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Suppose I want to find all the continuous solutions to the functional equation $$f(x)+f(x+1)=f(2x+1),\tag{E1}$$where $f$ is a continuous and $2$-periodic function defined on the dyadic rationals.

I know that if I have any continuous function $g$ satisfying $$g(x)+g(x+1) = 0,\tag{E2}$$ then I can build a continuous solution for the original equation as the uniformly convergent series $$f(x) = g(x) - g(2x)/2 - g(4x)/4 - \cdots$$

Indeed, we get $f(x)+f(x+1)-f(2x+1) = g(x)+g(x+1)-g(2x)-g(2x+1) = 0$.

Conversely, from a solution $f$ for $(\rm E1)$ I can build the corresponding solution $g$ for $(\rm E2)$, as the simply convergent series $$g(x) = f(x) + f(2x)/2 + f(4x)/2 + \cdots - f(1)/2.$$ For every dyadic number $x$, eventually $2^nx$ is an even integer and then $f(2^n x) = f(0) = 0$, so this is why the series converges.

And again, we get $$\begin{align*} g(x)+g(x+1) &= f(x)+f(x+1)+f(2x)+f(4x)+\cdots -f(1) \\ & = f(2x)+f(2x+1)+f(4x)+\cdots-f(1) \\ & = \cdots = f(1)-f(1) \\ & = 0. \end{align*}$$

But this time, because $g$ is not uniformly convergent, there is no guarantee that $g$ will be continuous.

These two operations are inverse of each other, so when looking at non-continuous solutions, there is a one-to-one correspondence between solutions of the two equations. But one direction preserves continuity, while the other doesn't.

Thus, my question is:

Are there continuous solutions for $(\rm E1)$ corresponding to non continuous solutions for $(\rm E2)$ ?

I don't see a priori why there wouldn't be, but I can't grasp a way to build one either. I also know that Fourier theory can be used to answer this kind of problems, but I don't know if it gives stronger results than what I have just exposed.

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Nice question +1 –  mick Jul 15 '13 at 21:36

1 Answer 1

up vote 8 down vote accepted

If you look at the Fourier side and write $f(x)=\sum_k c_ke^{\pi ikx}$, you'll see that your equation is equivalent to the relations $c_{2k}=(-1)^k\frac 12c_k$, so if $f$ is any continuous function satisfying E1, then $f(x)=g(x)-g(2x)/2-g(4x)/4-\dots$ where $g(x)=\sum_{k\text{ odd}}c_k e^{\pi ikx}=\frac 12(f(x)-f(x+1))$, so the $g$ to $f$ correspondence between continuous solutions is one-to-one.

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Hello fedja, welcome to math.SE, it's a nice surprise to see you around here! (We didn't have much interaction on MO but I read many of your answers over there) –  t.b. Jul 5 '11 at 21:13
    
Ah indeed, this is a much nicer way to write that inverse operation than the one I made up, everything is easier with it, too. Thanks ! –  mercio Jul 5 '11 at 21:49

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