This is a math problem I'm struggling on.
Show that there is no sequence of functions on $[0, 2 \pi]$ of the type
$$f_n(x) = a_n \sin(nx) +b_n \cos(nx)$$
which converges to the function $1$ almost everywhere on $[0, 2 \pi]$ and where $\lvert a_n \rvert + \lvert b_n \rvert \le 10$.
I haven't seen a problem like this before. I think it may have to do with fourier series, but I haven't learned that yet. Sorry about the poor formatting.
Thanks
Observe that if $f_n$ is of the given form, then
$|f_n(x)|\le |a_n|+|b_n|\le 10$. Therefore if $f_n\to 1$ a.e., $(f_n-1)^2 \to 0$ a.e. and is bounded above by the constant $121$. By Lebesgue's bounded convergence theorem:
$$ \int_0^{2\pi} (f_n -1)^2(x) dx \to 0.$$
But this is clearly impossible (orthogonality of $(1,\cos(nx),\sin(nx):n\ge 1)$). To see through direct computation, observe that
$$\int_0^{2\pi} (f_n -1)^2 dx = \int_0^{2\pi} f_n^2 - 2f_n +1 dx .$$ Now $\int_0^{2\pi} f_n dx =0$. Therefore, since $f_n^2\ge 0$, we have
$$ \int_0^{2\pi} (f_n-1)^2 dx \ge \int_0^{2\pi} 1 dx =2\pi,$$
a contradiction.