Xrfgtjtk

I'm trying to understand an article of Reinsch (1967), on smoothing spline functions. The author uses some rules that unfortunately I couldn't find.

The minimized functional is:
$$
int_{x_0}^{x_n} g''(x)^2 dx + p leftlbrace sum_{i=0}^nleft(frac{g(x_i)-y_i}{delta y_i}right)^2 + z^2 -S rightrbrace
$$

With this, the author deduces that:
$$
forall i, f^{(3)}(x_i)_{-} - f^{(3)}(x_i)_{+} = 2p frac{f(x_i)-y_i}{delta y_i}
$$
where $f^{(3)}$ is undefined in each $x_i$, $f^{(3)}(x_i)_{-}$ is the inferior limit in $x_i$, and $f^{(3)}(x_i)_{+}$ the superior limit.

This is the point I don't get. I understand how to compute the derivative of a functional when it is formulated as an integral, but this one isn't. I guess there is some rules I missed.

The author cite a book (Variationsrechnung und ihre Anwendung in Physik und Technik, Funk, 1962), but unfortunately I can't read German, and I couldn't find any source to corroborate Reinsch in his reasoning.

What is the derivation rule I missed? Is there a source (in English or in French) to corroborate the author computation?

Thanks!

Consider the functional
$$
F(g)=int_{x_{0}}^{x_{n}}(g^{primeprime}(x))^{2}dx+psum_{i=0}^{n}left(
frac{g(x_{i})-y_{i}}{delta y_{i}}right) ^{2}
$$
and let $f$ be a minimum over all function $gin C^{2}([x_{0},x_{n}])$. Then
taking $f+th$, for $tinmathbb{R}$ and $hin C^{2}([x_{0},x_{n}])$, you have
that
$$
F(f+th)geq F(f)
$$
and so the one variable function $k(t)=F(f+th)$ has a minimum at $t=0$. Hence,
$k^{prime}(0)=0$. So if we now differentiate under the integral sign, we get
begin{align*}
k^{prime}(t) & =frac{d}{dt}(F(f+ht))=frac{d}{dt}int_{x_{0}}^{x_{n}
}(f^{primeprime}(x)+th^{primeprime}(x))^{2}dx\&quad+pfrac{d}{dt}sum_{i=0}
^{n}left( frac{f(x_{i})+th(x_{i})-y_{i}}{delta y_{i}}right) ^{2}\
& =int_{x_{0}}^{x_{n}}2(f^{primeprime}(x)+th^{primeprime}(x))h^{prime
prime}(x),dx\&quad+2psum_{i=0}^{n}frac{(f(x_{i})+th(x_{i})-y_{i})h(x_{i}
)}{(delta y_{i})^{2}}.
end{align*}
Taking $t=0$ gives
$$
0=k^{prime}(0)=int_{x_{0}}^{x_{n}}2f^{primeprime}(x)h^{primeprime
}(x),dx+2psum_{i=0}^{n}frac{(f(x_{i})-y_{i})h(x_{i})}{(delta y_{i})^{2}}.
$$
This is true for all $hin C^{2}([x_{0},x_{n}])$. We now play with $h$. Fix
$i$ and consider functions $h$ which are zero except on $(x_{i-1},x_{i})$.
Then
$$
0=int_{x_{i-1}}^{x_{i}}f^{primeprime}(x)h^{primeprime}(x),dx.
$$
By Weyl's lemma, this implies that $f^{primeprime}$ has two derivatives in
$(x_{i-1},x_{i})$ and that $f^{primeprimeprimeprime}(x)=0$ in each
interval $(x_{i-1},x_{i})$. Thus, $f^{primeprimeprime}$ is constant in each
interval $(x_{i-1},x_{i})$ but can jump at each $x_{i}$. To find the
constants, fix $1<i<n$ and take $hin C^{4}([x_{0},x_{n}])$ which is zero outside
of $(x_{i}-delta,x_{i}+delta)$, where $delta<min{x_{i}-x_{i-1}%
,x_{i+1}-x_{i}}$. Then
begin{align*}
0 & =int_{x_{i}-delta}^{x_{i}+delta}f^{primeprime}(x)h^{primeprime
}(x),dx+pfrac{(f(x_{i})-y_{i})h(x_{i})}{(delta y_{i})^{2}}\
& =int_{x_{i}-delta}^{x_{i}}f^{primeprime}(x)h^{primeprime}
(x),dx+int_{x_{i}}^{x_{i}+delta}f^{primeprime}(x)h^{primeprime
}(x),dx+pfrac{(f(x_{i})-y_{i})h(x_{i})}{(delta y_{i})^{2}}.
end{align*}
Integrating by parts twice both integrals and using the fact that
$f^{primeprimeprime}=0$ in each open interval and that $h$ and its
derivatives up to order 4 are zero at $x_{i}pmdelta$ we get
begin{align*}
int_{x_{i}-delta}^{x_{i}}&f^{primeprime}(x)h^{primeprime}(x),dx
=-int_{x_{i}-delta}^{x_{i}}f^{primeprimeprime}(x)h^{prime}
(x),dx+f^{primeprime}(x_{i})h^{prime}(x_{i})-0\
& =int_{x_{i}-delta}^{x_{i}}f^{primeprimeprimeprime}(x)h(x),dx+0-f_{-}
^{primeprimeprime}(x_{i})h(x_{i})+f^{primeprime}(x_{i})h^{prime}
(x_{i})\
& =0-f_{-}^{primeprimeprime}(x_{i})h(x_{i})+f^{primeprime}(x_{i}
)h^{prime}(x_{i}).
end{align*}
Similarly,
begin{align*}
int_{x_{i}}^{x_{i}+delta}&f^{primeprime}(x)h^{primeprime}(x),dx
=-int_{x_{i}}^{x_{i}+delta}f^{primeprimeprime}(x)h^{prime}
(x),dx+0-f^{primeprime}(x_{i})h^{prime}(x_{i})\
& =int_{x_{i}}^{x_{i}+delta}f^{primeprimeprimeprime}(x)h(x),dx-0+f_{+}
^{primeprimeprime}(x_{i})h(x_{i})-f^{primeprime}(x_{i})h^{prime}
(x_{i})\
& =0+f_{+}^{primeprimeprime}(x_{i})h(x_{i})-f^{primeprime}(x_{i}
)h^{prime}(x_{i}).
end{align*}
Hence, if we combine the last three equations we get
begin{align*}
0 & =-f_{-}^{primeprimeprime}(x_{i})h(x_{i})+f^{primeprime}
(x_{i})h^{prime}(x_{i})+f_{+}^{primeprimeprime}(x_{i})h(x_{i}
)-f^{primeprime}(x_{i})h^{prime}(x_{i})\&quad+2pfrac{(f(x_{i})-y_{i})h(x_{i}
)}{(delta y_{i})^{2}}\
& =-f_{-}^{primeprimeprime}(x_{i})h(x_{i})+f_{+}^{primeprimeprime}
(x_{i})h(x_{i})+2pfrac{(f(x_{i})-y_{i})h(x_{i})}{(delta y_{i})^{2}}.
end{align*}
Now take $h$ such that $h(x_{i})=1$ and you get
$$
0=-f_{-}^{primeprimeprime}(x_{i})+f_{+}^{primeprimeprime}(x_{i}
)+2pfrac{f(x_{i})-y_{i}}{(delta y_{i})^{2}}.
$$
For $i=1$ and $i=n$ you do something similar.