Let $(X, \mathcal B, m)$ be a measure space. For $1 \leq p < q \leq \infty$, under what condition is it true that $L^q(X, \mathcal B, m) \subset L^p(X, \mathcal B, m)$ and what is a counterexample in the case the condition is not satisfied?
$L^p$ and $L^q$ space inclusion
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0Related: http://math.stackexchange.com/questions/1371017/are-there-relations-between-elements-of-lp-spaces/1371051#1371051 – 2016-06-06
3 Answers
Theorem Let $X$ be a finite measure space. Then, for any $1\leq p< q\leq +\infty$ $L^q(X, \mathcal B, m) \subset L^p(X, \mathcal B, m).$ The proof follows from Hölder inequality. Note that $\frac{1}{p}=\frac{1}{q}+\frac{1}{r}$, with $r>0$. Hence $\|f\|_{L^p}\leq\text{meas }(X)^{1/r}\|f\|_{L^q}.$
The case reported on the Wikipedia link of commenter answer follows from this, since of course, if $X$ does not contain sets of arbitrary large measure, $X$ itself can't have an arbitrary large measure.
For the counterexample: $f(x)=\frac{1}{x}$ belongs to $L^2([1,+\infty))$, but clearly it does not belong to $L^1([1,+\infty)).$
ADD
I would like to add other lines to this interesting topic. Namely, I would like to prove what is mentioned in Wikipedia, hope it is correct:
Theorem Suppose $(X,\mathcal B,m)$ is a measure space such that, for any $1\leq p
$L^q(X, \mathcal B, m) \subset L^p(X, \mathcal B, m).$ Then $X$ doesn't contain sets of arbitrarily large measure.
Indeed it is well defined the embedding operator $G:L^q(X, \mathcal B, m) \to L^p(X, \mathcal B, m)$, and it is bounded.
Indeed the inclusion $L^q(X,\mathcal B,m)\subset L^p(X,\mathcal B,m)$ is continuous. Convergence in $L^p$ and in $L^q$ imply convergence almost everywhere and we can conclude by the closed graph theorem.
By Hölder inequality, $\|f\|_{L^p}\leq\text{meas }(X)^{1/r}\|f\|_{L^q}.$ This means $\|G\|\leq \text{meas}(X)^{1/r}=\text{meas}(X)^{1/p-1/q}.$ But, considering $f(x)=\chi_X(x)$, one sees that $\|G\|=\text{meas }(X)^{1/r}<+\infty \Leftrightarrow \text{meas }(X)<+\infty.$ Now we can proceed by density of the vector space of the simple functions in both $L^p(X,\mathcal B,m)$ and $L^q(X,\mathcal B,m).$
Theorem Let $(X,\mathcal B,m)$ be a measure space. Then $X$ doesn't contain sets of arbitrarily small measure if and only if for any $1\leq p
, one has $L^p(X, \mathcal B, m) \subset L^q(X, \mathcal B, m).$
Let us suppose that, for any subset $Y\subseteq X,\quad Y\in\mathcal B$, we have $0<\alpha\leq\text{meas}(Y)$.
It sufficies to prove the statement for simple functions. Pick now $f(x) =\sum_{j=1}^n a_j\chi_{E_j},$ where $\{E_j\}_{j=1,\dots,n}$ is a collection of disjoint subsets of $\mathcal B.$ Then $\|f\|_{L^q} \le \sum_{j=1}^n \|a_j\text{meas}(E_j)\|_{L^q} = \sum_{j=1}^n a_j\text{meas}(E_j)^{1/q}=\sum_{j=1}^n a_j\text{meas}(E_j)^{1/q-1/p+1/p}\leq\frac{1}{\alpha^{1/p-1/q}}\|f\|_{L^p}.$
The first inequality is due to Minkowski inequality.
For the converse of the theorem note that again it is well defined the embedding operator $G:L^p(X,\mathcal B,m)\to L^q(X,\mathcal B,m)$, and the operator is bounded. Now consider that, for any subset $Y\subset X$, $Y\in\mathcal B$, the function $g_Y(x)=\frac{\chi_Y(x)}{(\text{meas(Y)})^{1/p}}$ satisfies $\|g_Y\|_{L^q}= \frac{1}{(\text{meas}(Y))^{1/p-1/q}}. $ But then, for any $Y\subset X$, $Y\in\mathcal B$, we have $\frac{1}{(\text{meas}(Y))^{1/p-1/q}}\leq \|G\|,$ which means $0<\frac{1}{\|G\|^{1/p-1/q}}\leq \text{meas}(Y).$ Hence the result is proved.
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0@juaninf Most probably, $\operatorname{meas}X=m(X)$ is the $m$-measure of $X$. – 2017-05-02
In Rudin's book Real an complex analysis, we can find the following result, shown by Alfonso Villani:
Let $(X,\mathcal B,m)$ be a $\sigma$-finite measure space, where $m$ is a non-negative measure. Then the following conditions are equivalent:
- We have $L^p(X,\mathcal B,m)\supset L^q(X,\mathcal B,m)$ for some $p,q$ with $1\leqslant p
.- $m(X)<\infty$.
- We have $L^p(X,\mathcal B,m)\supset L^q(X,\mathcal B,m)$ for all $p,q$ with $1\leqslant p
.
We only have to show that $1.\Rightarrow 2.$ and $2.\Rightarrow 3.$ since $3.\Rightarrow 1.$ is obvious.
$1.\Rightarrow 2.$: the inclusion $L^q(X,\mathcal B,m)\subset L^p(X,\mathcal B,m)$ is continuous. Indeed, let $\{f_n\}$ be a sequence in $L^q$ which converges to $f$ for the $L^q$ norm, and to $g$ for the $L^p$ norm. We extract a subsequence which converges almost everywhere to $f$ and $g$ (first extract a subsequence $\{f_{n_j}\}$ which converges to $f$ almost surely; this sequence still converges to $g$ for the $L^p$ norm; now extract from this sequence a subsequence which converges to $g$ almost surely), hence $f=g$, and by the closed graph theorem we get the conclusion since both $L^p$ and $L^q$ are Banach spaces.
Therefore, we can find $C>0$ such that $\lVert f\rVert_p\leqslant C\lVert f\rVert_q$. Since $X$ can be written as an increasing union of finite measure sets $A_n$, we get that $m(A_n)^{\frac 1p}\leqslant Cm(A_n)^{\frac 1q}$, hence $m(A_n)^{\frac{q-p} {pq}}\leqslant C$ and since $p\neq q$: $m(A_n)\leqslant C^{\frac{pq}{p-q}}$. Now we take the limit $n\to\infty$ to get $m(X)\leqslant C^{\frac{pq}{p-q}}$.
$2.\Rightarrow 3.$: let $1\leqslant p and $f\in L^q$. We put $E_n:= \left\{x\in X: \frac 1{n+1}\leqslant |f(x)|\lt\frac 1n\right\}$ for $n\in\mathbb N^*$. The sets $\{E_n\}$ are pairwise disjoint and by $2.$ we get $\displaystyle\sum_{n=1}^{\infty} m(E_n)<\infty$. The function $f$ is integrable because \begin{align*} \int_X |f|^pdm &=\int_{\{|f|\geq 1\}}|f|^pdm+\sum_{n=1}^{+\infty}\int _{E_n}|f|^pdm\\\ &\leqslant\int_X |f|^qdm+\sum_{n=1}^{+\infty}\frac 1{n^p}m(E_n)\\\ &\leqslant \int_X |f|^qdm+\sum_{n=1}^{+\infty}m(E_n)<\infty. \end{align*} Now we look at the case $q=+\infty$. If $m(E)<\infty$, since for each $f\in L^q$ we can find $C_f$ such that $|f|\leqslant C_f$ almost everywhere, we can see $f\in L^p$ for all $p$. Conversely, if $L^{\infty}\subset L^p$ for a finite $p$, then the function $f=1$ is in $L^p$ and we should have $m(E)<\infty$.
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0Shouldn't the $p,q$ verify 1\leqslant p
at all steps instead of 1\leqslant p
? Bounded (essentially) function $f$ in a finite measure set implies $f \in L^p$ $\forall p \geq 1$. i.e. $L^{\infty} \subset L^p$.
– 2016-06-16
There is a easy way to show that. Suppose that $p and X a space measure finite. Take any $f\in L^q$. Then, the q-norm is finite. In this way, $\int_X|f|^p = \int_{f(x)<1}|f|^p + \int_{f\geq1}|f|^p \leq\int1+\int_{f\geq1}|f|^q\leq\mu(X)+||f||_q^q<\infty$ A counter example just take $f(x)=\frac{1}{x}$ for $x\in(0,\infty)$ and Lebesgue measure. Then $f$ belongs to $L^2$ (integral is 1) but not $L^1$.
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4You mean $x\in (1,\infty)$ here don't you? – 2017-01-10