# Sampling uniformly from a unit sphere in Reproducing Kernel Hilbert Space

In finite dimensions, one can sample uniformly from a unit sphere Sd−1={x∈Rd:‖x‖=1}\mathcal{S}^{d-1}=\{x\in\mathbb{R}^d : \|x\|=1\}, by simply generating dd independent standard normal variables and normalizing (to make it unit norm).

Now suppose we have a positive definite kernel function k:Rd×Rd→Rk : \mathbb{R}^d\times \mathbb{R}^d \rightarrow \mathbb{R}, that gives rise to a unique reproducing kernel Hilbert space (RKHS) HK\mathcal{H}_K, where, k(x,y)=⟨k(x,⋅),k(y,⋅)⟩HKk(x,y)=\langle k(x,\cdot), k(y, \cdot)\rangle_{\mathcal{H}_K}. Let A={f:‖f‖HK=1}A=\{f: \|f\|_{\mathcal{H}_K}=1\}. Is there a simple way to sample uniformly from AA, analogous to the finite dimensional case? Can anyone please point me to relevant literature? Thank you.

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I don’t get how to go from functional analysis (RKHS) to probability theory. Can you formulate it without any random variable and probability ?
– user1952009
2 days ago

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What exactly do you mean by “sample uniformly”? In finite dimensions, the “uniform distribution” on the unit sphere is the (unique) rotation invariant measure. Do you intend the same thing for the infinite dimensional case? And what would you replace rotations with, unitary operators? In any case, you will run into issues with defining a finite measure on the sphere of a HIlbert space. See Halmos’ A Hilbert Space Problem Book problem 18.
– J. Loreaux
2 days ago

@user1952009: OP wants to know if there exists a probability measure μ\mu on the Borel σ\sigma-algebra of AA satisfying some conditions (“uniform”), but exactly what those conditions are, has not been made clear. The answer is that it probably doesn’t exist. Asking how to “sample from” this measure usually means something like: is there a convenient explicit map FF from [0,1]ω[0,1]^\omega to AA, such that the pushforward of the infinite product of Lebesgue measure on [0,1]ω[0,1]^\omega is μ\mu? But it’s moot if μ\mu doesn’t exist.
– Nate Eldredge
2 days ago

Indeed, if “uniformly” means “invariant under the action of every unitary operator”, then you can stop right now as no such probability measure exists. Proof: for any sufficiently small open set U⊂AU \subset A, there are unitary operators T1,T2,…T_1, T_2, \dots such that {TiU}\{T_i U\} are pairwise disjoint. So choose any UU with positive measure. This forces AA to have measure infinity. Actually, a similar argument forces every nonempty open subset of UU to have measure infinity.
– Nate Eldredge
2 days ago

@J.Loreaux: Note that “measure” for Halmos in that problem means “translation invariant measure”. Without requiring translation invariance, there are plenty of finite measures on the unit sphere of a Hilbert space. But they typically don’t have nice invariance properties.
– Nate Eldredge
2 days ago

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