Find the norm of the operator that is given by
$T\colon L^1[0,1]\to L^1[0,1]$ such that $T(f)=g f$ where $g$ is essentially bounded.
I proved that it's bounded linear operator and $ \lVert T\rVert \leqslant\lVert g\rVert_{\infty}$
but i didn't find a function that make the inequality be equality.
**I think that we could use something related to the duality of $L^p$ spaces ($L^1$ and $L^{\infty}$) to prove the inverse inequality but i don't know if that is possible or it would make the answer harder.
For a fixed $\varepsilon \in (0,\lVert g\rVert_\infty)$, there exists a set of positive measure $A$ such that $g(x)\geqslant\lVert g\rVert_\infty-\varepsilon$ for almost every $x\in A$. Define $f(x):=\operatorname{sign}(g(x))\mathbf 1_A(x)$. Then $\lVert f\rVert_1=\lambda(A)$ and $$\lVert T(f)\rVert_1=\int_{[0,1]}\left|g\right|\mathbf 1_A\geqslant \left(\lVert g\rVert_\infty-\varepsilon\right)\lambda(A)=\left(\lVert g\rVert_\infty-\varepsilon\right)\lVert f\rVert_1$$ hence $\lVert T\rVert\geqslant \lVert g\rVert_\infty-\varepsilon$. Since $\varepsilon$ is arbitrary, we get the wanted inequality.