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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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    $\geq$ is straight forward. I don't see how Holders Inequality applies?2011-04-05
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    @GWu: It is the special case of Hölder's inequality, which happens to be the most straightforward one, that $\int|fg|\leq\|f\|_\infty\|g\|_1$.2011-04-05
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    By the way, this isn't true if there are sets in your space of infinite measure with no subsets of positive finite measure. The reason is that if $A$ is such a set, then $f=\chi_A$ has $\|f\|_\infty=1$, but every $L^1$ function must be zero a.e. on $A$ and hence $M(f)=0$. It is true in every reasonable measure space.2011-04-05
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    Ah..., that's right.2011-04-05

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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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    Thanks Jonas. Could you explain what you mean by "concentrated"2011-04-05
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    Take $\phi$ to be a constant multiple of indicator function of $E$ --- $\phi$ is concentrated on $E$.2011-04-05
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    @Jenny: Not necessarily a constant multiple of the indicator function of $E$, but that is roughly the idea. $\phi$ will have its support on $E$. It turns out that you have to also worry about signs. Assuming you're dealing with real valued functions, you could either consider a positive measure subset of $E$ on which $f$ has constant sign, or you can let $\phi$ be a linear combination of two indicator functions corresponding to different signs of $f$ on $E$.2011-04-05
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    @Jenny: By the way, you can use the $\mathrm{sgn}$ function to make the writing down of $\phi$ a bit easier without explicitly fussing about the sign of $f$.2011-04-05
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    @Jonas: How should I scale my $\phi$, apart from the sgn?2011-04-05
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    @Jenny: You want to make $M(f)$ as big as possible while keeping the requirement $\|\phi\|_1\leq 1$, so you might as well scale $\phi$ to make $\|\phi\|_1$ as big as possible, namely $1$.2011-04-05
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    hmm do I get to use the Lebesgue differentiation theorem?2011-04-05
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    Oops, I meant "...make $|\int f\phi|$ as big as possible..."2011-04-05
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    @Jenny: There is no need for that; it is much simpler. It is also more general, because this would work in any measure space as long as all sets of positive measure have subsets of positive finite measure. But it is true that in $\mathbb{R}^n$ you could deduce this from the Lebesgue differentiation theorem.2011-04-05
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    @Jonas, can you tell me an explicit construction of the $\phi$?2011-04-05
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    @Jenny: Do you not have any guesses or starts from my and GWu's suggestions? If you start with $\chi_E$, this will have $\|\chi_E\|_1=\mu(E)$ (assuming $\mu$ is the measure), so to get $L^1$ norm $1$ you better divide by $\mu(E)$. This $\phi$ would get $|\int f\phi|$ close to $\|f\|_\infty$ if $f$ had constant sign, but it might not, so adjustments should be made accordingly. Please inform if you get the solution, or where you're stuck otherwise.2011-04-05
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    @Jenny: In any case, I am done for the night just so you know, so if you'd like to attract the attention of others who might help, you could perhaps edit to give an update on your progress and bump the question to the top of the main page.2011-04-05
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    @Jonas: I tried using your construction but I always end up using the Lebesgue Differentiation Thm to argue $\lim_{n\to \infty} \int \phi_n f = lim_{n\to \infty} \frac{1}{u(E_n)}\int_{E_n} f=||f||_\infty$. How do I avoid this?2011-04-05
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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