Get ready Part 11

Assuming that $|f(x)| \le e^{-\sqrt{x}}$ for $x \ge 0$ , which is enough to ensure that the moments $\mu_n$ all exist, the differential equation for $f$ tells us that $-n\mu_{n-1} \; = \; -3\mu_n + 3\times2^{-n}\mu_n \hspace{2cm} n \ge 1$ and hence $\mu_n \; = \; \frac{n!}{3^n}\left(\prod_{k=1}^n (1 - 2^{-k})\right)^{-1}\mu_0 \hspace{2cm} n \ge 0$ and so $\frac{3^n}{n!}\mu_n \; = \; \left(\prod_{k=1}^n (1 - 2^{-k})\right)^{-1}\mu_0 \hspace{2cm} n \ge 0$ Since $\sum_{k=1}^\infty 2^{-k} < \infty$ , the infinite product $\prod_{k=1}^\infty (1 - 2^{-k})$ converges to a nonzero limit, and so $\lim_{n \to\infty} \tfrac{3^n}{n!}\mu_n$ always exists, and the limit equals $0$ if and only if $\mu_0=0$ .

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Note that $-f(0) \; = \; \int_0^\infty f'(x)\,dx \; = \; -3\int_0^\infty f(x)\,dx + 6\int_0^\infty f(2x)\,dx \; = \; -3\mu_0 + 3\mu_0 = 0$ and this tells us that $f'(0) = -3f(0) + 6f(0) = 0$ . Indeed, a simple induction tells us that $f^{(n)}(0)= 0$ for all $n \ge 0$ . Thus the Maclaurin series of $f$ is identically zero.

This raises the equation as to whether the only solution of the differential equation is $f(x) \equiv 0$ . There are nontrivial functions with zero Maclaurin series, such as $g(x) \; = \; \left\{\begin{array}{lll} e^{-x^{-2}} & \hspace{1cm} & x \neq 0 \\ 0 & & x = 0 \end{array}\right.$ so the question is still open. If a nonzero function $f$ does exist, however, it will have to be pretty strange.