Find continuous $f$ with period $1$ such that $f(x) =\int_0^1 f(x-t)f(t) dt$ 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.
 A: 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.
A: 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, $\varepsilon_k = 0$. 
