Fix a $c \in (0,1)$. For $0 \leq t\leq c$, since $f(0) =0$ we have
$$|f(t)| = |\int_0^t f'(s) ds|\leq \int_0^t |f'(s)| ds.$$
Therefore, by Fubini theorem, we have
$$ \int_0^c |f(t)| dt \leq \int_0^c \int_0^t |f'(s)| ds dt = \int_0^c |f'(s)| (c-s) ds.$$
Similarly, we get
$$ \int_c^1 |f(t)| dt \leq \int_c^1 |f'(s)| (s-c) ds.$$
Hence, by summing two inequalities, we obtain
$$\int_0^1 |f(t)| dt \leq \int_0^1 |f'(s)| |c -s| ds.$$
Applying Cauchy-Schwartz, we have
$$(\int_0^1 |f(t)| dt)^2 \leq (\int_0^1 f'(s)^2 ds )(\int_0^1(c-s)^2 ds) =(c^2 -c + \frac13) \int_0^1 f'(s)^2 ds,$$
for any $c\in (0,1)$. Since $c^2-c + \frac13 \geq \frac1{12}$ for any $c\in (0,1)$ and attains at $c = \frac12$. By taking $c =\frac12$, we get the desired inequality.
This is essentially following the steps in my answer to a quasi-similar question.
I'm not going to explain how I find the function $g(x)$ below.
Let $X = \mathcal{C}^2[0,1]$ and $P,Q,C : X \to \mathbb{R}$ be functionals over $X$ defined by
$$P(f) = \int_0^1 f''(x)^2 dx,\quad Q(f) = \int_0^1 f(x)dx\quad\text{ and }\quad C(f) = \int_{1/3}^{2/3} f(x) dx$$
The question can be rephrased as
Given $f \in X$ with $C(f) = 0$, how to verify $\;P(f) \ge \frac{4860}{11} Q(f)^2$?
Since both the inequality and constraint is homogeneous in scaling of $f$ by a constant. We can restrict our attention to those $f$ which satisfies $C(f) = 0$ and $Q(f) = 1$.
Consider following functions
$$\phi(x) = x^4 - \frac12 x^2 + \frac{29}{6480}
\quad\text{ and }\quad
\psi(x) = \begin{cases}
\left(\frac13-x\right)^4, & x \le \frac13\\
0, & \frac13 \le x \le \frac23\\
\left(x - \frac23\right)^4, & x \ge \frac23
\end{cases}
$$
Combine them and define another function $g(x)$ by
$$g(x) = -\frac{405}{11}\left[ \phi\left(x-\frac12\right) - \frac32 \psi(x) \right]$$
It is not hard to check
- $g(x) \in \mathcal{C}^3[0,1] \subset X$.
- $C(g) = 0$, $Q(g) = 1$.
- $g''(0) = g'''(0) = g''(1) = g'''(1) = 0$
- $g''''(x) = \frac{4860}{11}$ for $x \in [0,\frac13)\cup (\frac23,1]$
- $g''''(x) = -\frac{9720}{11}$ for $x \in (\frac13,\frac23)$
- $P(g) = \frac{4860}{11}$.
For any $f \in X$ with $C(f) = 0, Q(f) = 1$, let $\eta = f - g$, we have
$$\begin{align}
& P(f) - P(g) - P(\eta)\\
= & 2\int_0^1 g''(x)\eta''(x) dx\\
= & 2\int_0^1 ( g''(x)\eta'(x))' - g'''(x)\eta'(x) dx\\
= & 2\int_0^1 ( g''(x)\eta'(x) - g'''(x)\eta(x))' + g''''(x)\eta(x)dx\\
= &2\left\{\left[ g''(x)\eta'(x) - g'''(x)\eta(x) \right]_0^1
+ \frac{4860}{11}(Q(\eta)-C(\eta)) -\frac{9720}{11}C(\eta)\right\}
\end{align}
$$
What's in the square bracket vanish because of $(3)$. The remain terms vanish
because
- $Q(\eta) = Q(f) - Q(g) = 1 - 1 = 0$.
- $C(\eta) = C(f) - C(g) = 0 - 0 = 0$.
Together with the fact $P(\eta)$ is non-negative, we obtain:
$$P(f) = P(g) + P(\eta) \ge P(g) = \frac{4860}{11}$$.
Best Answer
(The following is inspired by Integral inequality with a function twice differentiable: Integrating by parts transforms the integral with $f$ to an integral with $f'$. The condition $\int_0^1 f(x) \, dx = 0$ is used to add a term to the first integral so that the $u(b)v(b)-u(a)v(a)$ term vanishes. Cauchy-Schwarz then helps to estimate the integral containing $f'$ by an integral containing $f'^2$.)
Integrating by parts we get $$ \int_0^1 xf(x) \, dx = \frac 12 \int_0^1 (2x-1)f(x) \, dx = \frac 12 \int_0^1 x(1-x) f'(x) \,dx \\ = \frac 12 \int_0^1 \frac{x \sqrt{1-x}}{\sqrt{1+x}} \sqrt{1-x^2} f'(x) \, dx \, . $$ Now apply Cauchy-Schwarz: $$ \left( \int_0^1 xf(x) \, dx \right)^2 \le \frac 14 \int_0^1 \frac{x^2(1-x)}{1+x} \, dx \int_0^1 (1-x^2) (f'(x))^2 \, dx \\ \le \frac 14 \int_0^1 x^2(1-x) \, dx \int_0^1 (1-x^2) (f'(x))^2 \, dx \\ = \frac{1}{48 }\int_0^1 (1-x^2) (f'(x))^2 \, dx $$ which is better than the desired estimate by a factor of $2$.
Using the exact value $\int_0^1 \frac{x^2(1-x)}{1+x}\, dx = 2 \ln(2) - 4/3$ we get the sharp estimate $$ \int_0^1 (1-x^2) (f'(x))^2 \, dx \ge C \left( \int_0^1 xf(x) \, dx \right)^2 $$ with $$ C = \frac{2}{\ln(2)-2/3} \approx 75.53 \, . $$ Equality holds if equality holds in the Cauchy-Schwarz inequality, and that is if $$ f'(x) = \text{const} \cdot \frac{x}{x+1} $$ so that the integrands are linearly dependent. Together with the condition $\int_0^1 f(x) \, dx = 0$ this gives (up to a multiplicative constant) $$ f(x) = x - \ln(x) + 2 \ln(2) - \frac 23 \, . $$