It is well-known that Brownian motion is a Gaussian process with mean function $\mu(t) = 0$ and covariance kernel $k(s, t) = \min(s, t)$. Considering that we consider the time domain of the stochastic process to be $[0, 1]$, we solve the following covariance eigenvalue problem: \begin{align} \int_{0}^{1} k(s, t) f(s) ~\mathrm{d}s = \lambda f(t) \end{align} We guess for the eigenfunction the following formula: \begin{align} \phi_n(s) = \sqrt{2} \sin((n - 1/2) \pi s), \end{align} with eigenvalue $\lambda_n = 1 / (\pi^2 (n - 1/2)^2)$.

Let us write the eigenproblem as follows, breaking up $k(s, t) = \min(s, t)$ across its natural boundary at $s=t$:
\begin{align}
\int_{0}^t s f(s) ~\mathrm{d}s + \int_t^1 t f(s) ~\mathrm{d}s = \lambda f(t).
\end{align}
Now we differentiate both sides with respect to $t$, obtaining via two applications of Leibniz’s integral rule,
\begin{align}
\lambda f’(t) &= t f(t) + \int_t^1 f(s) ~\mathrm{d}s - t f(t) \\

&= \int_{t}^1 f(s) ~\mathrm{d}s.
\end{align}
Differentiating yet again yields,
\begin{align}
\lambda f''(t) = -f(t).
\end{align}
This is a second-order linear ordinary differential equation for which the solutions are of the form
\begin{align}
f(t) = c_1 \cos(t / \sqrt{\lambda}) + c_2 \sin(t / \sqrt{\lambda}).
\end{align}
We can obtain some boundary conditions for this differential equation by noting that
\begin{align}
\lambda f(0) &= \int_0^1 \min(s, 0) f(s) ~\mathrm{d} s \\

&= 0 \\

\lambda f’(1) &= \int_{1}^1 f(s)~\mathrm{d}s \\

&= 0.
\end{align}
Hence we conclude that $f(0) = 0$ and $f’(1) = 0$. From the first of these conditions we conclude that $c_1 = 0$, leaving us with candidate eigenfunctions of the form $f(t) = c_2 \sin(t / \sqrt{\lambda})$. From the second boundary condition we obtain,
\begin{align}
& f’(t) = \frac{c_2}{\sqrt{\lambda}} \cos(t / \sqrt{\lambda}) \\

\implies& f’(1) = \frac{c_2}{\sqrt{\lambda}} \cos(1 / \sqrt{\lambda}) = 0
\end{align}
For a non-trivial solution, we therefore see that we require
\begin{align}
& \cos(1 / \sqrt{\lambda}) = 0 \\

\implies& \sqrt{\lambda} = \frac{1}{\pi (n - 1/2)},
\end{align}
where $n\in \mathbb{Z}$. Clearly, there remains an additional degree of freedom in $c_2$, which we require not equal zero for a non-trivial solution. However, by integrating,
\begin{align}
\int_0^1 \abs{f(s)}^2 ~\mathrm{d}s &= \abs{c_2}^2 \int_0^1 \sin^2\left((n-1/2) \pi s\right) ~\mathrm{d}s \\

&= \abs{c_2}^2 \int_0^1 \left(\frac{1 - \cos((n-1/2)\pi s)}{2}\right) ~\mathrm{d}s \\

&= \abs{c_2}^2 \left(\frac{1}{2} + 0\right) \\

&= \frac{\abs{c_2}^2}{2}.
\end{align}
Therefore, we see that we can choose $c_2^2 = 2$ in order to obtain a unit eigenfunction.