A very weird integral equation

Question

I tried to come up with an integral equation for fun, and made this creature:$\def\d{\mathrm d}$ $$ f(x)-\int_x^{2x}f(t)\,\d t=0, $$ so I followed these steps: $$ \begin{aligned} &f(x)+\int_x^{2x}f(t)\,\d t=0\\ \implies\quad&\frac{\d f}{\d x}+2f(2x)-f(x)=0\\ \implies\quad&\frac{\d f}{\d x}-f(x)=-2f(2x)\\ \implies \quad&\left(\frac{\d}{\d x}-1\right)f(x)=-2f(2x)\\ \implies\quad&f(x)=\frac{2f(2x)}{1-\frac{\d}{\d x}}\\ \implies\quad&-f(x)=2f(2x)\sum_{n=0}^{\infty}\frac{\d^n}{\d x^n}=2f(2x)\left(1+\frac{\d}{\d x}+\frac{\d^2}{\d x^2}+\cdots\right)=2f(2x)\\ \implies\quad&f(x)=2f(2x). \end{aligned} $$

But since I am but a humble mathematical sailor, I don’t know functional analysis yet, so I couldn’t really prove the series expansion, but the function equations suggests that $f(x)=0$ is the solution to the integral equation.

Answer 1

You can assume that $f$ is a power series.

$$f(x) = a_0 + a_1 x + a_2 x^2 + a_3 x^3 + \dots = \sum_{n=0}^{\infty} a_n x^n$$

Then

$$\sum_{n=0}^{\infty} a_n x^n = \int_x^{2x} \left( \sum_{n=0}^{\infty} a_n t^n \right) dt$$

or that

$$a_0 + a_1 x + a_2 x^2 + \dots = a_0x + a_1 \frac{3}{2}x^2 + a_2 \frac{7}{3}x^3 + \dots$$

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The LHS has a constant term $a_0$, but the RHS has no constant term. So $a_0 = 0$. But the $x^1$ coefficient $a_1$ must be $0$ since $a_1x = a_0x = 0x$. The logic continues, and every coefficient $a_n$ must be zero. We have that $f(x) = 0$.

Answer 2

If the function only has to be defined on $[0,\infty)$, and if it doesn't have to be analytic, then there are nonzero solutions.

Consider the lacunary series $$f_q(x)\stackrel{\triangle}{=}\frac{1}{\prod_{k=1}^{\infty}(1-q^k)}\sum_{n=0}^{\infty}\frac{(-1)^n q^{\,_nC_2}}{\prod_{k=1}^n(1-q^k)}\mathrm{e}^{-q^{-n}x}$$ where $0<q<1$. It's uniformly convergent on $\Re x \geq 0$, so one can safely exchange summation and limiting operations to show that it's smooth and satisfies $$\begin{align}f_q(x) &= \int_x^{x/q}f_q(x')\mathrm{d}x'\\ \lim_{x\to\infty}\mathrm{e}^xf_q(x) &=\frac{1}{\prod_{k=1}^{\infty}(1-q^k)} \\ f_q^{(m)}(0^+) &= (-1)^m\sum_{n=0}^{\infty}\frac{(-1)^n q^{\,_nC_2-mn}}{\prod_{k=1}^n(1-q^k)} \end{align}$$ But this summation for $f_q^{(m)}(0^+)$ vanishes for all nonnegative integers $m$ by a $q$-binomial theorem, so $f_q$ cannot be analytic.

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More generally, the Laplace transform of the integral equation $$f(x)=\int_x^{x/q}f(x')\mathrm{d}x'$$ is $$\hat{f}(\kappa) = \frac{\hat{f}(q\kappa)-\hat{f}(\kappa)}{\kappa}\text{.}$$ Substituting $\hat{f}(\kappa) = \hat{f}_q(\kappa)\hat{\psi}(\kappa)$ where $\hat{f}_q(\kappa)=\prod_{n=0}^{\infty}(1+q^n\kappa)^{-1}$ is the Laplace transform of $f_q(x)$ above yields the constraint $$\hat{\psi}(q\kappa)=\hat{\psi}(\kappa)\text{.}$$ That is, $f$ satisfying the integral equation are convolutions of $f_q$ with $q$-multiplicatively periodic distributions $\psi$.

Answer 3

It seems like it is illegal, but if you define $T$ as the operator on $\mathcal{C}^{1}$, representing $$T[f]=\left(\dfrac{d}{dx}-\text{id}\right)f$$ Then dividing $1-\dfrac{d}{dx}$ is same as taking the inverse operator of $T$ on both sides, in particular, write $g(x)=-2f(2x)$, then $$T[f](x)=g(x)\implies (T^{-1}\circ T)[f](x)=T^{-1}[g](x).$$ But on $\mathcal{C}^{1}$, this operator likely does not admit an inverse as it is not injective, so you do it on the quotient space $\mathcal{C}^1/\sim$ where $f\sim g\iff f(x)=g(x)+Ce^x$ for some constant $C$.

That's why the function you solved out is generally an equivalent class just like you solve indefinite integral, you need to add $+C$ at the end in this case you should add $+ Ce^x$.

But finally, when you differentiate the integral equation, you lose some information as we cannot distinguish it come from which element in the equivalence class. So the original integral equation is actually giving you more information than $f'(x)+2f(2x)-f(x)=0$. And precisely it gives $f(0)=0$. Thus using this initial condition you can show that the above $C$ is actually $0$.

Finally from the functional equation, since the above process we explicitly assume $f$ is differentiable, $f$ is continuous, so for every $y\in\mathbb{R}$, $$f(y)=\dfrac{1}{2}f\left(\dfrac{y}{2}\right)=\cdots=\dfrac{1}{2^n}f\left(\dfrac{y}{2^n}\right)\quad\forall~n\in\mathbb{N}.$$ So continuity of $f$ forces $f(y)\equiv f(0)$ for every $y$.

Answer 4

Firstly, note that the operator $D-1$ acts on the left of $f$, that is why its inverse must be applied on the left too, hence $f(x) = (1-D)^{-1}2f(2x) \neq 2f(2x)(1-D)^{-1}$. Also, this kind of approach coming from operational calculus is most suitable for inhomogeneous equations, but it is rarely useful in the homogeneous case. Indeed, you could write an expression such as $(1-D-2S_2)f = 0$, where $S_2(f) := f(2x)$ is a scaling operator, hence $f = (1-D-2S_2)^{-1}0 = 0$, but it cannot produce another solution than the trivial zero solution, since the operator $(1-D-2S_2)^{-1}$ is linear.

Let's highlight that the equation satisfied by $f$, namely $f'(x) + f(x) - 2f(2x) = 0$ (watch out for the sign error), belongs to the family of pantograph delay differential equations. You could get rid of the middle term with the help of an integrating factor by setting $f(x) = g(x)e^{-x}$, hence $$ \frac{\mathrm{d}g(x)}{g(2x)} = 2e^{-x}\mathrm{d}x $$ after seperation of the variables. You can find here a few basic things about pantograph equations. They are intimately related to the deformed exponential function $\operatorname{Exp}_\alpha$, defined as the unique solution to the problem $y'(x) = y(\alpha x)$, with $y(0) = 1$, and described by the following series : $$ \operatorname{Exp}_\alpha(x) = \sum_{n\ge0} \alpha^{n(n-1)/2} \frac{x^n}{n!} $$ Then, the left-hand side of the above differential equation becomes $\operatorname{Log}_2(g)$ after integration, where $\operatorname{Log}_2 = \operatorname{Exp}_2^{-1}$ is the associated deformed logarithm, leading to $g(x) = \operatorname{Exp}_2(C-2e^{-x})$, where $C$ is a constant of integration, hence $f(x) = e^{-x}\operatorname{Exp}_2(C-2e^{-x})$ in the end. Finally, the initial condition $f(0) = 0$ is imposed by your original integral equation, so that the final solution is given by $$ f(x) = e^{-x} \sum_k \operatorname{Exp}_2(C_k - 2e^{-x}) $$ by superposition, where the constants $C_k$ are solutions to the equation $\operatorname{Exp}_2(C_k-2) = 0$. The said constants are thus related to the zeroes of the deformed exponential $\operatorname{Exp}_\alpha$. The latter are known to be negative when $0 < \alpha < 1$ and complex when $\alpha > 1$. Consequently, in the case of a real-valued $f$, the zero function might be the only solution, but I am not sufficiently knowledgeable about this topic in order to ensure such a statement.