vert_{L^2(\mu)} = 1$$

Because this is what I'd expect for the usual notation of $$\Vert x \Vert_E = \sup_{y\in E', \Vert y \Vert_{E'}=1} |y[x]|$$ With $$y[x] = \int_{\mathbb R^n} x(\xi) \overline{y(\xi)} d\xi$$

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Okay, a bit of thought put into it and using Parseval's identity I found the solution myself: $$\Vert f \Vert_{L^2(\mu)}^2 = \int f\bar{f} d\mu = \int \hat f \check{\bar f} d\mu = \int\hat f \overline{\hat f} d\mu = \Vert \hat f \Vert_{L^2(\mu)}^2$$

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