0

Plancherel's theorem states that if $f\in L^{1}(\mathbb{R}^{n})\cap L^{2}(\mathbb{R}^{n})$ and $\hat{f}$ is its Fourier transform, then $$\|f\|_{L^{2}(\mathbb{R}^{n})}= \|\hat{f}\|_{L^{2}(\mathbb{R}^{n})}.$$

Is it true that

$$\|f\|_{L^{2}(\mathbb{S}^{n-1})}= \|\hat{f}\|_{L^{2}(\mathbb{S}^{n-1})}\qquad (1)$$

where $\mathbb{S}^{n-1}$ is the Euclidean sphere in $\mathbb{R}^{n}$ ?

Plancherel's theorem is true on locally compact abelian groups in general. I don't know whether the $\mathbb{S}^{n-1}$ is a locally compact abelian group.

I think that (1) is false. If we reiterate the proof, we get a kernel of the form $$\int_{\mathbb{S}^{n-1}}\;e^{\dot{\imath}(x-y)\cdot \omega} d\sigma(\omega)$$

as opposed to

$$\int_{\mathbb{R}^{n}}\;e^{\dot{\imath}(x-y)\cdot \xi} d\xi$$ which is precisely $\delta(x-y)$.

Medo
  • 3,588
  • 14
  • 32

1 Answers1

3

The only Euclidean spheres that have a Lie group structure are $\mathbb{S}^0$, $\mathbb{S}^1$, and $\mathbb{S}^3$. (See here and here.) So assuming you're not willing to mash a non-intuitive group structure onto $\mathbb{S}^{n-1}$, that narrows the field down quite a bit on what we can even talk about.

Further, the Lie group structure on $\mathbb{S}^3$ isn't abelian. So again, standard harmonic analysis (Pontryagin duality, etc.) doesn't apply.

With $\mathbb{S}^1$, the Fourier transform maps to function on $\mathbb{Z}$, rather than functions on $\mathbb{S}^1$. So the question doesn't really make sense in that context either.

Finally, with $\mathbb{S}^0$, the Fourier transform does map to functions on $\mathbb{S}^0$, and Plancherel holds. But it's a relatively trivial case, since we're talking about a group with two elements.

We could also think about the question a little differently. Let's say we take a measure $d\mu = f d\sigma$, where $d\sigma$ is surface measure on $\mathbb{S}^{n-1}$ and $f \in L^{2}(d\sigma)$. Then we can take the Euclidean Fourier transform of this measure, which gives a continuous function on $\mathbb{R}^n$, and we can restrict it to $\mathbb{S}^{n-1}$. In this interpretation, Plancherel also fails to hold. One way of seeing this is to look at the family of modulations of a single $f$. For $\eta \in \mathbb{R}^n$, the modulated function $e^{-i \eta \cdot x} f(x)$ has the same $L^2(d\sigma)$ norm as $f$, while the Fourier transform of the modulation is the shift of $(f d\sigma)^{\wedge}$ by $\eta$. Since we can therefore translate $(f d\sigma)^{\wedge}$ arbitrarily before restricting to $\mathbb{S}^{n-1}$, it's not too hard to see that the Fourier transform won't preserve $L^2(d\sigma)$ norms in this sense, either.

Jason
  • 686