If $1 \le p \le \infty$ and let $g : \Omega \to \mathbb{R}$ be any measurable function s.t. $g \in L^\infty(\Omega,\mu)$, prove that the multiplication operator $M_g : f \to gf$ from $L^p(\Omega,\mu)$ into itself is bounded and has norm equal $\|g\|_{L^\infty}$.
Solution:
Let $\|g\|_{L^\infty} = K$, since $f\in {L^p}$ then to show that the operator is bounded
$$\|M_gf\|_{L^p} = \|gf\|_{L^p}=\Big(\int_\Omega |gf|^p\Big)^{\frac{1}{p}}\le \Big(\int_\Omega |Kf|^p\Big)^{\frac{1}{p}}\le K .\|f\|_{L^p} < \infty \ \ \ \ (\ast)$$
Also by def. we'd have
$$\|M_g\| = \inf \{c>0 : \|M_gf\|_{L^p}\le c . \|f\|_{L^p}\} = K = \|g\|_{L^\infty} \ \ \ \ \ \ \square $$
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Is above correct?
Is so, I saw in the solution that the $\|M_g\|= K = \|g\|_{L^\infty}$ was concluded as follows, which I cannot understand why it is better to do like below rather than the above:
1- $(\ast) \implies \|M_g\|\le K$
2- Trying to conclude that $\forall \epsilon>0$, we can show $\|M_gf\|_{L^p}\ge (K-\epsilon)\|f\|_{L^p} \implies \|M_g\|\ge (K-\epsilon)$
and by $1,2$ we conclude that $\|M_g\| = K$