Show that the norm of the multiplication operator $M_f$ on $L^2[0,1]$ is $\|f\|_\infty$

Hint: Fix $\varepsilon > 0$ and $S_{\varepsilon} := \{ x \in [0,1] \; \colon \; |f(x)| \geqslant N - \varepsilon \}$. Take $g$ to be $\chi_{S_{\varepsilon}}$, the indicator function of the set $S_{\varepsilon}$.

Fleshed out answer. We will prove that $\| M_f \| \geqslant N-\varepsilon$ for all $\varepsilon > 0$. Fix $\varepsilon > 0$ (s.t. $\varepsilon < N$) and set $$S_{\varepsilon} := \{ x \in [0,1] \; \colon \; |f(x)| \geqslant N - \varepsilon \} .$$ Finally, define $g$ by $$ g(x) = \begin{cases} 1, &x \in S_{\varepsilon}, \\ 0, &\text{otherwise}. \end{cases} $$ (The definition of $N$ implies that $g$ differs from zero in a positive measure set.) It is easy to check that $|fg| \geqslant (N-\varepsilon) \cdot |g|$ holds pointwise, from which it follows that $\| fg \|_2 \geqslant (N-\varepsilon) \cdot \|g \|_2$. Therefore, $\| M_f \| \geqslant N-\varepsilon$.