Minimizing the $L^2$ norm of the derivative (Schilder's theorem) We define the following functional on $C^0([0,1],\mathbb{R})$:
$$
I(f)=\begin{cases}
\frac{1}{2}\int_0^1(f')^2=:\frac{1}{2}\|f'\|_2^2&\text{ if $f$ is absolutely continuous with square-integrable derivative}\\
\infty&\text{ otherwise}
\end{cases}
$$
and we would like to minimise this over the set $C=\{f:\int_0^1f^2\geq 1\}$, i.e. find the quantity
$$
\inf_{f:\|f\|_2\geq 1}I(f)=\inf_{f:\|f\|_2\geq 1}\tfrac{1}{2}\|f'\|_2^2.
$$
This problem actually comes from p.17 in "Large Deviations" by Varadhan, where we're trying to apply Schilder's theorem for large deviations for Brownian motion to the set $C$ above. Varadhan just says this is an eigenvalue problem without elaborating, but I'd welcome some help on understanding how to solve this!
 A: This is mainly a consequence of Lagrange multipliers. First note that if $||f||_2 > 1$, you can rescale your energy with $f \mapsto \frac{1}{||f||_2}f$ and decrese $I(f)$ that way. So we can simplify the constraint to $||f||_2^2=1$.
For $I=(0,1)$ and $\langle \cdot, \cdot \rangle$ the standard $L^2(I)$ scalar product, the Euler-Lagrange associated to $I(f)$ is given by
$$
\langle f',\phi' \rangle=0  \; \; \forall \phi \in C_c^{\infty}(I)
$$
and for your constraint it is given by
$$
\langle f, \phi \rangle =0 \; \; \forall \phi \in C_c^{\infty}(I)
$$
Using the methods of Lagrange multipliers, you know that a solution of your variational problem satisfies
$$
\langle f',\phi' \rangle + \lambda \langle f,\phi \rangle=0 \; \; \forall \phi \in C_c^{\infty}(I)
$$
for some $\lambda \in \mathbb{R}$. Now you can apply regularity theory to deduce that $f \in C^2(I)$ and then integrate by parts to arrive at
$$
\langle f''-\lambda f, \phi \rangle=0 \; \; \forall \phi \in C_c^{\infty}(I)
$$
from which follows
$$
f''=\lambda f
$$
This is an eigenvalue problem for the linear operator defined by $Tf=f''$.
Solving this problem can be done via abstract existence theory for eigenvalues or the direct method of the calculus of variations - there could also be a more elementary approach to this problem which I can not recall as of now. There are many routes to take, just pick one that you are familiar with.
