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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.

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    $\begingroup$ The constant map $f=1$ is what you are looking for. $\endgroup$ – Crostul Nov 27 '16 at 18:46
  • $\begingroup$ @Crostul it gives me the 1norm of g not the infinity norm $\endgroup$ – Sara Suradi Nov 27 '16 at 18:47
  • $\begingroup$ I don't think such a function exists in general. You need a sequence of functions with support in smaller and smaller regions. The support should where g is large. $\endgroup$ – Simon Nov 27 '16 at 18:52
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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.

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  • $\begingroup$ Very obvious and clear explanation,a clever answer ,,thanks alot "شكراً لك كثيراً" $\endgroup$ – Sara Suradi Nov 29 '16 at 15:58

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