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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?


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up vote 6 down vote accepted

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, $\| M_f \| \geqslant N-\varepsilon$.

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That's a really nice idea! – Alex Dec 5 '11 at 23:46
Apart from being isometric, a further important property of the map $$ \begin{array}{rccc} M: & L^\infty[0,1] & \longrightarrow & B(L^2[0,1]) \\ & f & \longmapsto & M_f \end{array} $$ is that $M_{fg} = M_f \circ M_g$, $M_{1} = 1_{L^2}$, and $M_{\overline{f}} = (M_{f})^\ast$, that is, it is a $*$-homomorphism. This is exploited in some formulations of the spectral theorem and is important in the study of von Neumann algebras. – t.b. Dec 6 '11 at 5:01
t.b.'s comment made me think of an amusing (to me) overkill way to approach this. Because $M$ is a $*$-homomorphism between C* algebras, to show that it is isometric (and thereby answer the question), it suffices to show that it is injective. This general property of C* algebras follows from the spectral radius formula. But I don't think that showing that $M$ is injective is significantly easier than showing that $M$ is isometric directly. – Jonas Meyer Dec 6 '11 at 5:29

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