# $\mathbb{L}^{p}$ vs $\mathbb{L}^{q}$

Let $(X,\mathcal{A},\mu)$ be a measure space  and $1\leq p,q\leq\infty$. Generally there is no connection between $\mathbb{L}^{p}(\mu)$ and $\mathbb{L}^{q}(\mu)$ as sets. However, for some special measures, there is an interesting relationship between them. A few examples:

1. 1.

If $\lambda^{n}$ is the Lebesgue measure  on $\mathbb{R}^{n}$ and $p\neq q$, then $\mathbb{L}^{p}(\lambda^{n})\not\subseteq\mathbb{L}^{q}(\lambda^{n})$ for all $n\in\mathbb{N}^{+}$. Here is an example for $n=1$ and $1\leq p. Let

 $f(x):=\begin{cases}x^{\frac{-1}{p}},&x>1\\ 0,&x\leq 1\end{cases}$

and

 $g(x):=\begin{cases}x^{\frac{-1}{q}},&x\in(0,1)\\ 0,&x\not\in(0,1).\end{cases}$

This gives $\|f\|_{q}^{q}=\frac{p}{q-p}$, $\|g\|_{p}^{p}=\frac{q}{q-p}$ and $\|f\|_{p}=\|g\|_{q}=\infty$. So $f\in\mathbb{L}^{q}\setminus\mathbb{L}^{p}$ and $g\in\mathbb{L}^{p}\setminus\mathbb{L}^{q}$. For the $\infty$-norm, $\chi_{\mathbb{R}}\in\mathbb{L}^{\infty}\setminus\mathbb{L}^{q}$, where $\chi$ is the characteristic function     , and also $f\not\in\mathbb{L}^{\infty}$.

2. 2.

If $p then $l^{p}\subseteq l^{q}$. This is trivial if $q=\infty$. Now let $x=(x_{0},x_{1},\dots)\in l^{p}$ and $q<\infty$. Then

 $\|x\|_{q}^{q}=\sum_{n=0}^{\infty}|x_{n}|^{q}=\sum_{n=0}^{\infty}|x_{n}|^{p}|x_% {n}|^{q-p}\leq\sum_{n=0}^{\infty}|x_{n}|^{p}\|x\|_{\infty}^{q-p}=\|x\|_{\infty% }^{q-p}\|x\|_{p}^{p}<\infty,$

so $x\in l^{q}$.

3. 3.

If $\mu(X)$ is finite and $p, then $\mathbb{L}^{q}\subseteq\mathbb{L}^{p}$. This is easy if $q=\infty$, because $|f|\leq\|f\|_{\infty}$ almost everywhere, so $\|f\|_{p}^{p}=\int|f|^{p}d\mu\leq\int\|f\|_{\infty}^{p}d\mu=\|f\|_{\infty}^{p}% \mu(X)<\infty$. Now let $q<\infty$, thus

 $\displaystyle\|f\|_{p}^{p}$ $\displaystyle=\int|f|^{p}d\mu$ $\displaystyle=\int_{|f|>1}|f|^{p}d\mu+\int_{|f|\leq 1}|f|^{p}d\mu$ $\displaystyle\leq\int_{|f|>1}|f|^{q}d\mu+\int_{|f|\leq 1}d\mu$ $\displaystyle\leq\|f\|_{q}^{q}+\mu(X)$ $\displaystyle<\infty.$

Finally, we prove an interesting property for $p$-norms: if $X$ is a finite measure space, then for any measurable function  $f$ on $X$ the equality $\lim\limits_{p\to\infty}\|f\|_{p}=\|f\|_{\infty}$ holds. We have already seen that $\|f\|_{p}\leq\|f\|_{\infty}\mu(X)^{\frac{1}{p}}$. Now for any $\varepsilon\in(0,\|f\|_{\infty})$ define $A_{\varepsilon}:=\{x\in X:|f(x)|\geq\|f\|_{\infty}-\varepsilon\}$, $\delta_{\varepsilon}:=\mu(A_{\varepsilon})>0$ and $g:=(\|f\|_{\infty}-\varepsilon)\chi_{A_{\varepsilon}}$. Since $|g|\leq|f|$, we have $\|g\|_{p}=(\|f\|_{\infty}-\varepsilon)\delta_{\varepsilon}^{\frac{1}{p}}\leq\|% f\|_{p}\leq\|f\|_{\infty}\mu(X)^{\frac{1}{p}}$. Now we take $\liminf$ on the left and $\limsup$ on the right side: $\|f\|_{\infty}-\varepsilon\leq\liminf\limits_{p\to\infty}\|f\|_{p}\leq\limsup% \limits_{p\to\infty}\|f\|_{p}\leq\|f\|_{\infty}$. Taking $\varepsilon\downarrow 0$ gives $\lim\limits_{p\to\infty}\|f\|_{p}=\|f\|_{\infty}$.

Title $\mathbb{L}^{p}$ vs $\mathbb{L}^{q}$ mathbbLpVsmathbbLq 2013-03-22 15:22:05 2013-03-22 15:22:05 yark (2760) yark (2760) 11 yark (2760) Topic msc 28A25