Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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 $$y[x] = \int_{\mathbb R^n} x(\xi) \overline{y(\xi)} d\xi$$

share|cite|improve this question
up vote 2 down vote accepted

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

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.