${𝕃}^p$ vs ${𝕃}^q$

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

1. If ${\lambda }^n$ is the Lebesgue measure on ${ℝ}^n$ and $p\ne q$,then ${𝕃}^p({\lambda }^n)⊈{𝕃}^q({\lambda }^n)$for all $n\in {\mathbb{N}}^{+}$.Here is an example for $n=1$ and .Let

   and

   This gives ${\parallel f\parallel }_q^q=\frac{p}{q-p}$, ${\parallel g\parallel }_p^p=\frac{q}{q-p}$and ${\parallel f\parallel }_p={\parallel g\parallel }_q=\mathrm{\infty }$.So $f\in {𝕃}^q\setminus {𝕃}^p$ and $g\in {𝕃}^p\setminus {𝕃}^q$.For the $\mathrm{\infty }$-norm,${\chi }_{ℝ}\in {𝕃}^{\mathrm{\infty }}\setminus {𝕃}^q$,where $\chi$ is the characteristic function,and also $f\notin {𝕃}^{\mathrm{\infty }}$.

2. If then $l^p\subseteq l^q$.This is trivial if $q=\mathrm{\infty }$.Now let $x=(x_0,x_1,\mathrm{\dots })\in l^p$ and .Then

   so $x\in l^q$.

3. If $\mu (X)$ is finite and ,then ${𝕃}^q\subseteq {𝕃}^p$.This is easy if $q=\mathrm{\infty }$,because $|f|\le {\parallel f\parallel }_{\mathrm{\infty }}$ almost everywhere,so .Now let ,thus

   ${\parallel f\parallel }_p^p$$={\displaystyle \int {|f|}^p𝑑\mu }$

   $={\displaystyle {\int }_{|f|>1}}{|f|}^p𝑑\mu +{\displaystyle {\int }_{|f|\le 1}}{|f|}^p𝑑\mu$

   $\le {\displaystyle {\int }_{|f|>1}}{|f|}^q𝑑\mu +{\displaystyle {\int }_{|f|\le 1}}𝑑\mu$

   $\le {\parallel f\parallel }_q^q+\mu (X)$

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 $\underset{p\to \mathrm{\infty }}{lim}{\parallel f\parallel }_p={\parallel f\parallel }_{\mathrm{\infty }}$ holds.We have already seen that ${\parallel f\parallel }_p\le {\parallel f\parallel }_{\mathrm{\infty }}\mu {(X)}^{\frac{1}{p}}$.Now for any $\epsilon \in (0,{\parallel f\parallel }_{\mathrm{\infty }})$define $A_{\epsilon }:=\{x\in X:|f(x)|\ge {\parallel f\parallel }_{\mathrm{\infty }}-\epsilon \}$,${\delta }_{\epsilon }:=\mu (A_{\epsilon })>0$and $g:=({\parallel f\parallel }_{\mathrm{\infty }}-\epsilon ){\chi }_{A_{\epsilon }}$.Since $|g|\le |f|$,we have ${\parallel g\parallel }_p=({\parallel f\parallel }_{\mathrm{\infty }}-\epsilon ){\delta }_{\epsilon }^{\frac{1}{p}}\le {\parallel f\parallel }_p\le {\parallel f\parallel }_{\mathrm{\infty }}\mu {(X)}^{\frac{1}{p}}$.Now we take $lim\; inf$ on the left and $lim\; sup$ on the right side:${\parallel f\parallel }_{\mathrm{\infty }}-\epsilon \le \underset{p\to \mathrm{\infty }}{lim\; inf}{\parallel f\parallel }_p\le \underset{p\to \mathrm{\infty }}{lim\; sup}{\parallel f\parallel }_p\le {\parallel f\parallel }_{\mathrm{\infty }}$.Taking $\epsilon \downarrow 0$gives $\underset{p\to \mathrm{\infty }}{lim}{\parallel f\parallel }_p={\parallel f\parallel }_{\mathrm{\infty }}$.