a non-decreasing sequence of functions with bounded L^p-norm is a Cauchy sequence in L^p space

Question

Let $\{f_k\}_{k=1}^\infty$ be a sequence in $L^p(\mathbb{R})$ for $1\leq p<\infty$. Suppose $f_1\leq f_2\leq\cdots$ and $\sup ||f_k||_p<\infty$. Prove that $\{f_k\}_{k=1}^\infty$ converges in $L^p$-norm.

Case 1. $p=1$.

Let $g_n=f_n-f_1$. Then $g_n$ non-negative, increasingly converges to a function $g$ pointwise. By Levi's Theorem, $\int g_n\to \int g$. Let $f=g+f_1$. Then $f_n$ converges pointwise to $f$ and $\int f_n\to \int f$. $$ \int|g|=\lim_{n\to\infty}\int|g_n|=\sup\int|g_n|\leq \sup \int |f_n|+\int |f_1|<\infty. $$ By Lebesgue dominated convergence theorem, $\lim_{n\to\infty}\int |g-g_n|d\lambda=0$. Thus $$\lim_{n\to\infty}\int|f_n-f|d\lambda=\lim_{n\to\infty}\int|g_n-g|d\lambda=0. $$

Case 2. $1<p<\infty$.

The above argument cannot be applied in this case. What should I do?

Answer 1

The measure of the set of point that is unbounded is null, otherwise the norm would be unbounded on the sequence. Thus you can define a limit of pointwise convergence almost everywhere.

Now we separate out the set of point that is eventually positive and those that never be. For point that is eventually positive then eventually $|f_{n}|$ will be increasing, by monotone convergence, the norm of that limit is the limit of the norm. For the rest, then $|f_{n}|$ must be decreasing, you can use dominated convergence (dominated by $|f_{1}|$). Since the norm of the sequence is increasing and bounded, it does indeed have a finite limit and thus is in $L^{p}$.

Hence the sequence converge in the space and thus must be Cauchy.

Answer 2

Following your construction of $g_n$, for $1<p<\infty$, we use the fact that $L^p$ is reflexive Banach spaces. Given $\|g_n\|_p$ is bounded, $g_n \rightarrow g$ pointwise a.e. imples $g_n\rightharpoonup g$ in $L^p$. And from $0\leq g_n \leq g_2 \leq...$ we have $$\limsup\|g_n\|\leq \|g\|,$$ together with weak convergence we get $g_n \rightarrow g$ strongly in $L^p$. Thus $f_n \rightarrow g+f_1$ strongly in $L^p$.