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Let $K$ be a positive definite integral operator with continuous kernel $K(x,y)$ defined by $$ Kf(x) = \int_0^1 K(x,y) f(y) \, dy. $$ Let $K_n$ denote the matrix $K(x_i, x_j)$ with $x_i = i /n$ and the operator defined by $$ K_n f(x) = \frac{1}{n} \sum_j K(x_i, x_j) f(x_j) \quad \text{for} \quad x = x_i $$ and by linear interpolation for $x$ between $x_i$ and $x_{i+1}$. As $n\to\infty$, the grid gets dense.

Assuming that $K$ has an unbounded inverse, is there a way to show that $K_n^{-1/2} f \to K^{-1/2} f$ for some class of functions $f$ without explicitly finding $K_n^{-1/2}$? Does this hold for all $f \in \operatorname{Dom} K^{-1/2}$?

I tried doing the following: $$ \begin{aligned} | \langle ( K^{-1} - K_n^{-1} ) g, g \rangle | & = | \langle K^{-1} ( K_n - K ) K_n^{-1} g, g \rangle | \\[3pt] & = | \langle ( K_n - K ) K_n^{-1} g, K^{-1} g \rangle | \\[3pt] & \leq \| K_n - K \| \cdot \| K_n^{-1} g \| \cdot \| K^{-1} g \| \end{aligned} $$ and since $K_n$ converges to $K$ uniformly (unless I am mistaken), we have that $$ \langle ( K^{-1} - K_n^{-1}) g, g \rangle \to 0 \quad \text{for all} \quad g \in \operatorname{Dom} ( K_n^{-1} ) \cap \operatorname{Dom} ( K^{-1} ). $$ But I'm not sure whether something good follows from this argument...

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  • $\begingroup$ If the grid is not uniform (or something like that), I think you cannot even even get $K_n\approx K$. $\endgroup$
    – Iosif Pinelis
    Commented Apr 19 at 13:17
  • $\begingroup$ Maybe you're right. Since general grid is not what I'm after, I'll just add an assumption to the question. Thanks! $\endgroup$
    – tsnao
    Commented Apr 19 at 13:30
  • $\begingroup$ Although my intuition tells me that it shouldn't matter since the kernel is continuous. But I may be wrong and in the end it's not that important. $\endgroup$
    – tsnao
    Commented Apr 19 at 13:33
  • $\begingroup$ In order to get $K_n\approx K$, the uniform probability distribution on the set of the $x_i$'s should be close to the uniform distribution over $[0,1]$. So, the grid should be approximately uniform. But, of course, this does not matter to you. $\endgroup$
    – Iosif Pinelis
    Commented Apr 19 at 14:14
  • $\begingroup$ @IosifPinelis, a simple model example is when $K$ is the covariance operator of a Brownian motion ($K(x, y) = \min(x, y)$). Then $K^{-1/2}$ is the derivative operator and $K_n^{-1/2}$ is something like discrete derivative. $\endgroup$
    – tsnao
    Commented Apr 19 at 14:21

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