# Plancherel Theorem (Fourier transform)

1. Oct 12, 2015

### Incand

I'm having a hard time understand this theorem in our book:

The Plancherel Theorem
The Fourier transform, defined originally on $L^1\cap L^2$ extends uniquely to a map from $L^2$ from $L^2$ to itself that satisfies
$\langle \hat f, \hat g \rangle = 2\pi \langle f,g\rangle$ and $||\hat f||^2= 2\pi||f||^2$
for all $f,g\in L^2$.

I don't really understand the formulation here. Some questions:
What does it mean "extends uniquely" here?
When am I allowed to use the theorem? Can I use the formula for every $L^2$ function?
Is the Fourier transform of an $L^2$ function always an $L^2$ function as well (even if the function is both in $L^1$ and $L^2$)?

2. Oct 13, 2015

### mathman

Extends uniquely means that there is only one Fourier transform image for each function in the extension.

From Wikipedia
A more precise formulation is that if a function is in both L1(R) and L2(R), then its Fourier transform is in L2(R), and the Fourier transform map is an isometry with respect to the L2 norm. This implies that the Fourier transform map restricted to L1(R) ∩ L2(R) has a unique extension to a linear isometric map L2(R) → L2(R). This isometry is actually a unitary map. In effect, this makes it possible to speak of Fourier transforms of quadratically integrable functions.
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Second and third questions - answer is yes for both.

3. Oct 13, 2015

### micromass

Staff Emeritus
What do you mean with "the formula"?

4. Oct 14, 2015

### Incand

Thanks! Especially thanks for the link to the article about unitary operators. I understand it a bit better now.

I meant the $\langle \hat f, \hat g \rangle = 2\pi \langle f,g\rangle$ and $||\hat f||^2= 2\pi||f||^2$ formula. I.e. my question was if I have two $L^2$ functions and I know the Fourier transform of these is the "formula" always true. Which I believe mathman answered was true.

5. Oct 14, 2015

### mathman

It is a standard theorem in Fourier analysis (Plancherel or Parseval).