Properties of functions provided that the integral equation $\int_x^1 a(y-x)a(y) \, dy = \int_x^1 b(y-x)b(y) \, dy$ holds for $x \in [0,1]$

Question

Let $a,b : [0,1] \to \mathbb R$ are two functions (e.g. suppose that they are in $L^2[0,1]$ or are $N$-times continuously differentiable). Now suppose that $$ \int_x^1 a(y-x)a(y) \, dy = \int_x^1 b(y-x)b(y) \, dy $$ for every $x \in [0,1]$. I'm interested in properties of $a,b$ and how they are related, provided that $a,b$ satisfy the above equation of integrals for every $x \in [0,1]$. Clearly, they do not have to be equal since if $a=-b$ then the above equation holds. I'm wondering of this is the only thing which can happen. Did someone of you came across such a problem before?

Thanks in advance!

Answer 1

There are plenty of examples of such pairs of functions. First, extend your $a$ and $b$ to the whole real line by setting them to $0$ outside the interval $[0,1]$. So your relation becomes $$\int_{-\infty}^\infty a(y-x)a(y)dy=\int_{-\infty}^\infty b(y-x)b(y)dy.$$ Your condition is that this must hold only for $x\in[0,1]$, but I will construct many examples in which this will hold for all real $x$. Suppose first that this equation holds for all real $x$. Now let $f(x)=a(-x),\; g(x)=g(-x)$. Then your equation is equivalent to $$a\star f=b\star g,$$ where $*$ is the convolution. Taking Fourier transforms and using $\hat{f}=\hat{a}(-s)$, and similar for $\hat{g}$, we obtain $$\hat{f}(s)\hat{f}(-s)=\hat{g}(s)\hat{g}(-s),\quad\quad\quad\quad\quad\quad (1)$$ and this is equivalent to the assumption that your equation holds for all real $x$. Now $\hat{f},\hat{g}$ are entire functions of exponential type, bounded in the lower half-plane (by the Wiener-Paley theorem), so $$\hat{f}(s)=e^{cs}\prod_j\left(1-\frac{s}{s_j}\right)e^{s/s_j},$$ and they are determined by their zeros and the exponential factor in front. In equation (1), the exponential factors cancel, so you already have plenty of examples: take $a(x)$ with some small support on $(0,1)$ and let $b$ be a small shift: $b(x)=a(x-\epsilon)$ so that it also has support on $(0,1)$.

But there is much more. Let $Z$ be the set of zeros of $\hat{f}$. Then zeros of $\hat{f}(s)\hat{f}(-s)$ are $Z\cup(-Z)$, and by decomposing this union in some other way: $Z\cup(-Z)=Z_1\cup(-Z_1)$ you obtain a function $\hat{g}$ for which (1) holds. Typically $\hat{f}$ will have infinitely many zeros, so such decomposition can be done in infinitely many ways, by redistributing, say, finitely many of zeros.

To make sure that all these $\hat{f},\hat{g}$ are really Fourier transforms of some functions supported on $[-1,0]$ one uses the Wiener-Paley theorem.

Answer 2

comment
I have seen something similar. But not the same. The set of continuous real-valued functions $C[0,\infty)$ with pointwise addition and convolution for multiplication $$ (f * g)(x) = \int_0^x f(x-t)g(t)\;dt $$ is an integral domain (without unit). Consequently, such a function can have at most $2$ square-roots... $$ f*f = g*g \\ f*f-g*g = 0 \\ (f-g)*(f+g)=0 \\ f-g = 0 \text{ or } f+g = 0 \\ f=g \text{ or } f=-g $$

That "integral domain" result is due to Titchmarsh .

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For the OP question, a similar approach would be to let $c=a-b, d=a+b$ and then ask: is it true that $$ \int_x^1 c(y-x)d(y)\;dy = 0\quad\text{for all } x \in [0,1] \\ \Downarrow\\ \big[c(x) = 0 \quad\text{for all } x \in [0,1]\big]\quad\text{ or }\quad \big[d(x) = 0 \quad\text{for all } x \in [0,1]\big] $$