I am trying to prove the following statement for 2 sets of conditions:
$$\int_{a}^{b}[s''(x)]^2dx \leq \int_{a}^{b}[f''(x)]^2dx$$ for any $f \in C^2[(a,b)]$
For the following 2 conditions
natural conditions : $s''(a)=0$ and $s''(b)=0$
different set of conditions : $s'(a) = f'(a), s'(b) = f'(b) $.
I have tried the following as a first shot at this proof, and think its a common start to both conditions:
$$\int_a^b|f''(x)-s''(x)|^2dx=\int_a^b|f''(x)|^2dx-2\int_a^bf''(x)s''(x)dx+\int_a^b|s''(x)|^2dx$$ Rearranging this i got: $$=\int_a^b|f''(x)|^2dx-2\int_a^b(f''(x)-s''(x)) s''(x) dx-\int_a^b|s''(x)|^2dx$$
I think I should now integrate by parts (according to a tip I got), twice starting with the middle integral but I don't know how to do so