Let $I=[0,\pi]$ and $f\in L^2(I)$. Is it possible to have simultaneously $\int_I(f(x)-\sin x)^2 dx\leq \frac{4}{9}$ and $\int_I(f(x)-\cos x)^2 dx\leq \frac{1}{9}$?
I don't understand what this problem asked to do. By "possible", does it ask to give an example for $f\in L^2(I)$ so that this is true or does it ask to show that this is not true for some $f\in L^2(I)$? Can anyone please help me to understand it?
On
Brute force:
One approach is to pick $f$ such that $(f(x)-\cos x)^2+(f(x)-\sin x)^2$ is minimized for all $x$. If this $f$ results in $\int_I ((f(x)-\cos x)^2+(f(x)-\sin x)^2)dx > { 5 \over 9}$ then we know that no such function exists.
Let $L$ be the line $L = \{(t,t)\}$. Note that distance from $(x,y) $ to $L$ is given by $\sqrt{{1 \over 2} (x-y)^2}$.
Hence $\|(\alpha -\cos x, \alpha - \sin x)\|^2 \ge {1 \over 2} (\sin x - \cos x)^2$.
It follows that for any function $f$, we have $\|(f(x) -\cos x, f(x) - \sin x)\|^2 \ge {1 \over 2} (\sin x - \cos x)^2$.
Integrating gives $\int_0^\pi \|(f(x) -\cos x, f(x) - \sin x)\|^2 dx \ge { \pi \over 2}$.
Does there exist an $f$ so that $$||f - \sin x||_2 \le \sqrt{\frac{4}{9}}=\frac{2}{3}\\ ||f - \cos x||_2 \le \sqrt{\frac{1}{9}}=\frac{1}{3}$$ ? Like in euclidian geometry: $\ f$ exists if and only if $$||\sin x- \cos x||_2 \le \frac{2}{3}+\frac{1}{3}=1$$ $\tiny{\text{the two balls must intersect.}}$ But we calculate $$||\sin x- \cos x||_2^2= \int_0^{\pi} (\sin x - \cos x)^2 = \int_0^{\pi} (1 - \sin 2 x)=\pi$$ so $||\sin x- \cos x||_2= \sqrt{\pi} > 1$. No, such an $f$ does not exist.