1
$\begingroup$

We know that every distribution or $L^1$ function $f$ over space $\mathcal{X}$ (e.g., $R^d$) can be embedded to an RKHS $\mathcal{H}$ with a $1$-bounded kernel $\mathcal{K}$ (e.g., the RBF kernel) through the linear embedding operator $\mathbb{K}$ $$ (\mathbb{K}f)(y)=\int_{\mathcal{X}} \mathcal{K}(y,x)f(x){\rm d}x, $$ since for any $f$ such that $\|f\|_1 < \infty$, we have $$ \| \mathbb{K}f \|^2_{\mathcal{H}} =\int_{\mathcal{X}^2} \mathcal{K}(y,x)f(x)f(y){\rm d}x{\rm d}y \le \|f\|_1^2 < \infty. $$

Then, it seems we can define an inner product $\langle \cdot,\cdot \rangle$ over $L^1(\mathcal{X})$ by $$ \langle f_1, f_2 \rangle := \langle \mathbb{K}f_1, \mathbb{K}f_2 \rangle_{\mathcal{H}}. $$ When $\mathcal{K}$ is a characteristic kernel, we have $\|\mathbb{K}f\|_{\mathcal{H}}=0$ if and only if $\|f\|_{1}=0$, which implies the inner product defined above is positive definite.

However, $L^1(\mathcal{X})$ should not be an inner product space or Hilbert space. What is wrong here? What is the difference between $L^1(\mathcal{X})$ and $\mathbb{K}(L^1(\mathcal{X}))$?

$\endgroup$
1
  • $\begingroup$ any thoughts or comments? $\endgroup$
    – epsilon
    Commented May 17, 2022 at 20:39

0

You must log in to answer this question.