Let $F: C[0,L] \to C[0,L]$ be a linear operator defined as $Fu(x) = \int_0^x u(y)dy$. We wish to show this operator is continuous, one-to-one, find it's norm, and find it's eigenvalues.
I've already proved the first three properties, but I have never found the eigenvalues of functions before, only matrices.
I've set up the problem as $$Fu(x) = \lambda u(x)$$
$$\int_0^x u(y)dy = \lambda u(x)$$
however I don't feel confident when I try to touch this any further.
You have to distinguish between two cases. First, assume that $\lambda \neq 0$. Then the equation
$$ (Fu)(x) = \int_0^x u(t) \, dt = \lambda u(x) $$
together with the fact that $f$ is continuous and the fundamental theorem of calculus implies that if $u$ is an eigenvector (often called eigenfunction, for obvious reasons) of $F$ then $u$ must be continuously differentiable. Thus, we can differentiate both sides and obtain the differential equation
$$ u(x) = \lambda u'(x) $$
whose solutions are given by $u(x) = Ae^{\frac{x}{\lambda}}$ for $A \in \mathbb{R}$. By plugging in $x = 0$ into the original equation, we see that we must have $\lambda u(0) = \int_0^0 u(t) \, dt = 0$ so $u(0) = 0$ which implies that $A = 0$ and so $u \equiv 0$. Hence, there are no eigenfunctions for $\lambda \neq 0$.
Now, assume that $\lambda = 0$. By looking at the equation above, we can again differentiate both sides and obtain $u(x) \equiv 0$ which shows that there are no eigenfunctions for $\lambda = 0$.