<s>[INST] Find $\min\int_0^\pi {f^2(x)dx}$, assume $\int_0^\pi{f(x)\sin xdx} = \int_0^\pi{f(x)\cos xdx} = 1.$ $f(x)$ is continuous on $[0,\pi]$ and $\int_0^\pi{f(x)\sin xdx} = \int_0^\pi{f(x)\cos xdx} = 1.$
Find $\min\int_0^\pi {f^2(x)dx}.$
I try to solve this problem by this:
$$\begin{array}{l}
{\left( {\int\limits_0^\pi  {f(x)\sin xdx} } \right)^2} \le \left( {\int\limits_0^\pi  {{f^2}(x){{\sin }^2}xdx} } \right)\left( {\int\limits_0^\pi  {dx} } \right) \le \pi \int\limits_0^\pi  {{f^2}(x){{\sin }^2}xdx} \\
{\left( {\int\limits_0^\pi  {f(x)\cos xdx} } \right)^2} \le \pi \int\limits_0^\pi  {{f^2}(x){{\cos }^2}xdx} \\
 \Rightarrow \pi \int\limits_0^\pi  {{f^2}(x)\left( {{{\sin }^2}x + {{\cos }^2}x} \right)dx}  \ge 1 + 1 = 2
\end{array}$$
The thing is I can't find $f(x)$ to let the equation happens. Any help? Thank you in advance.
 [/INST] Let us consider the Fourier series of $f$
\begin{align}
f(x) = \frac{1}{2}a_0+\sum^\infty_{n=1} a_n \cos nx+ \sum^\infty_{n=1} b_n \sin nx
\end{align}
then that means
\begin{align}
\int^\pi_0 f^2(x)\ dx= \frac{1}{4}a_0^2+\frac{\pi}{2}\sum^\infty_{n=1}(a_n^2+b_n^2).
\end{align}
Since
\begin{align}
\int^\pi_0 f(x) \cos x\ dx = 1
\end{align}
then this means $a_1= 2/\pi$ and likewise $b_1 = 2/\pi$. It follows
\begin{align}
\int^\pi_0 f^2(x)\ dx \geq \frac{4}{\pi}= \frac{\pi}{2}(a_1^2+b_1^2).
\end{align}
</s>