I'm having some (hopefully small) issues computing the norm of an operator. Firstly, the problem,
For $f\in L^\infty[0,1]$, define $M_f: L^2[0,1]\to L^2[0,1]$ by $M_f(g)(x) = f(x)g(x)$. Show that $M_f$ is a bounded linear operator and $\|M_f\| = \|f\|_\infty$.
What I've done so far:
To see that $M_f$ is linear, let $g_1,g_2\in L^2[0,1]$ and $\lambda\in\mathbb{R}$. Then $ M_f(g_1 + \lambda g_2)(x) = f(x)(g_1(x) + \lambda g_2(x)) = f(x)g_1(x) + \lambda f(x)g_2(x) = M_f(g_1)(x) + \lambda M_f(g_2)(x).$
As $f\in L^\infty$ we know there is some minimal $N\in \mathbb{R}$ such that $|f(x)| \leq N$ almost everywhere (i.e., $\|f\|_\infty = N \lt \infty$). The fact that $M_f$ is bounded comes straight this assupmtion, as $ \|M_f\| = \sup \frac{\|fg\|_2}{\|g\|_2} \leq \sup \frac{\|Ng\|_2}{\|g\|_2} = N\sup \frac{\|g\|_2}{\|g\|_2} = N < \infty.$ almost everywhere in $[0,1]$, for all non-zero $g\in L^2[0,1]$. So $M_f$ is a bounded linear operator on $L^2[0,1]$.
To prove equality, we must show there is some $g\in L^2[0,1]$ so that $\frac{\|fg\|_2}{\|g\|_2} = N$. Now I want to say something along the lines of pick $g = 1$, but this won't have $\|g\|_2 = 1$ for all measures, so this won't work. Is there any simple way of picking a $g$ that does what we want? Or am I farther off than I'm expecting?
Thanks!