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I have formally derived a solution to a PDE as a power series $$ u = \sum_{n=0}^\infty \epsilon^n u_n . $$ I would like to show that the radius of convergence for is $\mathbb{R}$. I assume that the easiest way to do that is to show $$ \lim_{n \to \infty} |\frac{u_{n+1}}{u_n}| = 0 . $$ The difficulty in showing this is that the $u_n$ are of the form $$ u_n = \int_\mathbb{R} f_n(x) dx $$ So far, my best strategy is to introduce $$ g_{n+1} := \frac{f_{n+1}}{f_n} $$ so that $$ |\frac{u_{n+1}}{u_n}|^2 = \frac{| \int g_{n+1}f_n dx |^2}{|\int f_n dx|^2} \leq \frac{\int | g_{n+1} |^2 dx \cdot \int | f_n |^2 dx}{|\int f_n dx|^2} $$ and $$ \lim_{n \to \infty} |\frac{u_{n+1}}{u_n}|^2 \leq \limsup_{n \to \infty} \int | g_{n+1} |^2 dx \cdot \lim_{n \to \infty} \frac{ \int | f_n |^2 dx}{|\int f_n dx|^2} $$ $$ \leq \int \limsup_{n \to \infty} | g_{n+1} |^2 dx \cdot \lim_{n \to \infty} \frac{ \int | f_n |^2 dx}{|\int f_n dx|^2} $$ Thus I need to show $$ \limsup_{n \to \infty} | g_{n+1} |^2 = 0 \qquad (A) $$ and $$ \lim_{n \to \infty} \frac{ \int | f_n |^2 dx}{|\int f_n dx|^2} < \infty \qquad (B) $$ This seems more complicated to me than I think it should be. Intuitively, it seems to me like I should just need to show (A). Anyway, if anybody has a simpler strategy, I would greatly appreciate it.

* Additional Information not in my original post *

The integrals written above in terms of $x$ in the notation of my paper are $$ \int_\mathbb{R} f_n(\lambda) d\lambda $$ The specific form of $f_n(\lambda)$ is as follows $$ f_n(\lambda) = \left( \sum_{k=0}^n \frac{e^{t \phi_{\lambda-ik\beta}}} {\prod_{j\neq k}^n (\phi_{\lambda-ik\beta}-\phi_{\lambda-ij\beta})} \right) \left( \prod_{k=0}^{n-1} \chi_{\lambda-ik\beta}\right) (\psi_\lambda, h) \psi_{\lambda} $$ where $$ \phi_\lambda =\frac{1}{2} a_0^2 \( -\lambda^2 - i \lambda \) - \int_\mathbb{R} \nu_0(dz) \( e^{z} - 1 - z \) i \lambda + \int_\mathbb{R} \nu_0(dz) \( e^{ i \lambda z} - 1 - i \lambda z \) - c_0 $$ and $$ \chi_\lambda =\frac{1}{2} a_1^2 \( -\lambda^2 - i \lambda \) - \int_\mathbb{R} \nu_1(dz) \( e^{z} - 1 - z \) i \lambda + \int_\mathbb{R} \nu_1(dz) \( e^{ i \lambda z} - 1 - i \lambda z \) - c_1 $$ the $a_i$ and $c_i$ are real positive constants. The $\nu_i$ are Levy measures. $$ \psi_\lambda = e^{i \lambda y} \qquad (\psi_\lambda,h) = \int_\mathbb{R} \psi_\lambda(y) h(y) dy $$ The domain of $f(\lambda)$, $\psi_\lambda$ and $\chi_\lambda$ are $\mathbb{C}$. For simplicity, assume that $h$ is $L^2(\mathbb{R},dy)$

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2 Answers 2

I think you should offer more information about the mathematical objects you mentionend (e.g. $f_n$ and $u_n$) such like their forms, domain and range.

For example, as you didn't say anything about $f_n$, what if $f_n = constant$, then it is obvious $(A)$ can't be hold.

If these necessary details are omitted, nobody can help you.

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Note that $$ \limsup\int |g_n|^2 \leq \int\limsup |g_n|^2 $$ does not hold for arbitrary $g_n$.

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Right. Yes, I should check that there is an integrable function that bounds the $|g_n|^2$ –  psyduck Jul 15 '12 at 1:04

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