Show that there is at least a nonzero function $f$, differentiable on $[0,+\infty)$, satisfying $$f'(x)=2f(2x)-f(x) \qquad \forall x>0 $$ $$M_n:=\int_{0}^{\infty}x^nf(x)dx<\infty \qquad \forall n\in \mathbb{N} $$

My best idea so far is to assume that the solution is a power series, i.e. $$ f(x)=\sum_{n=0}^{\infty}a_nx^n\qquad \forall x>0$$ Then the equation becomes $$ \sum_{n=0}^{\infty}na_nx^{n-1}=2\sum_{n=0}^{\infty}a_n2^nx^n-\sum_{n=0}^{\infty}a_nx^n$$ equating all the coefficients of the same degree I get $$na_n=(2^{n}-1)a_{n-1}\qquad \forall n\geq 1$$ So setting $a_0=1$, I get $$a_{n}=\frac{1}{n!}\prod_{k=1}^{n}(2^k-1) \qquad \forall n$$ But does the power series actually converge? Using Hadamard's formula, and that $2^{k}-1\geq 2^{k-1}$, $$ |a_n|^{1/n}\geq\frac{1}{(n!)^{1/n}}\left[2^{n(n-1)/2}\right]^{1/n}\sim\frac{e}{n(2\pi n)^{1/2n}}2^{(n-1)/2}\to \infty$$ so the radius of converge of the series is $0$, so it doesn't actually define a solution on $[0,+\infty)$.


Well, it looks like your best idea (which by itself was not bad at all) fails precisely for the reasons you so comprehensively described. So what shall we do?

Guess what?

We turn it the other way around. Let's look for a function in the form of $$f(x)=\sum_{n=0}^\infty a_ne^{-2^nx},\quad x\geqslant0$$

You might wonder what's with $2^n\cdot x$ up there in the exponent. Well, it's simple: I was going to write $e^{-nx}$ when I realized we won't be needing most of those terms, so let it be this way.

Now, the equation becomes $$-\sum_{n=0}^{\infty}2^na_ne^{-2^nx}=2\sum_{n=0}^{\infty}a_ne^{-2^{n+1}x}-\sum_{n=0}^{\infty}a_ne^{-2^nx}$$ or $$\sum_{n=0}^{\infty}2^na_ne^{-2^nx} = -2\sum_{n=1}^{\infty}a_{n-1}e^{-2^nx}+\sum_{n=0}^{\infty}a_ne^{-2^nx}$$ which gives $$(2^n-1)a_n=-2a_{n-1},\quad n\geqslant1$$

So continuing in your footsteps, I set $a_0=1$ and get $$a_n=\frac{2^n}{\prod_{k=1}^{n}(2^k-1)}, \quad n\geqslant1$$ which makes my series converge pretty fast, for the same reason why your series fails to do so.

Now that's our solution.

  • $\begingroup$ Thanks. That leaves the question of the convergence of the integrals $M_n$. I need to exchange a series and integral sign so for that I need the uniform convergence of the series $\sum_{n=0}^{\infty}a_n x^n e^{-2^nx}$. My idea would be using the fact that $a_nx^n e^{-2^{n}x}\leq a_nx^n$, and that the series $\sum_{n=0}^{\infty}a_nx^n$ converges uniformly on $[0,+\infty)$ because $(a_n)^{1/n}\to 0$. Is this correct? Another question: couldn't a solution also be of the form $\sum_{n=0}^{\infty}a_nx^{-n}$? Did you jump to inverse exponentials just to make sure it would converge quickly enough? $\endgroup$ – Lorenzo Quarisa Dec 6 '17 at 19:32
  • $\begingroup$ No. True, $\sum_{n=0}^{\infty}a_nx^n$ does converge for all $x$, but it definitely does not converge uniformly on $(0,+\infty)$. Then again, how the combination $a_nx^n$ came about in the first place? $\endgroup$ – Ivan Neretin Dec 6 '17 at 19:39
  • $\begingroup$ $x^{-n}$ behave badly at 0. But your guess is also true: yes, that's to make sure it converges quickly. $\endgroup$ – Ivan Neretin Dec 6 '17 at 19:40
  • $\begingroup$ I see, convergence of power series only implies uniform convergence on compact subsets of the convergence disk, and $[0,+\infty)$ is not compact so that doesn't work. Actually I just need the fact that $x^ne^{-2^nx}$ is bounded (uniformly in $x$ and $n$), so I can bound the series with $C\sum_{n=0}^{\infty}a_n$ which clearly converges and doesn't depend on $x$, so the first series must converge uniformly. $\endgroup$ – Lorenzo Quarisa Dec 6 '17 at 19:52
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    $\begingroup$ You don't have the combination $x^ne^{-2^nx}$. The $n$ in the definition of your $M_n$ and another $n$ in the definition of my series are two totally different, unrelated things. $\endgroup$ – Ivan Neretin Dec 6 '17 at 20:05

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