Cubic interpolating spline

• Consider calculating a cubic interpolating spline with the additional boundary conditions $s''(x_0)=0$ and $s''(x_n)=0$. Show that $$\int_{x_0}^{x_n}[s''(x)]^2dx \leq \int_{x_0}^{x_n}[g''(x)]^2dx$$ for any $g \in c^2[x_0,x_n]$ that satisfy the interpolating conditions $g(x_i)=y_i$, $i=0,...,n$.

Attempt at solution:

Let $k(x)=s_c(x)-g(x)$, and $$\int_a^b|g''(x)|^2dx=\int_a^b|s_c''(x)-k''(x)|^2dx=\int_a^b|s_c''(x)|^2dx-2\int_a^bs_c''(x)k''(x)dx+\int_a^b|k''(x)|^2dx$$ then by integration by parts, and using the interpolating conditions, we have $$\int_a^bs_c''(x)k''(x)dx=0$$ and thus $$\int_a^b|g''(x)|^2dx=\int_a^b|s_c''(x)|^2dx +\int_a^b|s_c''(x)-g''(x)|^2dx.$$

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This is a classical result in spline theory. It first appeared in J. C. Holladay, A smoothest curve approximation, Math. Tables Aids Comput. 11(1957), 233-243. You can find a proof in many books about splines. Your argument seems correct, to me. The third term in your last equation is the integral of a non-negative function, so it's non-negative. This means that the integral of $\vert g''(x) \vert^2$ is greater than the integral of $\vert s_c''(x) \vert^2$, and you're done.