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Suppose I have a linear operator $L(x)$, $x\in X\subseteq \mathbb{C}$, and has a convergent Taylor series $L(x) = \sum_{k=0}^\infty M_kx^k$ with radius of convergence $R$, for appropriate linear operators $M_k$. It is well known that if $L=f$ is a function, then it satisfies Cauchy's integral formula: $$f^{(n)}(x) = \frac{n!}{2\pi i}\int_\Gamma \frac{f(z)}{(z-x)^{n+1}}dz, $$ which allows one to derive: $|f^{(n)}|\leq \frac{n!}{R^{n+1}}\max_x|f(x)| $.

If $L(x)$ is no longer a function, but a general linear operator (for example a matrix defined as in terms of $x$), is there an analogous formula that one can use to bound the $n^{th}$ of derivative of L with respect to $x$ in terms of some appropriate norm?

That is, does the exist a formula of the type: $$||L^{(n)}(x)||_X\leq \frac{n!}{R^{n+1}}\max_x ||L(x)||_X,$$ for an appropriate norm $||.||_X$?

If there does not exist a class of linear operators of the power-series form given above, then is there a class of operators which do satisfy a bound on the $n^{th}$ derivative? Is there some definition of holomorphic linear operators which satisfying this (whatever holomorphic means for linear operators in this context).

Hans Schmuber
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    There is, since the Cauchy integral formula holds for $L$. Most things in complex analysis generalize to holomorphic functions that take values in an arbitrary Banach space. see https://math.stackexchange.com/questions/4217820/reference-request-holomorphic-functions-with-values-in-banach-spaces – Mason Aug 29 '22 at 15:39

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Suppose $V$ is a complex Banach space, $\Omega\subset\Bbb{C}$ is open, and $f:\Omega\to V$ is any holomorphic map (i.e $\lim\limits_{h\to 0}\frac{f(z_0+h)-f(z_0)}{h}$ exists in the norm topology of $V$ for each $z_0\in \Omega$... or equivalently, $f$ admits a power series expansion $f(z_0+h)=\sum_{n=0}^{\infty}a_nh^n$ in a neighborhood of $z_0$, where $a_n\in V$). Note that this covers both the case that $V=\Bbb{C}$ (the usual case), or $V=\mathcal{L}(\Bbb{C}^n,\Bbb{C}^m)$ a space of linear maps which seems to be what you're interested in, or really any complex Banach space.

One can still establish the Cauchy integral formula; for finite-dimensional $V$ this follows easily by using a basis and checking component-by-component, and in the general case it follows easily by the Hahn-Banach theorem (which is roughly speaking a fancy way of checking component by component). Once you have the Cauchy-integral formula, you just take the norm of both sides, and put the norm inside the integral (which is really just a glorified triangle inequality).

peek-a-boo
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