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Let $$f(s)=\sum_0^\infty c_vs^v$$ be some entire function. We say that the differential operator $f(d/dx)=\sum_0^\infty c_vd^v/dx^v$ is defined in some fundamental space $\varPhi$, if for any $\varphi \in \varPhi$, the series $$f(\frac{d}{dx})\varphi(x)=\sum_0^\infty c_v \varphi^{(v)}(x)$$ is again a fundamental function.

My questions :

1) What guarantees me to represent f as an infinite series as defined here.
2) I know that $f(d/dx)$ is an operator acting on some test function from a fundamental space $\varPhi$, what does the notation $f(d/dx)$ convey in particular ?
3) How am I able to replace $s$ by $d/dx$ ?

Thank you for your help.

Reference: Generalized Functions.. Volume 2 by I.M. Gelfand and G.E. Shilov

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1) $f$ is an entire function, so it has a power series representation centered at $0$ that converges on the whole complex plane.

2) The notation $f(d/dx)$ denotes exactly what you have written: it is an operator defined by $$ f\left(\frac{d}{dx}\right)\varphi(x) = \sum_{v=0}^\infty c_v \varphi^{(v)}(x) $$ where as part of the definition the right-hand side is assumed to converge. (The convergence might not hold if we take a different function $g$, but this is beside the point when we're just taking the definition.)

3) Replacing $s$ by $d/dx$ to write $$ f\left(\frac{d}{dx}\right) = \sum_{v=0}^\infty c_v\frac{d^v}{dx^v} $$ is purely notation. There is a calculus associated with this sort of notation (the holomorphic functional calculus) that makes it particularly convenient. But you shouldn't read too far into it. The above sum shouldn't be interpreted as literally the partial sums $$ \sum_{v=0}^N c_v\frac{d^v}{dx^v} $$ converging; any convergence should be interpreted against a suitable fundamental space of test functions.

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  • $\begingroup$ Thanks for your answer. It has given me a little better understanding. But one small doubt. The underlying space for this operator is a space of test functions on which The Fourier Transform is defined. For the third part of the question, is there any justification that $s$ be replaced by $d/dx$ because of this $\endgroup$ – creative Jan 20 '16 at 5:28
  • $\begingroup$ Nice answer. I didn't know you were active on SE! $\endgroup$ – user98602 Jan 20 '16 at 5:35
  • $\begingroup$ @Hirak If you're looking for a statement like "the partial sums of the series converge to the operator $f(d/dx)$ (as defined by the action on test function)," then you need to specify the space of test functions and the type of convergence. For instance, convergence in operator norm won't make sense since $f(d/dx)$ should be treated as a distribution and spaces of distributions typically won't have a Banach space structure. But weak-$*$ or bounded convergence might be suitable. continued... $\endgroup$ – Gyu Eun Lee Jan 20 '16 at 7:50
  • $\begingroup$ As for what makes the series for $f(d/dx)\varphi(x)$ converge: without any conditions on $f$ and $\varphi$ I'm not sure. The definition doesn't claim that $f(d/dx)\varphi(x)$ is well-defined for every $x$ in the underlying space and $\varphi\in\Phi$. All the definition says is that if $f$ is such an entire function that the series $f(d/dx)\varphi(x)$ does converge for every $\varphi$, then the operator $f(d/dx)$ is defined by that action on test functions. It may well be that for a particular choice of $f$ there is a test function $\varphi$ that fails this convergence. $\endgroup$ – Gyu Eun Lee Jan 20 '16 at 7:58
  • $\begingroup$ The underlying space is the Gelfand-Shilov space of type $\mathcal{S}$. Ref: Generalized Function, Volume 2 by I.M. Gelfand and G.E. Shilov $\endgroup$ – creative Jan 20 '16 at 8:14

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