# 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?

Theorem Let $$XX$$ be a finite measure space. Then, for any $$1\leq p< q\leq +\infty1\leq p< q\leq +\infty$$ $$L^q(X, \mathcal B, m) \subset L^p(X, \mathcal B, m).L^q(X, \mathcal B, m) \subset L^p(X, \mathcal B, m).$$ The proof follows from Hölder inequality. Note that $$\frac{1}{p}=\frac{1}{q}+\frac{1}{r}\frac{1}{p}=\frac{1}{q}+\frac{1}{r}$$, with $$r>0r>0$$. Hence $$\|f\|_{L^p}\leq\text{meas }(X)^{1/r}\|f\|_{L^q}.\|f\|_{L^p}\leq\text{meas }(X)^{1/r}\|f\|_{L^q}.$$

The case reported on the Wikipedia link of commenter answer follows from this, since of course, if $$XX$$ does not contain sets of arbitrary large measure, $$XX$$ itself can't have an arbitrary large measure.

For the counterexample: $$f(x)=\frac{1}{x}f(x)=\frac{1}{x}$$ belongs to $$L^2([1,+\infty))L^2([1,+\infty))$$, but clearly it does not belong to $$L^1([1,+\infty)).L^1([1,+\infty)).$$

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 $$(X,\mathcal B,m)(X,\mathcal B,m)$$ is a measure space such that, for any $$1\leq p $$L^q(X, \mathcal B, m) \subset L^p(X, \mathcal B, m).L^q(X, \mathcal B, m) \subset L^p(X, \mathcal B, m).$$ Then $$XX$$ doesn't contain sets of arbitrarily large measure.

Indeed it is well defined the embedding operator $$G:L^q(X, \mathcal B, m) \to L^p(X, \mathcal B, m)G:L^q(X, \mathcal B, m) \to L^p(X, \mathcal B, m)$$, and it is bounded.

Indeed the inclusion $$L^q(X,\mathcal B,m)\subset L^p(X,\mathcal B,m)L^q(X,\mathcal B,m)\subset L^p(X,\mathcal B,m)$$ is continuous. Convergence in $$L^pL^p$$ and in $$L^qL^q$$ imply convergence almost everywhere and we can conclude by the closed graph theorem.

By Hölder inequality, $$\|f\|_{L^p}\leq\text{meas }(X)^{1/r}\|f\|_{L^q}.\|f\|_{L^p}\leq\text{meas }(X)^{1/r}\|f\|_{L^q}.$$ This means $$\|G\|\leq \text{meas}(X)^{1/r}=\text{meas}(X)^{1/p-1/q}.\|G\|\leq \text{meas}(X)^{1/r}=\text{meas}(X)^{1/p-1/q}.$$ But, considering $$f(x)=\chi_X(x)f(x)=\chi_X(x)$$, one sees that $$\|G\|=\text{meas }(X)^{1/r}<+\infty \Leftrightarrow \text{meas }(X)<+\infty.\|G\|=\text{meas }(X)^{1/r}<+\infty \Leftrightarrow \text{meas }(X)<+\infty.$$ Now we can proceed by density of the vector space of the simple functions in both $$L^p(X,\mathcal B,m)L^p(X,\mathcal B,m)$$ and $$L^q(X,\mathcal B,m).L^q(X,\mathcal B,m).$$

Theorem Let $$(X,\mathcal B,m)(X,\mathcal B,m)$$ be a measure space. Then $$XX$$ 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).L^p(X, \mathcal B, m) \subset L^q(X, \mathcal B, m).$$

Let us suppose that, for any subset $$Y\subseteq X,\quad Y\in\mathcal BY\subseteq X,\quad Y\in\mathcal B$$, we have $$0<\alpha\leq\text{meas}(Y)0<\alpha\leq\text{meas}(Y)$$.

It sufficies to prove the statement for simple functions. Pick now $$f(x) =\sum_{j=1}^n a_j\chi_{E_j},f(x) =\sum_{j=1}^n a_j\chi_{E_j},$$ where $$\{E_j\}_{j=1,\dots,n}\{E_j\}_{j=1,\dots,n}$$ is a collection of disjoint subsets of $$\mathcal B.\mathcal B.$$ Then $$\|f\|_{L^q} \le \sum_{j=1}^n \|a_j\text{meas}(E_j)\|_{L^q} = \sum_{j=1}^n a_j\text{meas}(E_j)^{1/q}=\sum_{j=1}^n a_j\text{meas}(E_j)^{1/q-1/p+1/p}\leq\frac{1}{\alpha^{1/p-1/q}}\|f\|_{L^p}.\|f\|_{L^q} \le \sum_{j=1}^n \|a_j\text{meas}(E_j)\|_{L^q} = \sum_{j=1}^n a_j\text{meas}(E_j)^{1/q}=\sum_{j=1}^n a_j\text{meas}(E_j)^{1/q-1/p+1/p}\leq\frac{1}{\alpha^{1/p-1/q}}\|f\|_{L^p}.$$

The first inequality is due to Minkowski inequality.

For the converse of the theorem note that again it is well defined the embedding operator $$G:L^p(X,\mathcal B,m)\to L^q(X,\mathcal B,m)G:L^p(X,\mathcal B,m)\to L^q(X,\mathcal B,m)$$, and the operator is bounded. Now consider that, for any subset $$Y\subset XY\subset X$$, $$Y\in\mathcal BY\in\mathcal B$$, the function $$g_Y(x)=\frac{\chi_Y(x)}{(\text{meas(Y)})^{1/p}}g_Y(x)=\frac{\chi_Y(x)}{(\text{meas(Y)})^{1/p}}$$ satisfies $$\|g_Y\|_{L^q}= \frac{1}{(\text{meas}(Y))^{1/p-1/q}}. \|g_Y\|_{L^q}= \frac{1}{(\text{meas}(Y))^{1/p-1/q}}.$$
But then, for any $$Y\subset XY\subset X$$, $$Y\in\mathcal BY\in\mathcal B$$, we have $$\frac{1}{(\text{meas}(Y))^{1/p-1/q}}\leq \|G\|,\frac{1}{(\text{meas}(Y))^{1/p-1/q}}\leq \|G\|,$$ which means $$0<\frac{1}{\|G\|^{1/p-1/q}}\leq \text{meas}(Y).0<\frac{1}{\|G\|^{1/p-1/q}}\leq \text{meas}(Y).$$ Hence the result is proved.