# $L^p$ and $L^q$ space inclusion

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?

• 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
• en.wikipedia.org/wiki/Lp_space#Embeddings – user16391 Sep 20 '11 at 10:23
• You should also consider the counting measure on a finite set - what happens in this case? – AD. Sep 20 '11 at 12:25
• The question has been effectively answered by the answers below in the case of finite spaces. For infinite spaces it may be interesting to read: math.stackexchange.com/questions/55170/… – Ragib Zaman Sep 20 '11 at 12:50
• – Watson Jun 6 '16 at 16:57

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 $$L^1([1,+\infty)).$$

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. Convergenge 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.

• A 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. Sep 20 '11 at 12:23
• In 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
• What's is a definition of meas(X)? – juaninf Nov 9 '12 at 23:06
• I'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
• @juaninf Most probably, $\operatorname{meas}X=m(X)$ is the $m$-measure of $X$. – 0xbadf00d May 2 '17 at 13:21

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:

1. We have $L^p(X,\mathcal B,m)\supset L^q(X,\mathcal B,m)$ for some $p,q$ with $1\leqslant p<q<\infty$.
2. $m(X)<\infty$.
3. 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 $m(A_n)^{\frac{q-p} {pq}}\leqslant C$ 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 $1\leqslant p<q<\infty$ 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 $m(E)<\infty$.

• 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 9 '16 at 14:25
• I have added details. – Davide Giraudo Apr 13 '16 at 9:33
• Shouldn't the $p,q$ verify $1\leqslant p<q \leq\infty$ at all steps instead of $1\leqslant p<q<\infty$? Bounded (essentially) function $f$ in a finite measure set implies $f \in L^p$ $\forall p \geq 1$. i.e. $L^{\infty} \subset L^p$. – D1X Jun 16 '16 at 19:05

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 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 $L^1$.

• You mean $x\in (1,\infty)$ here don't you? – user161518 Jan 10 '17 at 18:36