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The problem is to find all $f : \mathbb{R} \to \mathbb{C}$ that is continuous and has a period of $1$ (not necessarily smallest period) such that the following equality holds:

$$f(x) = \int_0^1 f(x-t)f(t) dt \quad \forall x \in \mathbb{R}$$

I believe the only answer is $f=0$ and $f=1$, but I don't know how to go about proving it.

Harry Peter
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user99185
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2 Answers2

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Quick look: since $f$ is continuous and periodic, it has a (unique) Fourier series of the form $f(x) = \sum_{k\in\mathbb Z} c_k e^{i2\pi kx}$.

Inject this form in the integral and you get (I don't even care about switching sums and integrals at this stage):

$$f(x) = \int_0^1 \sum_{k,m\in\mathbb Z} c_k c_m e^{i2\pi k(x-t)}e^{i2\pi mt}dt = \sum_{k,m\in\mathbb Z} c_k c_m e^{i2\pi kx} \int_0^1 e^{i2\pi(m-k)t}dt$$

the term under the integral is nil unless $m=k$, hence:

$$f(x) = \sum_{k\in\mathbb Z} c_k e^{i2\pi kx} = \sum_{k\in\mathbb Z} c_k^2 e^{i2\pi kx}$$

And you get: $\forall k, c_k = c_k^2$, i.e. $c_k \in \{0,1\}$.

As stated in the other (better) answer, Plancherel theorem then implies $\sum |c_k|^2 < \infty$, which can only happen if $c_k=0$ for all but a finite number of $k$. So $f$ is of the form $\displaystyle f(x) = \sum_{m=1}^N e^{i2\pi k_m x}$ with $k_m\in \mathbb Z$ all distinct.

Alexandre Halm
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  • Smart! Didn't think to use fourier series, but I guess it's natural since $f$ is continuous and periodic. – user99185 Mar 09 '15 at 06:07
  • How would the functions $f(x) = \sum c_k e^{i 2 \pi k x}$ with $c_k \in {0,1}$ look like? I suppose we can think of it as a sum of harmonics of period $1$ and magnitude $1$. $c_k$ being equal to $0$ or $1$ just tells us whether or not the harmonic is present. – user99185 Mar 09 '15 at 06:08
  • Also, is continuous and periodic strong enough to guarantee the convergence of the Fourier series to $f$? Do we not need also differentiable? I'm a bit rusty on my fourier series. – user99185 Mar 09 '15 at 06:13
  • Clearly the argument above is really short of justifications and is there just to give an intuition of what is happening. If I remember well, continuous implies point-wise converge almost everywhere. Ideally, you would want $f$ to be $C^2$, so that $c_k$ are $O(1/k^2)$ and all sums converge absolutely without question. – Alexandre Halm Mar 09 '15 at 06:19
  • What is meant by the notation $C^2$? – user99185 Mar 09 '15 at 06:44
  • $f''$ continuous. But anyway @MathJow995's shows that there is no need for that since Plancherel's theorem guarantees convergence. – Alexandre Halm Mar 09 '15 at 06:50
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Well, let's try to use a bit of Fourier Analysis.

As $f$ is continuous, 1-periodic, we can then calculate its Fourier Transform, which is defined by

$$ \hat{f}(n) = \int_0^1 f(t)e^{-2 \pi i t n} dt $$

Using this in the given formula, we have that

$$ \hat{f}(n) = (\hat{f}(n))^2 $$

Which implies that $\hat{f}(n) \in \{0,1\}$. But $f$ is continuous $\Rightarrow f \in L^2(0,1)$, which, by the Plancherel's Theorem, implies that

$$ \sum_n |\hat{f}(n)|^2 < \infty $$

But this is a sum of zeroes and ones. So, there must be an $n_0$ such that $|N| \ge n_0 \Rightarrow \hat{f}(N) = 0 $.

This shows that the solutions to the given problem are exactly the functions of the form

$$ f(x) = \sum_{k \in \mathbb{Z}} \varepsilon_k e^{2 \pi i k x} $$

Where $\varepsilon_k \in\{0,1\}$, and, for $|k|$ sufficiently large, $\varepsilon_k = 0$.

João Ramos
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  • Of course, good answer. I'll point to your answer in mine. – Alexandre Halm Mar 09 '15 at 06:43
  • Actually I think it's Parseval's theorem rather than Plancherel in this case. – Alexandre Halm Mar 09 '15 at 06:57
  • Well, I think you may use both of them, as Parseval's Theorem may stand for a more general functional analytic version, while Plancherel's Theorem is the "continuous" analogue to this one. – João Ramos Mar 09 '15 at 06:58
  • What is $L^2(0,1)$? Functions defined on $(0,1)$ with $L^2 $ norm? – user99185 Mar 09 '15 at 07:15
  • Yes! The Parseval's (or Plancherel's) Theorem states that, if a function $F:[0,1]\rightarrow \mathbb{C}$ is square-integrable, then the identity $\int_0^1 |F(x)|^2 dx = \sum_{n \in \mathbb{Z}} |\hat{F}(n)|^2$ holds. – João Ramos Mar 09 '15 at 07:17
  • @MathJow995 Why does that imply that $\sum |f(n)|^2$ is bounded? because $f$ is bounded? – user99185 Mar 09 '15 at 15:15
  • @user99185 Note that we only assert that $\sum |\hat{f}(n)|^2$ is finite, and this follows directly from the boundedness of $f$ and from the formula above. – João Ramos Mar 09 '15 at 15:22
  • @MathJow995 ah, thanks for the clarification – user99185 Mar 09 '15 at 15:23
  • @JoãoRamos Why is it justified to replace $f$ with its Fourier series? I thought that $f$ smooth implies that the Fourier series converges uniformly to $f$ (Dirichlet), but $f$ only continuous implies that the Fourier series is Cesaro summable to f (Fejer's theorem)? – user99185 Mar 09 '15 at 16:15
  • @user99185 Ok, I really did not write all the details, but you can do it this way: It is straightforward to check that a sum of the mentioned form is a solution to the functional equation.

    Conversely, if $f$ is a solution, by what we've made, then it has at most finitely many nonzero fourier coefficients, each of them being 1. By Fourier Inversion, as then $\sum_n |\hat{f}(n)| < \infty$, we have that $f(x) = \sum_{k \in \mathbb{Z}} \epsilon_k e^{2\pi i k x}$ for almost every $x \in [0,1]$, and, as the RHS and LHS are both continuous functions of $x$, we conclude that they agree everywhere.

     – João Ramos Mar 09 '15 at 16:30
  • Let us continue this discussion in chat. – user99185 Mar 09 '15 at 23:18