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Is there a function $f(x)$ on the real domain and real constants $a$ and $b\neq 0$ for which the following is true:

$$f(x)-f(x-\delta)+a+bx^2=0$$ for some real $\delta\neq 0$?

EDIT: I missed a very important constraint in the original posting of this question: $b\neq 0$. Sorry!

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For a trivial example take any constant function $f(x)$, any $\delta$ and $a=b=0$. – Dennis Gulko Sep 12 '12 at 19:00
Have you tried looking for one? – Erick Wong Sep 12 '12 at 19:01
May you have missed some constraint? As it is, the problem is somewhat trivial. Maybe $b \neq 0$? – Karolis Juodelė Sep 12 '12 at 19:03
Oops, yes, I did miss a constraint... $b\neq0$ – M.B.M. Sep 12 '12 at 20:09
up vote 3 down vote accepted


Let's develop a concrete solution, shall we? Observe that for any polynomial $p(x)$ of degree $n>0$ and any $\delta$, we have that $p(x)-p(x-\delta)$ is again a polynomial, having degree at most $n-1$. This suggests that we try for a degree $3$ polynomial. For the moment, we may as well assume the constant term is $0$, since if not, there'd be cancellation in our subtraction, anyhow. Let $f(x)=c_1x+c_2x^2+c_3x^3$, so we need $$-a-bx^2=f(x)-f(x-\delta)=(c_1\delta-c_2\delta^2+c_3\delta^3)+(2c_2\delta-3c_3\delta^2)x+3c_3\delta x^2.$$ Equating the coefficients of the terms of same degree, we see that: $$-a=c_1\delta-c_2\delta^2+c_3\delta^3$$ $$0=2c_2\delta-3c_3\delta^2\quad(\text{so}\: 0=2c_2-3c_3\delta,\:\text{since}\:\delta\neq 0)$$ $$-b=3c_3\delta$$ The third of these equations gives us $$c_3=-\frac{b}{3\delta}$$ since $\delta\neq 0$, and substituting $-b$ in for $3c_3\delta$ in the adapted second equation yields $$c_2=-\frac{b}2.$$ Finally, the first equation becomes $$-a=c_1\delta-c_2\delta^2+c_3\delta^3=c_1\delta+\frac{b\delta^2}2-\frac{b\delta^2}3=c_1\delta+\frac{b\delta^2}6\!,$$ so $$c_1=-\frac{a}\delta-\frac{b\delta}6=\frac{6a+b\delta^2}6.$$

Thus, we have in fact that for any $a,b$ and any $\delta\neq 0$, the polynomial family $$f(x)=d-\frac{6a+b\delta^2}6x-\frac{b}{2}x^2-\frac{b}{3\delta}x^3$$ satisfies the desired functional equation for all $d$.

Addendum It occurred to me (once I processed Karolis's comment below) that instead of the arbitrary $d$, and instead of requiring $f$ to be a polynomial, we can take any function $g$ with period $\delta$ (of which constant functions are merely an uninteresting family of examples), and then let $$f(x)=g(x)-\frac{6a+b\delta^2}6x-\frac{b}{2}x^2-\frac{b}{3\delta}x^3.$$ I'm not sure if this describes all $f$ that fit this equation, as I've never done any serious functional analysis, but it certainly gives a fairly general family of solutions.

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Hehe, our first two ideas were the simplest ones possible, I think :) – rschwieb Sep 12 '12 at 19:04
With the same logic any $f(x) = g(x) + kx$ where $g$ has period $\delta$ works. – Karolis Juodelė Sep 12 '12 at 19:05
Good observation, Karolis! – Cameron Buie Sep 12 '12 at 19:11
In fact constant functions are the degenerated special cases of the periodic functions with any period. – doraemonpaul Sep 13 '12 at 1:11
True. As I recall, a function is continuous on $\Bbb R$ and having every period iff it is constant. Probably the same works for any (separable) Banach space. – Cameron Buie Sep 13 '12 at 1:32

Take the function $f$ be a polynomial of degree $3$ and solve for the coefficients. $$f(x)=a_3x^3+a_2 x^2 +a_1x+a_0$$ let us take the difference in respect of the powers $$a_3x^3-a_3(x-\delta)^3=3a_3\delta x^2-3a_3x \delta ^2+\delta ^3 a_3$$ $$a_2x^2-a_2(x-\delta)^2=2a_2\delta x-\delta ^2 a_2$$ $$a_1x-a_1(x-\delta)=a_1 \delta$$ Hence $$f(x)-f(x-\delta)=3a_3\delta x^2+(2a_2\delta-3a_3 \delta ^2)x+\delta ^3 a_3-\delta ^2 a_2+a_1 \delta$$ Now since $\delta$ is not zero you can solve this system $$3a_3\delta=b$$ $$2a_2\delta-3a_3 \delta ^2=0$$ $$\delta ^3 a_3-\delta ^2 a_2+a_1 \delta=a$$ which is linear with unique solution. Also notice that $a_0$ can be arbitary (as will Erick Wong notified me)

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Of course the constant term of $f$ could be arbitrary rather than $0$. – Erick Wong Sep 12 '12 at 19:35
Thanks, I will add it. – clark Sep 12 '12 at 20:18



In fact this belongs to a functional equation of the form

The general solution of this functional equation is $f(x)=\Theta(x)+f_p(x)$ , where $\Theta(x)$ is an arbitrary periodic function with period $\delta$

Luckily we can find $f_p(x)$ by method of undetermined coefficients:

Let $f_p(x)=Ax^3+Bx^2+Cx$ ,

Then $f_p(x-\delta)=A(x-\delta)^3+B(x-\delta)^2+C(x-\delta)=Ax^3-3A\delta x^2+3A\delta^2x-A\delta^3+Bx^2-2B\delta x+B\delta^2+Cx-C\delta=Ax^3+(B-3\delta A)x^2+(3\delta^2A-2\delta B+C)x-\delta^3A+\delta^2B-\delta C$

$\therefore Ax^3+Bx^2+Cx-(Ax^3+(B-3\delta A)x^2+(3\delta^2A-2\delta B+C)x-\delta^3A+\delta^2B-\delta C)\equiv-bx^2-a$

$3\delta Ax^2+(2\delta B-3\delta^2A)x+\delta^3A-\delta^2B+\delta C\equiv-bx^2-a$

$\therefore\begin{cases}3\delta A=-b\\2\delta B-3\delta^2A=0\\\delta^3A-\delta^2B+\delta C=-a\end{cases}$


$\therefore f(x)=\Theta(x)-\dfrac{bx^3}{3\delta}-\dfrac{bx^2}{2}-\dfrac{b\delta x}{6}-\dfrac{ax}{\delta}$ , where $\Theta(x)$ is an arbitrary periodic function with period $\delta$

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The restrictions are extremely mild in this question: if $g_x(y)$ is any family of bijections on $\mathbb R$, then for any $\delta > 0$ there are infinitely many solutions to the functional equation $f(x-\delta) = g_x(f(x))$: just define $f$ arbitrarily on $[0,\delta)$ and use the functional equation to extend it in either direction ($g_x$ has an inverse for any $x$).

In your case you want $g_x(y) = y + a + bx^2$. No matter what $a$ and $b$ are, this is a bijection.

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Let $f(x)=x$, $b=0$ and $a=-\delta\neq 0$.

Edit: This was posted before the OP introduced the $b\neq 0$ condition.

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Uh oh! I guess you missed the restriction $b\neq 0$ that the OP added later, and someone downvoted you. I've upvoted you to compensate (as it was a fine answer based on the initial post). – Cameron Buie Sep 13 '12 at 0:37

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