# Operator norm on Lebesgue integrable functions

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 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 $$\frac {\pi} 2 -\delta . Let $$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$$.