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

Question

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?

Answer 1

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 $$\|G\|\leq \text{meas}(X)^{1/r}=\text{meas}(X)^{1/p-1/q}.$$ 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 $1\leq p<q\leq +\infty$, 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 $$\|f\|_{L^q} \le \sum_{j=1}^n \|a_j\text{meas}(E_j)\|_{L^q} = \sum_{j=1}^n a_j\text{meas}(E_j)^{1/q}=\sum_{j=1}^n a_j\text{meas}(E_j)^{1/q-1/p+1/p}\leq\frac{1}{\alpha^{1/p-1/q}}\|f\|_{L^p}.$$
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.

Answer 2

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 $m(E)<\infty$.

Answer 3

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 $$ L_{p}(X) \subset L_{q}(X). $$

Answer 4

This is to show that the restriction $1\leq p<q\leq\infty$ in the OP is not needed, and that the following result holds:

Theorem A: Suppose $(\Omega,\mathscr{F},\mu)$ is a $\sigma$-finite measure space. There exists $p,q$ with $0<p<q\leq\infty$ such that $L_q(\mu)\subset L_p(\mu)$ iff $\mu(X)<\infty$.

Sufficiency follows directly from Hölder's inequality. In fact, if $0<p<q\leq\infty$, $$\int_X|f|^p\,d\mu\leq \|f\|^p_q\big(\mu(X)\big)^{1-\tfrac{p}{q}}$$

Necessity can be shown using the simple and elegant argument described in Davide's posting. First some general observations about $L_p(\mu)$ spaces with $0<p<\infty$. To ease notation, we define $\|f\|_p=\Big(\int_X|f|^p\,d\mu\Big)^{1/p}$.

1. The function $d_p:L_p(\mu)\times L_p(\mu)\rightarrow[0,\infty)$ given by $$d_p(f, g)=\Big(\int_X|f-g|^p\,d\mu\Big)^{\min(1,1/p)}=\|f-g\|_p^{\min(p,1)}$$ defines a complete translation invariant metric on $L_p(\mu)$, and so, $(L_p(\mu),d_p)$ is an $F$-space. (Of course when $p\geq1$, $L_p(\mu)$ is a Banach space with norm $\|f\|_p=d_p(0,f)$; for $0<p<1$, see this posting).
2. The sets $B_p(0;r)=\{f\in L_p(\mu): d_p(0,f)<r\}$ form a local basis of neighborhoods for $(L_p(\mu),d_p)$.
3. For any $r>0$, $B_p(0;r)=r^{\max(1,1/p)}B_p(0;1)$.

Now recall the general versions of the Closed Graph Theorem in the setting of $F$-spaces.

Theorem: Suppose $X$ and $Y$ are $F$-spaces. If $\Lambda:X\rightarrow Y$ is linear and its graph $G=\{(x,\Lambda x):x\in X\}$ is closed in $X\times Y$, then $\Lambda$ is continuous (and thus bounded since $X$ and $Y$ are metrizable).

This can be sen for example in Rudin, W., Functional Analysis, McGraw-Hill, second edition, 1991, pp. 51

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Now we have all the ingredients to modify Davide's proof to the present setting. Suppose $L_q(\mu)\subset L_p(\mu)$ for some $0<p<q<\infty$. Consider the inclusion map $\iota: L_q(\mu)\rightarrow L_p(\mu)$. The continuity of $\iota$ follows from the Closed Graph Theorem just as explained by Davide: Suppose $(f_n:n \in\mathbb{N})\subset L_q(\mu)$ is a sequence which converges in $L_q(\mu)$ to some $f\in L_q(\mu)$, and which also converges in $L_p(\mu)$ to some $g\in L_p(\mu)$. Along a subsequence $f_{n'}$, $|f-f_{n'}|^q\rightarrow0$ $\mu$-a.s. Then, along a subsequence $f_{n''}$ of $f_{n'}$, $|f_{n''}-g|^p\rightarrow0$ $\mu$-a.s. It then follows that $f=g$ $\mu$-a.s. and so $\iota$ is bounded.

The key to extend the statement of the OP to the present setting is to understand how bounded sets in general $L_p(\mu)$ spaces look like.

Claim: For any $0<r\leq \infty$, $E\subset L_r(\mu)$ is bounded iff $$\sup\left\{f\in E: \|f\|_r\right\}<\infty$$

Proof: By definition (see Rudin, cit. op.) $E$ is bounded iff for any open neighborhood $U$ of $0$ in $L_p(\mu)$ there is $s>0$ such that $E\subset sU=\{sf: f\in U\}$. Notice that $$\|\lambda f\|^{\min(r,1)}_r=d_r(0,\lambda f)=\lambda^{\min(r,1)}\,d_r(0,f)=\lambda^{\min(r,1)}\|f\|^{\min(1,r)}_r$$ The conclusion of the claim follows immediately from observations (2) and (3), and by taking $U=B_r(0;1)$.

In particular, by observation (3) we have that any ball $\overline{B}_p(0;r)=\{f\in L_p(\mu):d_p(0,f)\leq r\}$ is bounded in $L_p(\mu)$.

By the continuity of $\iota$, $\overline{B}_q(0;1)$ is bounded in $L_p(\mu)$ since $\overline{B}_q(0;1)$ is bounded in $L_q(\mu)$. This means that $$c:=\sup_{f\in \overline{B}_q(0;1)}\|f\|_p<\infty$$

Notice that for any $f\in L_q(\mu)\setminus\{0\}$, $\frac{1}{\|f\|_q}f\in \overline{B}_q(0;1)$. Hence $$\|f\|_p\leq c\|f\|_q,\qquad f\in L_q(\mu)$$ Let $(A_n:n\in\mathbb{N})\subset\mathcal{F}$ such that $A_m\nearrow \Omega$. Then $$\big(\mu(A_n)\big)^{\tfrac{1}{p}-\tfrac{1}{q}}\leq c$$ whence we conclude that $\mu(X)\leq c^{\tfrac{pq}{q-p}}<\infty$.

Finally the case $0<p<q=\infty$ is simpler for if $L_\infty(\mu)\subset L_p(\mu)$, then as $f\equiv1\in L_\infty(\mu)$, $$\|f\|_p=\big(m(X)\big)^{1/p}<\infty$$