Operator norm on Lebesgue integrable functions

Question

Let $L_1([0,1],m)$ be the Banach space of $\mathbb{K}$-valued integrable functions with respect to Lebesgue measure $m$, where $\mathbb{K}$ is either $\mathbb{R}$ or $\mathbb{C}$. The norm on this space is defined like this: $||f||_1=\int_{[0,1]}|f| \ dm$. I have to show that:
$a)$ For $n \geq 2$ the operator $\varphi_n(f)=\int_{[0,1]}\ f g_n \ dm$, where $g_n(x)=n \sin(n^2x)$ for $x \in [0,1]$ is bounded with $||\varphi_n||=n$.
$b)$ Show that there exists $f \in L_1([0,1],m)$ such that $\lim_{n \to \infty} |\varphi_n( f)|=\infty$.
MY ATTEMPT:
$g_n$ is Lebesgue integrable on $[0,1]$ since it's Riemann integrable. Hence $fg_n \in L_1([0,1],m)$ and $|\int_{[0,1]}fg_n\ dm| \leq \int_{[0,1]}|fg_n| \ dm$. We also have that $||fg_n||_1 \leq ||f||_1||g_n||_\infty$. Thus $||\varphi_n(f)||=|\int_{[0,1]}fg_n\ dm| \leq \int_{[0,1]}|fg_n|\ dm=||fg_n||_1 \leq ||f||_1 ||g_n||_\infty$, i.e. $\varphi_n$ is bounded. Now $||g_n||_\infty=n$ since it's continuous on a bounded interval and the $essential$ $supremum$ is the same as the $max$. Now I would like to attain the equality with some function, and once that I find it I can use in part $b)$. Any ideas on the function?

Answer 1

Answer to part a): it is not necessary to get equality in $|\int_{[0,1]} f(x)n\sin(n^{2}x)\,dx | \leq \|f\|_1\|g\|_{\infty}$. Instead, we get an 'approximate equality' as follows: let $\epsilon >0$. Choose $\delta >0$ such that $\sin\, x>1-\epsilon$ for $\frac {\pi} 2 -\delta <x <\frac {\pi} 2 +\delta $. Let $f=\frac {n^{2}} {2\delta} I_A$ where $A=(\frac {\pi} {2n^{2}} -\frac {\delta} {n^{2}}, \frac {\pi} {2n^{2}} +\frac {\delta} {n^{2}})$. Simple calculations show that $\|f\|_1=1$ and $\phi_n(f) >n(1-\epsilon)$. Hence $\|\phi_n\| \geq n(1-\epsilon)$ for all $\epsilon >0$. Hence $\|\phi_n\|\geq n$.