I know that functions in $L^2$ space have finite norms by definition, but are they also bounded "almost everywhere" ?

So say for instance the following functions norm is finite but it is not bounded.

$$ f(x) = \frac{1}{(x-\frac{1}{2})^2} ~~ ; ~~~ f(x) \in L^2[(0,1),\mu] $$

$$ {\|f(x)\|}_2 = \left(\int_0^1{\frac{1}{(x-\frac{1}{2})^2}}\right)^{1/2}$$ $$ {\|f(x)\|}_2 = \left(\left. {\frac{1}{(x-\frac{1}{2})}}\right]_0^1\right)^{1/2}$$ $$ {\|f(x)\|}_2 = 2$$

  • $\begingroup$ Where is the $p$ in the integral? Are you already answering the question in the question? $\endgroup$ – Sergio Parreiras Dec 13 '13 at 5:48
  • $\begingroup$ I meant in general fo L^p space, but I have tried to give a specific example. $\endgroup$ – Michael T Mckeon Dec 13 '13 at 5:51
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    $\begingroup$ Michael, functions in $L^p$ spaces are equivalence classes of functions - you can change a function's values on a zero measure set and it stays in the same equivalence class. So you only need to question whether your function is bounded almost everywhere. $\endgroup$ – massy255 Dec 13 '13 at 5:51
  • $\begingroup$ OK, ill edit the question, I understand your point. $\endgroup$ – Michael T Mckeon Dec 13 '13 at 6:08
  • $\begingroup$ see this related post: math.stackexchange.com/questions/458437/… $\endgroup$ – massy255 Dec 13 '13 at 6:15

Let's fix the definitions first.

A function $f:(0,1)\to\mathbb R$ is called bounded if there is a number $M$ such that $|f(x)|\le M$ for all $x\in (0,1)$.

Since the elements of $L^p$ are not functions, but equivalence classes of functions, the concept of boundedness must be adjusted.

A function $f:[0,1]\to\mathbb R$ is called essentially bounded if there is a number $M$ such that $|f(x)|\le M$ for almost all $x\in (0,1)$. (That is, the inequality holds on some set $E$ such that $(0,1)\setminus E$ has zero measure.)

If two functions $f,g$ belong to the same equivalence class and $f$ is essentially bounded, then $g$ is also essentially bounded. Thus, the property of being essentially bounded makes sense for elements of $L^p$.

Does it hold for all elements of $L^p$? Yes if $p=\infty$ (by definition), no if $p<\infty$. For example, the function $f(x)=x^{-1/(p+1)}$ defines an element of $L^p((0,1))$ which is not essentially bounded. For every $M$, the set $\{x:|f(x)|>M\}$ contains an interval, and therefore is not a null set.

Remark. One often allows functions to take values in the extended real line $[-\infty, \infty]$. In this case, one can say that every element of $L^p$ is finite almost everywhere.

  • $\begingroup$ Concerning your remark, isn't any element of Lp(I) finite almost everywhere, whatever the interval I? Not only for the extended real line? $\endgroup$ – Stephane Jul 23 '19 at 16:03

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