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I have two real functions $f$ and $g$ defined by $$ f(x) =\begin{cases} \tfrac1x & \text{if } x \in (0,1] \\ 0 & \text{if } x \in \mathbb R\setminus (0,1] \end{cases} \qquad \text{and} \qquad g(x) = \begin{cases} \tfrac1x & \text{if } x \in (1,\infty] \\ 0 & \text{if } x \in (-\infty,1] \end{cases}.$$

I have to calculate $\{p \in [0,\infty] \colon f \in \mathcal L^p(\lambda)\}$ and $\{p \in [0,\infty] \colon g \in \mathcal L^P(\lambda)\}$.

So far, I've thought this: Since $\mathcal L^\infty(\lambda)\subseteq\mathcal L^r(\lambda)\subseteq\mathcal L^0(\lambda)$, where $r \in(0,\infty)$, I only have to calculate $\mathcal L^0(\lambda)$, which is defined like this: $\mathcal L^0(\lambda)={f\in\mathcal M(\mathcal E): \lim_{t\to \infty}\lambda({|f|\ge t})=0}$. I am, however, not at all sure how to calculate this. I would appreciate help a lot.

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What is the set where $|f|\ge t$? What is its measure? – Nate Eldredge Nov 15 '12 at 22:10
Assuming that $\lambda$ is Lebesgue measure, it is not true that $\mathcal L^\infty(\lambda) \subseteq \mathcal L^r(\lambda)$. Consider the function which is constantly equal to $1\ldots$ – kahen Nov 15 '12 at 22:11
I didn't edit your function $g$ which has some pretty obvious "double definition" problems in the interval $(0,1]$. I guess it's supposed to be $1/x$ for $x \in (1,\infty)$ (rather than $x \in (0,\infty]$)? – kahen Nov 15 '12 at 22:16
Yes, it was supposed to be $x\in(1,\infty)$, thank you for pointing that out. – Cathrine Nov 15 '12 at 22:22
But if it's not true that $\mathcal L^\infty(\lambda) \subseteq \mathcal L^r(\lambda)$, then what am I supposed to do instead? Then I honestly have no clue how to even get started on it. – Cathrine Nov 15 '12 at 22:35
up vote 0 down vote accepted

For both functions we can integrate explicitly to determine for which $p\in (0,\infty]$ we have $\int\vert \cdot\vert^p dx<\infty$. Take $f(x)$ for example:

$$ \int_{-\infty}^\infty \vert f\vert^p dx=\int_0^1 x^{-p}dx= (-p+1)^{-1}x^{-p+1}\Big\vert_0^1$$

so we see that in order for this integral to be finite we need $-p+1>0$ or $p<1$. Similarly for $g$:

$$\int_{-\infty}^\infty \vert g\vert ^p dx =\int_1^\infty x^{-p} dx =(-p+1)^{-1}x^{-p+1}\Big\vert_1^\infty$$ so now we need $-p+1<0$ or $p>1$.

It should be clear that $f\notin L^\infty$, but that $g$ is.

I'm a little unclear of your definition of $L^0$; typically $L^0$ denotes all measurable functions, but it looks like you're looking at some type of weak $L^1$ norm? If you clarify this, I might be able to suggest an approach.

Edit: For the $\mathcal{L}^0$ case, $g(x)$ should be clear since $\mathcal{L}^\infty\subset\mathcal{L}^0$. For $f(x)$, consider $\{\vert f\vert>t\}$; this is precisely $(0,1/t)$. As $t\rightarrow\infty$, the measure of these sets goes to 0. Thus $f\in \mathcal{L}^0$.

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In my book, $\mathtt L$ isn't the same as $\mathcal L$, so I just wanted to be sure we're talking about the same thing? But the definition of $\mathcal L^0$ I'm using is the one from my book, except in my book, it's $\mathcal L^0( \mu )$ instead of $\mathcal L^0(\lambda)$, but I figured it would be the same? I might be wrong though. – Cathrine Nov 15 '12 at 23:21
Ah, okay - what book are you using...? – icurays1 Nov 15 '12 at 23:23
Better yet, maybe you could give the definition of $\mathcal{L}^p$? – icurays1 Nov 15 '12 at 23:26
It's a book written by my professor for the class, so I'm afraid it's not very much of a help.. But $\mathtt L^P(\mu)={[f]:f\in\mathcal L^P(\mu)}$, if that's any help. – Cathrine Nov 15 '12 at 23:31
The definition: $\mathcal L^p=${$f\in\mathcal M(\mathcal E): \int_X|f|^Pd\mu \lt \infty$} – Cathrine Nov 15 '12 at 23:36

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