Periodic solutions of integral equations

Remarks

The same notion applies to systems, where $x$ takes values in $\R^n$, and to equations of Fredholm type over one period, such as $x(t) = f(t) + \int_0^T k(t - s) \, x(s) \, ds$ with a $T$-periodic kernel $k$.

Example: $T = 2\pi$, $C(t, s) = e^{-(t - s)}$, $g(s, x) = x$ and $f(t) = \frac{1}{2}(\sin t + \cos t)$. The function $x(t) = \sin t$ is continuous and $2\pi$-periodic. The integral $\int_{-\infty}^t e^{-(t - s)} \sin s \, ds$ converges absolutely and equals $\frac{1}{2}(\sin t - \cos t)$, as the imaginary part of $\int_{-\infty}^t e^{-(t - s)} e^{i s} \, ds = e^{i t} / (1 + i)$. Adding $f(t)$ gives $\sin t$, so all three clauses hold.

Non-example: periodic data do not force periodic solutions. The equation $x(t) = 1 + \int_0^t x(s) \, ds$ on $[0, \infty)$ has constant data, but its only continuous solution is $x(t) = e^t$, which fails clause 2 for every $T$.