TITLE: Prove that if $\int f^2$ and $\int( f'')^2$ converge, so does $\int (f')^2$
QUESTION [18 upvotes]: Question:
Let $f: [a,\infty) \to \Bbb R \in C^2$ and the two following integrals converge:
  $$\int _a^\infty (f''(x))^2\,dx ,~~~~~~~~~ \int _a^\infty (f(x))^2\,dx$$
Prove that $\int _a^\infty (f'(x)^2)\,dx$ converges as well. 

What we tried:
Taylor expansion, Lagrange mean value theorem, integration by parts, comparison test, limit comparison test, none really helped us get there...

REPLY [10 votes]: By Taylor's expansion with integral form of the remainder,
$$f(x+1)=f(x)+f'(x)+g(x),\tag{1}$$
where 
$$g(x)=\int_x^{x+1}(x+1-t)f''(t)~dt.\tag{2}$$
By $(1)$, Minkowski's inequality and $\int_a^\infty (f(x))^2dx<\infty$, to show 
$\int_a^\infty (f'(x))^2dx<\infty$, it suffices to show that
$$\int_a^\infty (g(x))^2dx<\infty.\tag{3}$$
By $(2)$ and Cauchy-Schwarz inequality,
$$(g(x))^2\le \int_x^{x+1}(x+1-t)^2dt\cdot \int_x^{x+1}(f''(t))^2dt=\frac{1}{3}\int_x^{x+1}(f''(t))^2dt.\tag{4}$$
By $(4)$, Fubini's theorem and $\int_a^\infty (f''(x))^2dx<\infty$,
$$3\int_a^\infty (g(x))^2dx\le\int_a^\infty\left(\int_x^{x+1}(f''(t))^2dt\right)dx\le \int_a^\infty\left(\int_{t-1}^tdx\right)(f''(t))^2dt<\infty,$$
which completes the proof of $(3)$.