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.


1 Answer 1


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.

  • $\begingroup$ That's brilliant! Thanks! Btw, the fact that $f\in L^{p,r}$ can also be deduced from $|f|\leq g$ and $g\in L^{p,r}$. $\endgroup$
    – Feng
    Oct 11, 2019 at 15:53
  • $\begingroup$ Yes. It's just that the estimate is needed anyway to show that $f$ is the limit. $\endgroup$ Oct 11, 2019 at 16:30

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