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Let $f$ be a continuous function on the measure space $\mathbb{R}^n,\mathcal{L},\lambda$(Lebesgue measure). Prove that

$\|f\|_\infty = \sup\{|f(x)|$ $|$ $x \in \mathbb{R}^n\}$

I saw same problem but I couldn't understand the answer.

My idea is this: Let $\|f\|_\infty=K$. Since $f$ is continuous, a small neighborhood of any $x\in \mathbb{R}^n$, say $p$ should satisfy $|f(p)|\leq M,$ for $M>K$. So I can take $supremum$ over p.

But I don't know how to proceed rigorously.

Any help will be thankful.

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SW PARK
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2 Answers2

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Let $S\equiv\sup_{x\in\mathbb R^n}|f(x)|$. By definition, the essential supremum norm is defined as follows: $$\|f\|_{\infty}=\inf_{c\geq 0}\big\{\lambda(\{x\in\mathbb R^n\,|\,|f(x)|>c\})=0\big\}.$$ In words, $\|f\|_{\infty}$ is the infimum of such non-negative numbers above which the function $|f|$ takes values only on a measure-zero set. Intuitively, $\|f\|_{\infty}$ is the least “almost upper bound” on the values of $|f|$ (hence the term “essential supremum”), in the sense that if you take it as an upper bound, you'll only miss a set of measure zero and this is the least number with this property.

First, I will show that $\|f\|_{\infty}\leq S$. To see this, note that for all $x\in\mathbb R^n$, one has $|f(x)|\leq S$. Hence, the set $\{x\in\mathbb R^n\,|\,|f(x)|>S\}$ is empty and thus has measure zero. By the definition of infimum, this readily yields that $\|f\|_{\infty}\leq S$.

Now to show that $\|f\|_{\infty}\geq S$ (which will complete the proof), assume, for the sake of contradiction, that $\|f\|_{\infty}< S$. By the definitions of infimum and $\|f\|_{\infty}$, there is some $\varepsilon>0$ such that $$\|f\|_{\infty}+\varepsilon<S$$ and the set $$V\equiv\{x\in\mathbb R^n\,|\,|f(x)|>\|f\|_{\infty}+\varepsilon\}$$ has zero measure: $\lambda(V)=0$. On the other hand, by the definitions of supremum and $S$, there is some $x^*\in\mathbb R^n$ such that $\|f\|_{\infty}+\varepsilon<|f(x^*)|\leq S$. Since the function $|f|$ is continuous (it is the composition of $f$ and the absolute-value function, both being continuous), the set $$V=\{x\in\mathbb R^n\,|\,|f(x)|>\|f\|_{\infty}+\varepsilon\}$$ is open—since it is the pre-image of the open interval $(\|f\|_{\infty}+\varepsilon,\infty)$—and, as one has seen, it contains $x^*$. Since non-empty open subsets of $\mathbb R^n$ have positive Lebesgue measure, it follows that $\lambda(V)>0$. This is a contradiction and the proof is complete.

triple_sec
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    Thank you very much. It is impressive using the fact that inverse image of open set of continuous function is open. – SW PARK Jun 09 '15 at 10:49
  • @SWPARK This is the general topological definition of continuity. The amazing fact that this property is equivalent to a function being continuous at every point in the usual $\varepsilon$-$\delta$ sense is quite a powerful tool in many complicated proofs, indeed. – triple_sec Jun 09 '15 at 10:52
  • The property was sleeping somewhere in my head. ha ha.. – SW PARK Jun 09 '15 at 11:01
  • By the way, is the existence of $x*$ always guaranteed? By the definition of supremum? – SW PARK Jun 09 '15 at 11:02
  • @SWPARK Yes. For the sake of contradiction, suppose that there is no $x^\in\mathbb R^n$ such that $|f|_{\infty}+\varepsilon<|f(x^)|\leq S$. Then, for all $x\in \mathbb R^n$, one has $|f(x)|\leq|f|{\infty}+\varepsilon$. Hence, $|f|{\infty}+\varepsilon$ is an upper bound on $|f|$, so that $S\leq|f|{\infty}+\varepsilon$ (supremum = least upper bound), but this contradicts $|f|{\infty}+\varepsilon<S$. – triple_sec Jun 09 '15 at 11:09
  • Sorry for bothering you. I should have done that by myself. – SW PARK Jun 09 '15 at 11:11
  • @SWPARK No worries; I'm happy to help. – triple_sec Jun 09 '15 at 11:11
  • Why there is such $\epsilon>0$ such that $|f|{\infty}+\varepsilon<S$ and $V\equiv{x\in\mathbb R^n,|,|f(x)|>|f|{\infty}+\varepsilon}$ has measure zero? – mac Aug 05 '17 at 19:41
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    @mac Suppose that $|f|{\infty}<S$. Now look at the definition of $|f|{\infty}$; it is defined as an infimum. Since this infimum is less than $S$, there exists some $c$ such that $|f|{\infty}\leq c<S$ and $\lambda(V)=0,$ where $$V\equiv{x\in\mathbb R^n,|,|f(x)|>c}.$$ Now, if $|f|{\infty}<c<S$, then define $\varepsilon\equiv c-|f|{\infty}>0$ and you’re done. If $c=|f|{\infty}$, then choose $\varepsilon>0$ so small that $|f|{\infty}+\varepsilon<S$ and the “new” set $${x\in\mathbb R^n,|,|f(x)|>|f|{\infty}+\varepsilon}$$ still has measure $0$ being a subset of $V$. – triple_sec Aug 06 '17 at 01:53
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Since the Lebesgue measure is $\sigma$-finite we have that $$\| {f}\|_{\infty}=\inf\left\{M\geq 0:\,\lambda(\left\{x: |f(x)| > M\right\})=0 \right\}$$

Clearly $\lambda(\left\{x: |f(x)| > \sup_{x \in \mathbb{R}^{d}}|f(x)|\right\})=0$

Now suppose there is a constant $M < \sup_{x \in \mathbb{R}^{d}}|f(x)|$, that satisfies the above critera. It should be fairly easy to reach a contradiction due to continuity of f

TheOscillator
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