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?
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Do you want $L^q(X,\mathcal B,m)$ to be any subset of $L^p(X,\mathcal B, m)$, or a *proper* subset? Also did you mean $p \leq q$ or something else in the question? – Srivatsan Sep 20 '11 at 10:23

3http://en.wikipedia.org/wiki/Lp_space#Embeddings – Sep 20 '11 at 10:23

1You should also consider the counting measure on a finite set  what happens in this case? – AD  Stop Putin  Sep 20 '11 at 12:25

1The question has been effectively answered by the answers below in the case of finite spaces. For infinite spaces it may be interesting to read: http://math.stackexchange.com/questions/55170/isitpossibleforafunctiontobeinlpforonlyonep – Ragib Zaman Sep 20 '11 at 12:50

1Related: http://math.stackexchange.com/questions/1371017/arethererelationsbetweenelementsoflpspaces/1371051#1371051 – Watson Jun 06 '16 at 16:57
4 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 $1\leq p<q\leq +\infty,$ $$L^q(X, \mathcal B, m) \subset L^p(X, \mathcal B, m).$$ 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 $$\G\\leq \text{meas}(X)^{1/r}=\text{meas}(X)^{1/p1/q}.$$ 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 $1\leq p<q\leq +\infty$, 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 $$\f\_{L^q} \le \sum_{j=1}^n \a_j\text{meas}(E_j)\_{L^q} = \sum_{j=1}^n a_j\text{meas}(E_j)^{1/q}=\sum_{j=1}^n a_j\text{meas}(E_j)^{1/q1/p+1/p}\leq\frac{1}{\alpha^{1/p1/q}}\f\_{L^p}.$$
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 $$\g_Y\_{L^q}= \frac{1}{(\text{meas}(Y))^{1/p1/q}}. $$ But then, for any $Y\subset X$, $Y\in\mathcal B$, we have $$\frac{1}{(\text{meas}(Y))^{1/p1/q}}\leq \G\,$$ which means $$0<\frac{1}{\G\^{1/p1/q}}\leq \text{meas}(Y).$$ Hence the result is proved.
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5A slightly better counterexample is given by $f_p(x)=x^{1/p}$; this function belongs to $L^q(1,\infty)\setminus L^p(1,\infty)$ provided $q>p$. – AD  Stop Putin  Sep 20 '11 at 12:23

4In the second theorem (the first after the add), how do you show that the embedding $G\colon L^q\to L^p$ is bounded? – Davide Giraudo Sep 20 '11 at 16:02


6I'm sure you made a mistake in the proof of the last theorem. Now suppose $f$ is a simple function, i.e. $f(x)=\sum_{j=1}^n a_j\chi_{E_j}$, where $E_j$ are disjoint, the $L^q$ norm of $f$ should be $$\f\_{L^q}=\left(\sum_{j=1}^n a_j^q\mu(E_j)\right)^{1/q}$$. – Xiang Yu Oct 16 '15 at 10:24

1@juaninf Most probably, $\operatorname{meas}X=m(X)$ is the $m$measure of $X$. – 0xbadf00d May 02 '17 at 13:21

When you say sets of "arbitrarily small measure" I'm assuming this is disregarding measure $0$ sets? – AIM_BLB Jan 13 '20 at 12:47

1But how can you apply Holder inequality if measure of $X$ isn't finite?! – Hrit Roy Oct 27 '20 at 12:05


I think your proof of the first theorem after the ADD, as stated, is incorrect. You wrote in the proof: "By Hölder inequality, $$\f\_{L^p}\leq\text{meas }(X)^{1/r}\f\_{L^q}."$$ But, by applying Hölder inequality in this way, you are (implicitly) assuming that $\chi_X \in L^q$, which means $X$ has finite measure. However, the fact that $X$ has finite measure is **exactly** what you want to prove. – Ramiro Dec 12 '20 at 17:44
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 nonnegative 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 $1\leqslant p<q<\infty$.
 $m(X)<\infty$.
 We have $L^p(X,\mathcal B,m)\supset L^q(X,\mathcal B,m)$ for all $p,q$ with $1\leqslant p<q<\infty$.
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 $m(A_n)^{\frac{qp} {pq}}\leqslant C$ and since $p\neq q$: $m(A_n)\leqslant C^{\frac{pq}{pq}}$. Now we take the limit $n\to\infty$ to get $m(X)\leqslant C^{\frac{pq}{pq}}$.
$2.\Rightarrow 3.$: let $1\leqslant p<q<\infty$ 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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I don't see how you can just pick one subsequence in step 1 that converges to bot $f$ and $g$. It seems you can you just pick one for $f$ (or $g$), using Theorem 4.9 in Breviz, but how can you pick one that converges to both? – csss Apr 09 '16 at 14:25


There is a easy way to show that. Suppose that $p<q$ and X a space measure finite. Take any $f\in L^q$. Then, the qnorm is finite. In this way, $$\int_Xf^p = \int_{f(x)<1}f^p + \int_{f\geq1}f^p \leq\int1+\int_{f\geq1}f^q\leq\mu(X)+f_q^q<\infty$$ 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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Alternatively one can take into account this proposition.
$\textbf{Proposition. }$Let $(X, \mathcal{A}, \mu)$ a finite measure space. Let $f$ be $\mathcal{A}$measurable function and let $E_{n} = \{x \in X: (n  1) \leq \lvert f (x) \rvert < n\}$. Then $f \in L^{p}(X)$ if and only if $$\sum_{n = 1}^{\infty} n^{p} \mu(E_{n}) < + \infty.$$
Using this proposition, we have that for $1 \leq q < p$, if $f \in L^{p}(X)$ then $$ \sum_{n = 1}^{\infty} n^{q} \mu(E_{n}) \leq \sum_{n = 1}^{\infty} n^{p} \mu(E_{n}) < + \infty. $$ Hence $f \in L^{q}(X)$. So we have the inclusion $$ L_{p}(X) \subset L_{q}(X). $$
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