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let $f \in L^\infty$

and let $M(f)=sup\{\left|\int \phi f\right| : \phi \text{ simple, } ||\phi||_1 \le 1 \}$

i wish to show that $||f||_\infty=M(f)$

i was able to show $||f||_\infty \ge M(f)$ using Hölders inequality but I am having trouble showing $\le$. Any tips?

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    Ah..., that's right.2011-04-05

1 Answers 1

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Since you just want tips, I'll try not to give everything away.

You want to show that $\phi$ can be chosen so that $|\int\phi f|$ is arbitrarily close to $\|f\|_\infty$. To do so, you want to choose $\phi$ to be "concentrated" where $f$ is close to its essential supremum. More explicitly, there is a set $E$ of positive finite measure where $|f|>\|f\|_\infty - \epsilon$, and you can use this to find suitable $\phi$. (Some rescaling and sign adjustments may be necessary.)

If my attempt to give tips is completely opaque, please let me know.

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    @Jenny: (I guess I lied, I wasn't done quite yet :)). I don't know what your $\phi_n$ is, but all you need is this: for each \varepsilon>0, there exists a simple function $\phi$ such that $\|\phi\|_1\leq 1$ and |\int f\phi|>\|f\|_\infty -\varepsilon. To do so, take $E$ such that |f(x)|>\|f\|_\infty -\varepsilon for all $x\in E$, with 0<\mu(E)<\infty, which is possible by def of $\|f\|_\infty$ & the fact that positive meas sets have positv finite meas subsets. Take $\phi=\frac{1}{\mu(E)}\mathrm{sgn}(f)\chi_E$. Note that this $\phi$ takes only 2 values, so is simple, & check it works.2011-04-05