TITLE: An unpleasant measure theory/functional analysis problem
QUESTION [9 upvotes]: I am currently taking a functional analysis course, and at the moment every student on the course is stumped by a specific question. We're looking at the bounded linear map 
$$
\varphi_n:L^1\left([0,1]\right)\to \mathbb{R},\quad \varphi_n(f) = \int^1_0 f(x)(n\sin(n^2 x)) dx,\quad\|\varphi_n\|=n.
$$
We are attempting to prove the existence of an $f\in L^1$ with respect to the Lebesgue measure, so that $\lim_{n\rightarrow \infty} |\varphi_n(f)|\rightarrow \infty$, and from an earlier assignment we know that this function cannot be continuously differentiable. However, nobody has been able to make even the slightest progress. I, and all of us would appreciate a hint or a sketch of proof!

REPLY [5 votes]: @supinf showed us a nice set of examples. I think the proof can be simplified. So suppose $1/2<p<1$ and let $f(x)=x^{-p}.$ Let $x=y/n^2$ to see
$$\tag 1 n\int^1_0 x^{-p}\sin(n^2 x)\, dx = n^{2p-1}\int_0^{n^2} y^{-p}\sin(y)\, dy.$$
If we show $\int_0^{\infty} y^{-p}\sin(y)\, dy >0,$ then it will follow that $(1)\to \infty$ on the order of $n^{2p-1}.$ But this integral equals
$$\sum_{k=0}^{\infty}\int_0^{\pi}\sin y\left(\frac{1}{(2\pi k+y)^p} - \frac{1}{(2\pi k+\pi +y)^p}\right)\,dy.$$
Each of the integrands here is positive on $(0,\pi),$ hence so is the sum, and we're done.