# Essential Supremum

For $f\in L^\infty[a,b]$, show that $$\|f\|_\infty = \min \big\{M : m\{x \in [a, b] : |f(x)|>M\} = 0\big\}\;,$$ and if, furthermore, $f$ is continuous on $[a, b]$, that $\|f\|_\infty = \|f\|_{\max}$.

I am having trouble getting started, but particularly, decomposing $$\|f\|_\infty = \min \big\{M : m\{x \in [a, b] : |f(x)|>M\} = 0\big\}\;.$$ It equals the minimum $M$ that $|f(x)|$ is always greater than such that the measure over $[a, b]$ equals $0$, correct? How do I start this? Is there another definition of $\|f\|_\infty$ that I should be working with as well?

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Where did the problem come from? A class, a book, notes, etc.? I ask in part because the answer to the question of what definition you should be using is to be found there, not from us. – Jonas Meyer Jan 28 '13 at 2:52
The second part follows immediately from the Extreme Value Theorem. The first part depends on what your definition of $\Vert f\Vert_\infty$ is... – Clayton Jan 28 '13 at 3:12
Jonas, that is a main reason that I am confused. I cannot find any other definition of $||f||_\infty$. My book is Royden's Real Analysis 4th – Jake Casey Jan 28 '13 at 18:51

For each $M\in\Bbb R$ let $A_M=\{x\in[a,b]:f(x)>M\}$; clearly $m(A_M)\ge m(A_N)$ if $M\le N$. That is, the sets $A_M$ get smaller as $M$ gets bigger. In particular, if $m(A_M)=0$ and $M\le N$, we must have $m(A_N)=0$ as well. Let $Z=\{M\in\Bbb R:m(A_M)=0\}$; these are the cutoffs $M$ with the property that $f(x)\le M$ almost everywhere, and we’ve just seen that if $N\ge M\in Z$, then $N\in Z$.
The claim that you’re asked to prove is that $\|f\|_\infty=\min Z$ and further that if $f$ is continuous, then $\|f\|_{\max}=\min Z$ as well. The second part is rather easy: if $f$ is continuous, there is an $x_0\in[a,b]$ such that $f(x_0)=\max_{x\in[a,b]}f(x)$, and it’s not hard to use the continuity of $f$ to show that if $M<f(x_0)$, then $M\notin Z$.