# The composition space of $L^p$ spaces is complete.

Here is a homework problem from real analysis class.

We take as our underlying space as the product space $$\{(x,t)\}=\mathbb R^d\times \mathbb R$$, with the product measure $$dxdt$$, where $$dx$$ and $$dt$$ are Lebesgue measures on $$\mathbb R^d$$ and $$\mathbb R$$, repectively. We define $$L_t^r(L_x^p)=L^{p,r}$$ with $$1\leq p\leq\infty$$, $$1\leq r\leq\infty$$, to be the space of equivalence class of jointly measurable functions $$f(x,t)$$ for which the norm $$\|f\|_{L^{p,r}}=\left(\int_\mathbb R\left(\int_{\mathbb R^d} |f(x,t)|^p\,dx\right)^{\frac rp}\right)^{\frac1r}$$ is finite when $$p<\infty$$ and $$r<\infty$$ and an obvious variant when $$p=\infty$$ and $$r=\infty$$. What I need to do is to verify that $$L^{p,r}$$ with this norm is complete and hence is a Banach space.

My attempt: Suppose $$p<\infty$$ and $$r<\infty$$. Suppose $$\{f_n(x,t)\}_1^\infty$$ is a Cauchy sequence in $$L^{p,r}$$. Let $$g_n(t)=\|f_n(\cdot,t)\|_{L_x^p}$$ for all $$n\geq 1$$ then $$\{g_n\}_1^\infty\subset L_t^r$$ and by Minkowski's inequality \begin{align*} \|g_n-g_m\|_{L_t^r}&=\left(\int_\mathbb R\left|\|f_n(\cdot,t)\|_{L_x^p}-\|f_m(\cdot,t)\|_{L_x^p}\right|^r\right)^{\frac1r}\\ &\leq \left(\int_\mathbb R\left|\|f_n(\cdot,t)-f_m(\cdot,t)\|_{L_x^p}\right|^r\right)^{\frac1r}\\ &=\|f_n-f_m\|_{L^{p,r}}, \end{align*} $$\{g_n(t)\}_1^\infty$$ is a Cauchy sequence in $$L_t^r$$ and thus there is $$g(t)\in L_t^r$$ such that $$g_n\to g$$ in $$L_t^r$$. But how can we find the $$L^{p,r}$$ limit of $$f_n$$ from this?

Another thought: I want to establish a quasi-Chebyshev's inequality under this norm and then we can deduce that $$\{f_n\}$$ is Cauchy in measure from the assumption that it is Cauchy in $$L^{p,r}$$, and then find an a.s.-convergent subsequence of $$\{f_n\}$$ and go on from here. But I failed to establish the desired inequality.

Any help will be appreciated.

• Oct 11, 2019 at 15:01

You can repeat the usual proof of completeness of $$L^p$$. Namely, let $$\{f_n\}$$ be Cauchy; by choosing a subsequence we may assume that $$\|f_{n+1}-f_n\|_{L^{p,r}}<2^{-n}$$. Define $$g=|f_1|+\sum_j|f_{j+1}-f_j|,$$ this is a measurable function because it is a sum/limit of positive measurable functions. Now \begin{align} \left\||f_1|+\sum_{j=1}^n|f_{j+1}-f_j|\right\|_{L^{p,r}} &\leq\|f_1\|_{L^{p,r}}+\sum_j\|f_{j+1}-f_j\|_{L^{p,r}}\leq\|f_1\|_{L^{p,r}}+1. \end{align} As the bound does not depend on $$n$$ and the integrals on the left-hand-side are increasing on $$n$$, we get that $$g\in {L^{p,r}}$$. Then $$|g|<\infty$$ a.e. and $$\tag1 f=f_1+\sum_j(f_{j+1}-f_j)$$ is finite a.e.; as $$(1)$$ telescopes, we also have that $$f=\lim f_n$$ a.e. Fix $$\varepsilon>0$$; there exists $$m$$ such that $$\|f_n-f_m\|_{L^{p,r}}<\varepsilon$$ for all $$n\geq m$$. Applying Fatou's Lemma twice we get \begin{align} \int_{\mathbb R}\left(\int_{\mathbb R^d}|f(x,t)-f_m(x,t)|^p\,dx\right)^{\tfrac rp}\,dt &\leq\liminf_n\int_{\mathbb R}\left(\int_{\mathbb R^d}|f_n(x,t)-f_m(x,t)|^p\,dx\right)^{\tfrac rp}\,dt\\ \ \\ &=\liminf_n\|f_n-f_m\|_{L^{p,r}}^r<\varepsilon^r. \end{align} In particular $$f-f_m\in L^{p,r}$$, and so $$f\in L^{p,r}$$. Now the same above estimate shows that $$f_m\to f$$ in $$L^{p,r}$$.
The above shows that any Cauchy sequence in $$L^{p,r}$$ has a convergent subsequence. But then the sequence itself converges to the same limit.
• That's brilliant! Thanks! Btw, the fact that $f\in L^{p,r}$ can also be deduced from $|f|\leq g$ and $g\in L^{p,r}$.
• Yes. It's just that the estimate is needed anyway to show that $f$ is the limit. Oct 11, 2019 at 16:30