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I had to dive into the basic definitions and properties of $L_p$ spaces as part of a course project I am doing. (More specifically while I was trying to understand Barbalet's Lemma). It was my first time in this topic and I have got really confused now. It would be great if you could answer some of my questions with a numerical example if possible.

  1. What exactly is an essentially bounded function? The definition I know of is:-

    function $f:[0,1]→ℝ $ is called essentially bounded if there is a number $M $ such that $|f(x)|≤M $ for almost all $ x∈(0,1)$. (That is, the inequality holds on some set $E$ such that $(0,1)$∖E has zero measure.)

    Why is $f(x)=x^{−1/(p+1)}$ which defines an element of $L_p((0,1))$ not essentially bounded. I would also like to ask what is the use of equivalence class of function in such cases?

  2. Also, what would be the set consisting of "almost everywhere" in any appropriate example (maybe above one)?

  3. Does essentially bounded function imply it belongs to $L_{\infty}$ space?

  4. How can a function have $\|f\|_p = 0$ but $f \ne 0$.

Please explain as if I don't know anything. I am not very sure how much I have understood. I am asking the question only after reading many answers and online lecture notes.

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  • $\begingroup$ You can easily show that for continuous functions boundedness and essentially boundedness coincide. $\endgroup$
    – Red shoes
    Aug 15, 2017 at 18:48

1 Answer 1

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1) An essentially bounded function is exactly what you described. The $f$ you gave is not essentially bounded, because for any real number $M$, there exists $\varepsilon>0$ such that $f(x)>M$ for all $x\in (0,\varepsilon)$. This shows that it is not essentially bounded because $(0,\varepsilon)$ has positive measure.

2) You wrote in part (1) that "the inequality holds on some set $E$ such that $(0,1)\setminus E$ has zero measure". This means precisely that the inequality holds almost everywhere. Your example doesn't apply because that $f$ is not essentially bounded. Consider the function $g:(0,1)\to\mathbb{R}$ given by $$ g(x) = \begin{cases} n & \text{if}\ x=1/n\ \text{for some}\ n\in\mathbb{N},\,n\geq 2 \\ 0 & \text{otherwise}. \end{cases} $$ Then $g(x)=0\leq 1$ for all $x\in(0,1)\setminus\{1/n\mid n\in\mathbb{N},\,n\geq 2\}$ and $\{1/n \mid n\in\mathbb{N},\,n\geq2\}$ is a measure zero set. Thus $g=0$ a.e., which is another way of saying that the equivalence class of $g$ is equal to the equivalence class of the zero function $0$ under the "equal a.e." equivalence relation.

3) Yes, by definition $L_\infty(0,1)$ is the space of all essentially bounded functions from $(0,1)$ into $\mathbb{R}$. Some authors consider the quotient of this space by the "equal a.e." equivalence relation, but in practice the two are the same.

4) Do you mean $\|f\|_p=0$ but $f\ne0$? Consider the function $g$ given in (2). Then $\|g\|_p=0$, but $g\ne0$. Now we do have $g=0$ a.e., so some would write $g=0$ and it would be understood that the equality here is with respect to the "equal a.e." equivalence relation.

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  • $\begingroup$ Excellently explained!! Thanks. Yes, your modification in point 4 was exactly what I meant. $\endgroup$
    – Manish
    Aug 16, 2017 at 7:55

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