Functional calculus for compact operators using singular value decomposition and regularisation filters

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Notation: Let $X,Y$ be Hilbert spaces, $\mathcal{K}(X,Y)$ denote linear compact operators $X \to Y$, $L(X)$ the linear continuous operators $X \to X$, $\mathcal{N}(K)$ is the kernel of $K$, $\mathcal{R}$ its range and $P_A$ the orthogonal projection onto $A$.

Our lecture notes (functional analysis II) claim:

The singular value decomposition allows use to define functions of compact operators. Let $f: [0, \infty) \to \mathbb R$ be a piecewise continuous locally bounded function. For $K \in \mathcal{K}(X,Y)$ with singular system $((\sigma_n, u_n, v_n))_{n \in \mathbb N}$ define $$ f(K^* K): X \to X, \ x \mapsto \sum_{n \in \mathbb N} f(\sigma_n^2) \langle x, v_n \rangle v_n + f(0) P_{\mathcal{N}(K)} x. $$ The series converges in $X$, since $f$ is evaluated on the compact interval $[0, \sigma_1^2] = [0, \| K \|^2]$. We have $f(K^* K) \in L(X)$ as $$ \| f(K^* K) \| \le \sup_{n \in \mathbb N} | f(\sigma_n^2) | \le \sup_{0 \le x \le \sigma_1^2} | f(x) | < \infty. $$

Questions

  1. Aren't all piecewise continuous functions already locally bounded? If so is there a reason why the author wrote this down explicitly?
  2. What is the intuition behind $f(0) P_{\mathcal{N}(K)}$? What does it "do"?
  3. Later we defined a regularisation filter: a family of piecewise continuous and bounded functions $f_a: (0, \sigma_1^2] \to \mathbb{R}$, with $f_a(x) \xrightarrow{a \to 0} x^{-1}$ for all $x \in (0, \sigma_1^2]$ and $x | f_a(x) | \le C_f$ for some $C_f > 0$ and all $a > 0$, $x \in (0, \sigma_1^2]$. We then wrote for $y \in Y$ and $((\sigma_n, u_n, v_n))_{n \in \mathbb{N}}$ as above (recall that $((\sigma_n^2, v_n, v_n))_{n \in \mathbb{N}}$ is a singular system for $K^* K$) $$ f_a(K^* K) K^* y = \sum_{n \in \mathbb{N}} f_a(\sigma_n^2) \langle K^* y, v_n \rangle v_n $$ Why is this valid? Where did the $f(0) P_{\mathcal{N}(K)}$ term go? Does it vanish because of $\mathcal{N}(K) \perp \mathcal{R}(K^*)$?
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  1. No. Take $x \mapsto \max(0,1/(x-1))$. It is piecewise continuous, non negative, not locally bounded.
  2. We want to be able to select $f=Id$ and recover $A^*A$ (if the term with $f(0)$ wasn't there we couldn't do this)