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...
On
This solution is quite unsatisfactory to me for its seemingly unnecessary complexity, but I post it anyway.
We introduce two lemmas
Lemma 1. Assume $g(x) : [a, \infty) \to \Bbb{R}$ is absolutely integrable and continuous. Then $$ \lim_{s \to 0^{+}} \int_{a}^{\infty} g(x)e^{-sx} \, dx = \int_{a}^{\infty} g(x) \, dx. $$
and
Lemma 2. Assume $g(x) : [a, \infty) \to \Bbb{R}$ is non-negative and continuous. Then regardless of the convergence of the integral, $$ \lim_{s \to 0^{+}} \int_{a}^{\infty} g(x)e^{-sx} \, dx = \int_{a}^{\infty} g(x) \, dx. $$
By integrating by parts,
\begin{align*} \int_{a}^{R} \{ f'(x) \}^{2} e^{-sx} \, dx &= \Big[ f(x)f'(x)e^{-sx} \Big]_{a}^{R} - \int_{a}^{R} f(x)f''(x) e^{-sx} \, dx \\ &\quad + \left[ \frac{s}{2} f(x)^{2} e^{-sx} \right]_{a}^{R} + \frac{s^{2}}{2} \int_{a}^{R} f(x)^{2} e^{-sx} \, dx. \end{align*}
From CS-inequality,
$$ |f'(x)| \leq |f'(a)| + \left| \int_{a}^{x} f''(t) \, dt \right| \leq |f'(a)| + \sqrt{x - a} \left( \int_{a}^{\infty} f''(t)^{2} \, dt \right)^{1/2} $$
and hence both $f$ and $f'$ are of polynomial growth. So taking $R \to \infty$,
\begin{align*} \int_{a}^{\infty} \{ f'(x) \}^{2} e^{-sx} \, dx &= - f(a)f'(a)e^{-sa} - \int_{a}^{\infty} f(x)f''(x) e^{-sx} \, dx \\ &\quad - \frac{s}{2} f(a)^{2} e^{-sa} + \frac{s^{2}}{2} \int_{a}^{\infty} f(x)^{2} e^{-sx} \, dx. \end{align*}
Taking $s \to 0^{+}$, Lemma 1 and 2 show that
\begin{align*} \int_{a}^{\infty} \{ f'(x) \}^{2} \, dx = - f(a)f'(a)- \int_{a}^{\infty} f(x)f''(x) \, dx. \end{align*}
Since the right-hand side is finite, the same is true for the left-hand side and the proof is complete.
Addendum. (Proof of Lemmas)
Proof of Lemma 1. For any $\epsilon > 0$, choose $R$ such that $\int_{R}^{\infty} |g(x)| \, dx < \epsilon$. Since $g(x)e^{-sx}$ is also absolutely integrable, its integral is well-defined and $$ \left| \int_{a}^{R} g(x)(1 - e^{-sx}) \, dx \right| \leq \int_{a}^{R} |g(x)| (1 - e^{-sx}) \, dx + \epsilon. $$ Taking $\limsup$ as $s \to 0^{+}$, followed by $\epsilon \to 0$, we obtain the desired result. ////
and
Proof of Lemma 2. Let $I$ denote the integral on the right-hand side and $J$ denote the limit of the left-hand side. On the one hand, from the following inequality $$ \int_{a}^{\infty} g(x)e^{-sx} \, dx \leq \int_{a}^{\infty} g(x) \, dx, $$ we have $J \leq I$. On the other hand, for each fixed $R > a$ we have $$ \int_{a}^{R} g(x) \, dx = \lim_{s\to 0^{+}} \int_{a}^{R} g(x)e^{-sx} \, dx \leq J. $$ Thus taking $R \to \infty$ we obtain $I \leq J$ and the equality follows. ////
On
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 $(3)$.
Hint 1: Integration by parts gives $$ \begin{align} \int_a^\infty f'(x)^2\,\mathrm{d}x &=\int_a^\infty f'(x)\,\mathrm{d}f(x)\\ &=\lim_{b\to\infty}f'(b)f(b)-f'(a)f(a)-\int_a^\infty f(x)f''(x)\,\mathrm{d}x\tag{1} \end{align} $$ Hint 2: As Giraffe points out, if $\int_a^\infty f'(x)^2\,\mathrm{d}x$ diverges, then by $(1)$, $\lim\limits_{x\to\infty}f'(x)f(x)=\infty$. Since $$ f(b)^2=\int_a^bf'(x)f(x)\,\mathrm{d}x\tag{2} $$ we get that $\int_a^\infty f(x)^2\,\mathrm{d}x$ diverges.
By the comments, this seems to be a bit more involved than befits a hint, so I will explain in more detail. Note that what follows is not needed due to Hint 2, but the ideas used are more generally applicable, so I will leave it.
Proof: Suppose not; then, for some $\epsilon\gt0$ and all $x_0$, there is an $x\ge x_0$ so that $|f'(x)|\ge\epsilon$.
Since $\|f''\|_{L^2}\lt\infty$, we can choose a $b$ so that $$ \int_b^\infty f''(x)^2\,\mathrm{d}x\le\epsilon^4\tag{1} $$ Then, for any $x,y\ge b$ so that $|x-y|\le1$, Cauchy-Schwarz says $$ \begin{align} |f'(x)-f'(y)| &\le\int_x^y|f''(x)|\,\mathrm{d}x\\ &\le\left(\int_x^y|f''(x)|^2\,\mathrm{d}x\right)^{1/2}|x-y|^{1/2}\\[9pt] &\le\epsilon^2\tag{2} \end{align} $$ For any $x_0\ge b+1$, we can choose an $x\ge x_0$ so that $|f'(x)|\ge\epsilon$. If $f'(x)$ and $f(x)$ have the same sign, let $I=[x,x+1]$, otherwise, let $I=[x-1,x]$.
$\hspace{2cm}$
By $(2)$, for $t\in I$, $|f'(t)|\ge\epsilon-\epsilon^2$ and $|f(t)|\ge\left(\epsilon-\epsilon^2\right)|t-x|$. Thus, $$ \int_If(x)^2\,\mathrm{d}x\ge\frac13\left(\epsilon-\epsilon^2\right)^2\tag{3} $$ By supposition, we can find infinitely many points so that $|f'(x)|\ge\epsilon$. Therefore, $$ \int_b^\infty f(x)^2\,\mathrm{d}x\quad\text{diverges}\tag{4} $$ giving us a contradiction. QED
Proof: Suppose not; then, for some $\epsilon\gt0$ and all $x_0$, there is an $x\ge x_0$ so that $|f(x)|\ge\epsilon$.
Since $\displaystyle\lim_{x\to\infty}f'(x)=0$, we can choose a $b$ so that for all $x\ge b$, $$ |f'(x)|\le\epsilon^2\tag{5} $$ Then, for any $x,y\ge b$ so that $|x-y|\le1$, the Mean-Value Theorem says $$ \begin{align} |f(x)-f(y)| &\le\max_{t\in[x,y]}|f'(t)||x-y|\\ &\le\epsilon^2\tag{6} \end{align} $$ For any $x_0\ge b+1$, we can choose an $x\ge x_0$ so that $|f(x)|\ge\epsilon$. For any $t\in[x-1,x+1]$, $|f(t)|\ge\epsilon-\epsilon^2$. Thus, $$ \int_{x-1}^{x+1}f(t)^2\,\mathrm{d}t\ge2\left(\epsilon-\epsilon^2\right)^2\tag{7} $$ By supposition, we can find infinitely many points so that $|f(x)|\ge\epsilon$. Therefore, $$ \int_b^\infty f(x)^2\,\mathrm{d}x\quad\text{diverges}\tag{8} $$ giving us a contradiction. QED