My question arose while studying an article which finds the $K$-functional for the pair of spaces $L^1,L^\infty$, so it's related to interpolation theory, but I think it can be solved with some $\inf,\sup$ manipulations. I'm sorry if the tags aren't correct.

Consider $(X,\Sigma, \mu)$ an arbitrary measure space. For each measurable function $f: X \to \Bbb{C}$ and $\alpha \geq 0$ define $$ f_*(\alpha)=\mu(\{ x \in X : |f(x)|>\alpha \}),$$ the distribution function of $f$.

For any measurable function $f:X \to \Bbb{C}$ for which there exists $\alpha>0$ with $f_*(\alpha)<\infty$, define $f^* :(0,\infty) \to [0,\infty)$ by $$ f^*(t)=\inf \{ y>0 :f_*(y) \leq t\}.$$

It can be proved from the definitions that for every $t>0$ we have $$ f^*(f_*(t))\leq t,\ f_*(f^*(t))\leq t.$$

Moreover, the function $f^*$ is continuous from the right. Both of the functions $f_*,f^*$ are non increasing.

What I need to prove is that

$$\sup_{t>0} f^*(t)= \| f\|_\infty,$$ for every function $f \in L^\infty$.

The inequality $\leq $ is straight from the definition. The other one I can't seem to get, and the text says that it's pretty hard, but with "a little more effort" it can be done. I haven't succeeded.


Well, it is quite confusing notationwise, but let's just go through it one step at a time:

Suppose $\sup_{t > 0} f^\ast(t) < \|f\|_\infty$. Then there is $\epsilon > 0$ such that for all $t > 0$ we have $$ f^\ast(t) \le \|f\| - 2\epsilon $$ This means by definition $$\forall t > 0: \quad \inf\{y > 0| f_\ast(y) \le t\} \le \|f\| - 2\epsilon$$

Therefore $\forall t > 0$ there exists $y > 0$ such that $f_\ast(y) \le t$ and $y\le \|f\| - \epsilon$.

By definition this means that for all $t> 0$ there is a $y$ satisfying $\|f\|- \epsilon \ge y > 0$ such that $$\mu(\{x \in X |\, |f(x)| > y \}) \le t $$

But then, in particular we get that for all $t>0$ we have $$\mu(\{x \in X |\, |f(x)| > \|f\|_\infty -\epsilon \}) \le t $$

Since this is true independent of $t$, letting $t \to 0$, we get a contradiction to the definition of $\|f\| = \|f\|_\infty$.


For the inequality $\ge$ you'll have to show: $$ \forall \varepsilon>0 \,\exists\, t>0: f^*(t) \ge \|f\|_\infty - \varepsilon =: c_\varepsilon. $$ By the definition of $f^*$, the inequality $f^*(t) \ge c_\varepsilon$ means: $$ \text{for all $y>0$ satisfying}\ f_*(y)\le t\ \text{one has}\ y \ge c_\varepsilon. $$ By contraposition, the latter is equivalent to $$ \forall y\in(0,c_\varepsilon): f_*(y)>t. $$ Thus, given $\varepsilon>0$, you have to find $t>0$ such that the latter holds. This is easy: Take any $t \in (0,f_*(c_\varepsilon))\,$ (and use that $f_*(y) \ge f_*(c_\varepsilon)$ for $y\in(0,c_\varepsilon)$).

  • $\begingroup$ (Sam was quicker, but here you've got it without proof by contradiction.) $\endgroup$ – Hendrik Vogt Jun 24 '11 at 19:11
  • $\begingroup$ @Hendrik: If you dislike proofs by contradiction, you may of course read the proof backwards - avoiding all "ugliness". ;-) $\endgroup$ – Sam Jun 24 '11 at 20:24
  • $\begingroup$ Thank you for your proof. If I could, I would have accepted both answers. :) Anyway you have +1 from me. $\endgroup$ – Beni Bogosel Jun 24 '11 at 22:16
  • $\begingroup$ @Beni: A very naïve question: How can this argument be described as 'pretty hard, but with "a little more effort" it can be done', as was done in the text? (Hendrik: this is not meant to say anything against your nice answer, but the argument looks rather straightforward to me.) $\endgroup$ – t.b. Jun 25 '11 at 4:23
  • 2
    $\begingroup$ @Beni: Oh, I didn't doubt that! I also didn't mean to imply that you should have seen it (also, when someone tells you something is difficult, you tend to look further than necessary and doubt your own arguments, I know that too well from my own experience). It's a perfectly legitimate question! @Hendrik: You succeeded very well in showing that it's rather straightforward, but one needs to see that one can put it this way and there are various places to get entangled here. $\endgroup$ – t.b. Jun 25 '11 at 10:37

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