Note that
\begin{align*}
\int_{-\pi}^\pi f(x) \sin x \,dx
&= \int_0^{\pi/2} (f(x) - f(x - \pi) + f(\pi - x) - f(-x)) \sin x \,dx \\
&= \int_0^{\pi/2} (g(x) + h(x)) \sin x \,dx
\end{align*}
and similarly
\begin{align*}
\int_{-\pi}^\pi f(x) \cos x \,dx
&= \int_0^{\pi/2} (f(x) - f(x - \pi) - f(\pi - x) + f(-x)) \cos x \,dx \\
&= \int_0^{\pi/2} (g(x) - h(x)) \cos x \,dx
\end{align*}
where we define $g, h : [0, \pi/2] \to \mathbb{R}$ by $g(x) = f(x) - f(x - \pi)$ and $h(x) = f(\pi - x) - f(-x)$.
Then by Cauchy-Schwarz,
\begin{align*}
\left(\int_0^{\pi/2} (g(x) + h(x)) \sin x \,dx\right)^2
&\leq \int_0^{\pi/2} \sin^2 x \,dx \int_0^{\pi/2} (g(x) + h(x))^2 \,dx \\
&= \frac{\pi}{4} \int_0^{\pi/2} (g(x) + h(x))^2 \,dx
\end{align*}
and
\begin{align*}
\left(\int_0^{\pi/2} (g(x) - h(x)) \cos x \,dx\right)^2
&\leq \int_0^{\pi/2} \cos^2 x \,dx \int_0^{\pi/2} (g(x) - h(x))^2 \,dx \\
&= \frac{\pi}{4} \int_0^{\pi/2} (g(x) - h(x))^2 \,dx
\end{align*}
hence
\begin{align*}
\left(\int_{-\pi}^\pi f(x) \sin x \,dx\right)^2 &+ \left(\int_{-\pi}^\pi f(x) \cos x \,dx\right)^2 \\
&\leq \frac{\pi}{4} \int_0^{\pi/2} (g(x) + h(x))^2 \,dx + \frac{\pi}{4} \int_0^{\pi/2} (g(x) - h(x))^2 \,dx\\
&= \frac{\pi}{2} \int_0^{\pi/2} (g(x))^2 + (h(x))^2 \,dx \\
&\leq \frac{\pi}{2} \int_0^{\pi/2} (f(x))^2 + (f(x - \pi))^2 + (f(\pi - x))^2 + (f(-x))^2 \,dx \\
&= \frac{\pi}{2} \int_{-\pi}^\pi (f(x))^2 \,dx
\end{align*}
as desired, where the last inequality holds because $f$ is nonnegative.
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
We have $$\int_a^bf^2(x)dx=\int_a^b\left(\int_a^x f'(y)dy\right)^2dx$$ and by Cauchy-Schwarz $$\left(\int_a^x f'(y)dy\right)^2\leq \int_a^x[f'(y)]^2dy\int_a^x1dy=(x-a)\int_a^x[f'(y)]^2dy$$ thus we get $$\begin{align} \int_a^bf^2(x)dx &\leq \int_a^b(x-a)\int_a^x [f'(y)]^2dydx\\ &\leq \int_a^b(b-a)\int_a^b [f'(y)]^2dydx=(b-a)^2\int_a^b [f'(y)]^2dy \end{align}$$ as desired.