Define $E=C^2([0,1], \mathbb{R})$ and
$V=\{u \in E | u(0)=u(1)=0\}$ and
$ F=\{u \in E | u(0)=0 ,u(1)=1\}$
I want to evaluate the minimum of
$$ J = inf_{u\in F} \left(\int _{[0,1]} u^2+(u\prime)^2 \right) $$
in the proof, we consider the scalar product in $E$ :
$$<u,v>= \int_{[0,1]} uv+(u\prime v\prime)$$
and we state that $F$ is $(1)$ "an affine space of $E$ and direction $V$, and the $codim V=2$ "
and therefore $(2)$"the dimension of $V^{\perp}$ is at most 2 "
and given that $dimV^{\perp}\leq2 $ we prove that $V^{\perp}= Vect \{ch,sh\}$
And then they say $(3)$ "that the minimum $J$ occurs once and only once in $F \cap V^{\perp}$ "
which yields that minimum is at $x \rightarrow sh(x)/sh(1)$
I have no idea why these three bullet points are true, can someone explain?
Fix any $f\in F$ (say $f(x)=x$), then observe that $u\in F\iff u-f\in F$, or alternatively, $F=V+f$.
Both functionals $u\mapsto u(0)$ and $u\mapsto u(1)$ have $1$-codimensional kernel, so their intersection has codimension (at most) $2$.
Since $\bar V\oplus V^\perp =E$, we have $\dim V^\perp ={\rm codim}\, \bar V\le{\rm codim}\, V\le 2$.
We have an affine subspace and we are looking for its point with the smallest distance from the origin (i.e. with smallest norm).
The situation is totally analogous to having an affine plane $F=V+f$ (parallel to the plane $V$ which contains the origin) in the 3d space, and looking for its closest point to the origin: we just take the intersection $V^\perp\cap F$ which should consist of a single element, namely the orthogonal projection $f_{\perp V}$ of $f$.
And, actually the Pythagorean theorem - which holds in arbitrary inner product space - shows that any other $u\in F\setminus V^\perp$ has strictly greater norm than $f_{\perp V}$.