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This question is related to Corollary 3 of the paper: Dimensionality Reduction for Supervised Learning with Reproducing Kernel Hilbert Spaces by Kenji Fukumizu, et al.

Basicly they first defined the Covariance operator $\Sigma$ on some infinite-dimension separable Hilbert space $\mathcal{H}$, which is a linear, bounded, nonnegative, self-adjoint and trace-class operator.

At the beginning of Corollary 3, they said "Let $\Sigma^{-1}$ be the right inverse of $\Sigma$ on (Ker $\Sigma)^\bot \subset \mathcal{H}$".

My understanding is that $\Sigma^{-1}:(\text{Ker }\Sigma)^\bot\to (\text{Ker }\Sigma)^\bot$ is defined to be a linear operator such that $\Sigma\Sigma^{-1}f = f$ for any $f\in (\text{Ker }\Sigma)^\bot$.

I'm confused how this right inverse is defined explicitly? Also, since $(\text{Ker }\Sigma)^\bot = \overline{\text{Range }\Sigma}$, if $f\in (\text{Ker }\Sigma)^\bot\setminus \text{Range }\Sigma$, it seems impossible for $\Sigma\Sigma^{-1}f = f$ to hold.

Any help is appreciated!

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  • $\begingroup$ What is $(\text{Ker }\Sigma)^\top$? $\endgroup$
    – user8469759
    Commented Nov 17, 2023 at 4:25
  • $\begingroup$ Sorry it should be $\bot$, not $\top$ $\endgroup$
    – allen i
    Commented Nov 17, 2023 at 4:40
  • $\begingroup$ Use the spectral theorem to get that you can find an orthonormal basis of eigenvectors of $\Sigma$. Throw away the once that are associated to the kernel. However, on this remaining eigenbasis it is not hard to construct the inverse. Note that this inverse will in general not be bounded (unless your operator is finite rank). $\endgroup$
    – Severin Schraven
    Commented Nov 17, 2023 at 5:31
  • $\begingroup$ Our unbounded operator will be defined on $\Sigma(H)$, which is dense in $R$. $\endgroup$
    – Severin Schraven
    Commented Nov 17, 2023 at 5:37
  • $\begingroup$ Thanks! that explains how to define $\Sigma^{-1}f$ for $f\in \text{Range }\Sigma$, But I'm still not sure how to define $\Sigma^{-1}f$ for $f\in (\text{Ker }\Sigma)^\bot\setminus \text{Range }\Sigma$, which is needed in other part of that Corollary. $\endgroup$
    – allen i
    Commented Nov 17, 2023 at 6:02

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