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## Adobe Photoshop 2021 (Version 22.4.3) 2022 [New]

Q: Questions regarding the regularity of operator-valued functions Let $X$ be a complex Banach space and let $B$ be the open unit ball in $X$. Let $K\subset X$ be a compact set. Let $f$ be a function defined on $B$ such that $0\leq f\leq 1$. $f$ is analytic if $f$ is continuous and if $f$ is holomorphic, then $f$ is uniformly continuous. Let $T:K\to\mathcal L(X)$ be a function. The function $T$ is regular if for each $n\in\mathbb N$, there is a function $g_n$ which is analytic on $B$, with the property that $0\leq g_n\leq 1$, $g_n$ is analytic on $K$, and for all $x\in K$, $$T(x)=\sum_{n=0}^\infty x^n g_n(x)$$ It is well known that $T$ is regular if and only if $\lim_{n\to\infty}g_n$ exists uniformly on $K$. If $T$ is not regular, then $g_n$ can be taken to be a regular function on $K$. The following three statements seem to be true: If $T$ is regular, then for each $\epsilon>0$, there is $x\in K$ such that $g_n(x)>1-\epsilon$ for all $n$. If $T$ is not regular, then there is an increasing sequence of compact subsets $K_n$ of $K$ with $K=\bigcup_{n=1}^\infty K_n$. There is a sequence of compact subsets $K_n$ of $K$ such that $\lim_{n\to\infty}K_n=K$. But how can we show these statements? A: Each $K_n$ can be written as a countable intersection of closed balls of radius $1/n$, with centers in $K$, so $K = \lim_{n\to \infty} K_n$. By the Cauchy integral formula, each