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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0Do 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? – 2011-09-20
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3http://en.wikipedia.org/wiki/Lp_space#Embeddings – 2011-09-20
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0You should also consider the counting measure on a finite set - what happens in this case? – 2011-09-20
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0The 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/is-it-possible-for-a-function-to-be-in-lp-for-only-one-p – 2011-09-20
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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. 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.
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4A 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 $q>p$. – 2011-09-20
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3In the second theorem (the first after the add), how do you show that the embedding $G\colon L^q\to L^p$ is bounded? – 2011-09-20
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0What's is a definition of meas(X)? – 2012-11-09
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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 $$\|f\|_{L^q}=\left(\sum_{j=1}^n a_j^q\mu(E_j)\right)^{1/q}$$. – 2015-10-16
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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
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0I 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? – 2016-04-09
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0I have added details. – 2016-04-13
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0Shouldn't the $p,q$ verify $1\leqslant p
$f$
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
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4You mean $x\in (1,\infty)$ here don't you? – 2017-01-10