How prove this $\left(\int_{-\pi}^{\pi}f(x)\sin{x}dx\right)^2+\left(\int_{-\pi}^{\pi}f(x)\cos{x}dx\right)^2\le\frac{\pi}{2}\int_{-\pi}^{+\pi}f^2(x)dx$

Question

Prove or disprove:

if $f(x)\ge 0,\forall x\in [-\pi,\pi]$,show that $$\left(\int_{-\pi}^{\pi}f(x)\sin{x}dx\right)^2+\left(\int_{-\pi}^{\pi}f(x)\cos{x}dx\right)^2\le\dfrac{\pi}{2}\int_{-\pi}^{+\pi}f^2(x)dx$$

I can prove this if $2\pi$ takes the place of $\dfrac{\pi}{2}$

because use Cauchy-schwarz inequality we have $$\left(\int_{-\pi}^{\pi}f(x)\sin{x}dx\right)^2\le\int_{-\pi}^{\pi}\sin^2{x}dx\int_{-\pi}^{\pi}f^2(x)dx$$

$$\left(\int_{-\pi}^{\pi}f(x)\cos{x}dx\right)^2\le\int_{-\pi}^{\pi}\cos^2{x}dx\int_{-\pi}^{\pi}f^2(x)dx$$ add this two inequality, we have $$\begin{align*}\left(\int_{-\pi}^{\pi}f(x)\sin{x}dx\right)^2+\left(\int_{-\pi}^{\pi}f(x)\cos{x}dx\right)^2 &\le \int_{-\pi}^{\pi}f^2(x)dx\int_{-\pi}^{\pi}(\sin^2{x}+\cos^2{x})dx\\ &=2\pi\int_{-\pi}^{+\pi}f^2(x)dx\end{align*}$$

see this Discrete form of inequality：Prove this inequality with Cauchy-Schwarz inequality

So far, I haven't found any counterexamples，such $f(x)=1,\sin{x}+1$ it such this inequality

Answer 1

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.

Answer 2

I think so. We can assume that $f$ is $2\pi$-periodic over $\mathbb{R}$. The sum of integrals on the left-hand side is presented in the form $\left|\displaystyle\int\limits_{-\pi}^{\pi}f(x)e^{ix}\,dx\right|^2=|\rho e^{i\alpha}|^2=\rho^2$. But due to $2\pi$-periodicity, $$\rho=\rho e^{i\alpha}e^{-i\alpha}=\displaystyle\int\limits_{-\pi}^{\pi}f(x)e^{i(x-\alpha)}\,dx=\displaystyle\int\limits_{-\pi}^{\pi}f(t+\alpha)e^{it}\,dt.$$ Since $f$ is real-valued, $\rho=\displaystyle\int\limits_{-\pi}^{\pi}f(t+\alpha)\cos t\,dt$. Since $f\geq0$ we have $\rho\leq\displaystyle\int\limits_{-\frac{\pi}{2}}^{\frac{\pi}{2}}f(t+\alpha)\cos t\,dt$. Therefore, $$\rho^2\leq\displaystyle\int\limits_{-\frac{\pi}{2}}^{\frac{\pi}{2}}f^2(t+\alpha)\,dt\displaystyle\int\limits_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\cos^2 t\,dt\leq\frac{\pi}{2}\displaystyle\int\limits_{-\pi}^{\pi}f^2(t+\alpha)\,dt=\frac{\pi}{2}\displaystyle\int\limits_{-\pi}^{\pi}f^2(\tau)\,d\tau.$$

Answer 3

This is a consequence of parseval identity. Let $\mathcal{H}$ a Hilbert space (dont worry all you need to know is that you have an inner product and the concept of orthogonal vectors, i.e. when $\langle f, g\rangle = 0$ and that the norm is given by $\langle f, f \rangle = \lVert f \rVert^2$) and let $\{e_n\}$ be a numerable orthonormal set then if $f \in \mathcal{H}$ $$ \sum_n \lvert \langle e_n, f\rangle \rvert^2 \leq \lVert f \rVert^2. $$ with equality if $\{e_n\}$ is a basis. In this case $$ \lVert f \rVert^2 = \int_{-\pi}^{\pi} \lvert f(x) \rvert^2 \,dx $$ and $$ e_1(x) = A\cos(x), \quad e_2(x) = A\sin(x) $$ where $A$ is chosen so they are normal.

Parseval identity is deduced from the fact that for orthonormal (it also works with orthogonal, but the last equality is not true) vectors $u_n$ one has $$ \lVert \sum a_n u_n \rVert^2 = \sum \lVert a_n u_n \rVert^2 = \sum \lvert a_n \rvert^2 $$ and you can write $$ f = \langle e_1,f \rangle e_1 + \langle e_2, f\rangle e_2 + (f - \langle e_1,f \rangle e_1 + \langle e_2, f\rangle e_2) $$ where the 3 parts of the sum are orthogonal and you have that the norm of $f$ is bigger than the norm of the first 2 terms.

EDIT: it seems the constants are off. It does not coincide with the one in parseval's.