I am studying Roger Temam's Infinite-Dimensional Dynamical Systems in Mechanics and Physics (Second Edition) and am trying to work through section 4.1.2 - Eigenvalue of a Schrödinger Operator. For $\Omega = \left(-\frac{L}{2}, \frac{L}{2} \right)$, He defines the operator $K_0: D(K_0)\subseteq \dot L^2\left(\Omega\right) \to \dot L^2\left(\Omega\right) $ as $$ K_0 f = \nu D^4f -qf,$$ where $$q \in \left\{\ \phi\in C^\infty(\mathbb{R}): \phi(x+L) = \phi(x), \quad \phi(-x) = \phi(x), \quad \int_\Omega \phi dx = 0 \right\},$$ the space of infinitely differentiable, mean-free, even, and periodic functions, and $$ D(K_0) = \left\{f\in H^4(\Omega): f(x+L) = f(x), \quad f(-x) = -f(x), \quad \int_\Omega f dx = 0 \right\},$$ the space periodic odd functions which are mean free and are four times weakly differentiable. Temam also asserts that this operator is subject to boundary conditions $$\partial_j u\left(-\frac{L}{2}, t \right) = \partial_j u\left(\frac{L}{2}, t \right), \quad j = 0, 1,2,3.$$
Here is the part of this section that I am confused about. From here, Temam asserts that the first eigenvalue (first in the sense that it has the least value) is $$ \lambda_1 = \inf_{f\in D(K_0)}\frac{(K_0f, f)}{\| f\|^2_{L^2}}.$$
Now, if $\lambda$ is an eigenvalue of $K_0$ with associated eigenfunction $g$ then we see that $$ \frac{(K_0f, f)}{\| f\|^2_{L^2}} = \frac{(\lambda f, f)}{\| f\|^2_{L^2}} = \frac{\lambda ( f, f)}{\| f\|^2_{L^2}} = \lambda. $$ So I understand that $\lambda_1 \leq \lambda $ for any eigenvalue $\lambda$. From here, I do not know how to proceed. How can one find an eigenfunction which has $\lambda_1$ as its eigenvalue?