how to obtain Euler equation for smoothing spline minimization problem? The question might be trivial, but I don't understand why this minimization problem in Sobolev space
$$
\min_{g}\int_{0}^{1}\left\{ f(x)-g(x)\right\}^{2} dx+\lambda\int_{0}^{1}\left\{ g^{(q)}(x)\right\} ^{2}dx
$$
is equivalent to Euler equation
$$g(x)+\lambda (-1)^{q}g^{(2q)}(x)= f(x) $$
$$g^{(j)}(0)=g^{(j)}(1)=0,\   j=q,\ldots 2q-1.$$
 A: If you define the functional
$$
F\big(x,g(x),g^{(q)}(x)\big) = \int_0^1 \{f(x)-g(x)\}dx + \lambda \int_0^1 \{g^{(q)}(x)\}^2 dx
$$
and take test functions $h$ in the same space as the functions $g$, then
\begin{multline}
F\big(x,g + h, (g + h)^{(q)}\big) - F\big(x,g,g^{(q)}\big) = \\ 
\int_0^1\{f(x)- g(x) - h(x)\}dx + \lambda \int_0^1 \{g^{(q)}(x) + h^{(q)}(x)\}^2 dx \\
-\int_0^1\{f(x)- g(x)\}dx - \lambda \int_0^1 \{g^{(q)}(x)\}^2 dx,
\end{multline}
and then, the Frechet derivative is
$$
D F \cdot h = 2 \lambda \int_0^1 g^{(q)}(x) h^{(q)}(x) dx - \int_0^1 h(x) dx
$$
which leads me to believe you've missed a square in the first term, i.e.
$$
F\big(x,g(x),g^{(q)}(x)\big) = \int_0^1 \{f(x)-g(x)\}^\color{red}{2}dx + \lambda \int_0^1 \{g^{(q)}(x)\}^2 dx
$$
That being the case, the correct Frechet derivative is
$$
D F \cdot h = 2 \lambda \int_0^1 g^{(q)}(x) h^{(q)}(x) dx - 2 \int_0^1 \{f(x) - g(x)\}h(x) dx
$$
Integrating the first term by parts
\begin{multline}
D F \cdot h = 2 \lambda g^{(q)}(x) h^{(q)}(x) \Big|_0^1 \\- 2 \lambda \int_0^1 g^{(q+1)}(x) h^{(q-1)}(x) dx - 2 \int_0^1 \{f(x) - g(x)\}h(x) dx
\end{multline}
and so on, after $q-1$ integrations
\begin{multline}
D F \cdot h = 2 \lambda g^{(q)}(x) h^{(q)}(x) \Big|_0^1 - 2 \lambda g^{(q+1)}(x) h^{(q-1)}(x)\Big|_0^1 + \ldots \\ + 2 \int_0^1 \big\{\lambda (-1)^q g^{(2q)}(x) + g(x) - f(x)\big\}h(x) dx
\end{multline}
In order for the functional $F$ to have a critical point, the Frechet derivative $D F \cdot h = 0$ for all $h$ in the space, which leads to the Euler-Lagrange equation
$$
g(x) + \lambda (-1)^q g^{(2q)}(x) = f(x)
$$
with boundary conditions
$$
g^{(j)}(0) = g^{(j)}(1) = 0,\quad j=q,\ldots,2q-1.
$$
You only have to prove that the critical points are minima (hint: the functional is convex).
