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Suppose a function $f:[-1,1]^2\rightarrow \mathbb{C}$ has a singular value decomposition:

$$ f(x,y) = \sum_{k=1}^\infty \sigma_k u_k(y) v_k(x), \qquad \sum_{k=1}^\infty \sigma_k^2 <\infty, $$

where $\sigma_1,\sigma_2,\ldots,$ is a nonincreasing sequence of real numbers, and $\{u_1(y),u_2(y),\ldots,\}$ and $\{v_1(x),v_2(x),\ldots,\}$ are two sets of orthonormal functions.

I would like to understand how the smoothness of $f(x,y)$ effects the decay of the singular values $\sigma_1,\sigma_2,\ldots,$ and, in particular, find proofs of statements similar to the following:

  1. If $f(x,y)$ is an analytic function such that for every $a\in[-1,1]$ the univariate functions $f(a,y)$ and $f(x,a)$ are analytically continuable to bounded functions on a neighbourhood containing $[-1,1]$ then the singular values decay, at least, exponentially.

  2. If $f$ is $k$ times continuously differentiable, $f\in\mathcal{C}^k$, then the singular values decay, at least, algebraically with order $k$.

  3. If $f$ is $k-1$ times continuous differentiable and $f^{(k)}$ is Lipschitz smooth then the singular values decay, at least, algebraically with order $k+1$.

Thanks in advance.

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  • $\begingroup$ The idea is always the same: find an explicit linear space of functions such that the operator is small on the orthogonal complement. Let's say $f$ is analytic (with uniform bounds). Then if the test function $g$ is orthogonal to polynomials of degree up to $n$, the operator can spit out only something of order $e^{-cn}$ because only the tail of the Taylor series in each slice can give anything non-zero and it is exponentially small. For $C^k$, you need the polynomial approximations theorems instead of the Tailor decomposition, but the flavor is similar. $\endgroup$
    – fedja
    Commented Nov 26, 2012 at 0:04

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