L^pL^p and L^qL^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?

Answer

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.

Attribution
Source : Link , Question Author : Confused , Answer Author : kimchi lover

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