"Elementary proof" of "weak convergence implies norm boundedness in $L^p$"

Question

Suppose $p\in[1,\infty)$, $X$ $\sigma$-finite and $f_n\to f$ weakly in $L^p(X)$.

Can you give an "elementary proof" of the boundedness of $\{f_n\}$? By elementary I mean no uniform boundedness principle or functional analysis, only measure theoretic argument.

If the proof is long, a sketch or the main idea will be enough!

Answer 1

At least for $\ell^p(\mathbb N)$ one can give an elementary "gliding hump" argument - of course the proof by Uniform Boundedness is simpler, even if we include the proof of UB itself. It's not clear to me exactly how to adapt this to a general measure space.

An outline, leaving a lot of details to you:

Lemma 1. If the functions $f_n\in\ell^p$ have disjoint support and $||f_n||_p\to\infty$ then $f_n$ does not tend to $0$ weakly.

Proof: Replacing $(f_n)$ by a subsequence we have $E_n$ disjoint, $f_n$ supported on $E_n$, and $||f_n||_p\ge 2^n$. Choose $g_n\in\ell^{p'}$ so $g_n$ is supported on $E_n$, $||g_n||_{p'}=1/n^2$, $f_ng_n\ge0$, and $\sum_k f_n(k)g_n(k)=2^n/n^2$. Let $g=\sum g_n$; then $g\in\ell^{p'}$ and $\sum_kf_n(k)g(k)=\sum_kf_n(k)g_n(k)\to\infty$.

Here's the bit that's problematic for a general measure space:

Lemma 2 If $f_n\to0$ weakly in $\ell^p$ then $\lim_nf_n(k)=0$ for every $k$; hence $\lim_n\sum_{k\in F}|f_n(k)|^p=0$ for every finite set $F$.

Now suppose $f_n\to0$ weakly and $||f_n||_p\to\infty$. Using Lemma 2 we can find a subsequence $f_{n_j}$ such that there exist finite disjoint sets $F_j$ such that $\lim_j\sum_{k\in\mathbb N\setminus F_j}|f_{n_j}(k)|^p=0$.

(Having chosen $n_1,\dots n_j$: Choose $j+1$ so the norm of $f_{n_{j+1}}$ on $F_1\cup\dots \cup F_j$ is small; now choose $F_{j+1}$ disjoint from $F_1,\dots F_j$ so $F_{j+1}$ carries all but a small amount of the mass of $|f_{j+1}|^p$.)

Now let $g_j=f_{n_j}\chi_{F_j}$. Then Lemma 1 shows that $g_j$ does not tend to $0$ weakly, and since $||f_{n_j}-g_j||_p\to0$ it follows that $f_{n_j}$ does not tend to $0$ weakly either.

Answer 2

We assume that $f=0$. Assume that the sequence $\left(\left\lVert f_n\right\rVert_p\right)_{n\geqslant 1}$ is not bounded. We construct by induction an increasing sequence of integers $(n_j)$ such that

• $\left\lVert f_{n_{j}}\right\rVert_p\geqslant 2^j\left\lVert f_{n_{j-1}}\right\rVert_p^{2}$
• for each $j\geqslant 2$, $\left\lvert\int f_{n_j} \sum_{i=1}^{j-1}\operatorname{sgn}\left(f_{n_i}\right)\left\lvert f_{n_i}\right\rvert^{p-1} \left\lVert f_{n_i}\right\rVert_p^{1/2-p}\right\rvert\leqslant 1/j$.

Suppose that $n_1,\dots,n_{j-1}$ have been constructed. We know that $$\lim_{N\to +\infty} \int f_{N} \sum_{i=1}^{j-1}\operatorname{sgn}\left(f_{n_i}\right)\left\lvert f_{n_i}\right\rvert^{p-1} \left\lVert f_{n_i}\right\rVert_p^{1/2-p}=0 $$ hence for $N$ large enough, $$ \left\lvert\int f_{N} \sum_{i=1}^{j-1}\operatorname{sgn}\left(f_{n_i}\right)\left\lvert f_{n_i}\right\rvert^{p-1} \left\lVert f_{n_i}\right\rVert_p^{1/2-p}\right\rvert\leqslant 1/j. $$ Among these $N$'s, we can choose one such that $\left\lVert f_{N}\right\rVert_p\geqslant 2^j\left\lVert f_{n_{j-1}}\right\rVert_p^{1/2}$, by unboundedness and we choose $n_j$ equal to this $N$. Now define $$ g:=\sum_{i=1}^{+\infty}\operatorname{sgn}\left(f_{n_i}\right)\left\lvert f_{n_i}\right\rvert^{p-1} \left\lVert f_{n_i}\right\rVert_p^{1/2-p}. $$ Then $g$ belongs to $\mathbb L^{p'}$ and using Hölder's inequality, $$\left\lvert \int f_{n_j}g\right\rvert\geqslant \left\lVert f_{n_j}\right\rVert_p^{1/2}-\left\lvert\int f_{n_j} \sum_{i=1}^{j-1}\operatorname{sgn}\left(f_{n_i}\right)\left\lvert f_{n_i}\right\rvert^{p-1} \left\lVert f_{n_i}\right\rVert_p^{1/2-p}\right\rvert\\ -\left\lvert\int f_{n_j} \sum_{i=j+1}^{+\infty}\operatorname{sgn}\left(f_{n_i}\right)\left\lvert f_{n_i}\right\rvert^{p-1} \left\lVert f_{n_i}\right\rVert_p^{1/2-p}\right\rvert\\ \geqslant \left\lVert f_{n_j}\right\rVert_p^{1/2}-\frac 1j-\left\lVert f_{n_j}\right\rVert_p \sum_{i=j+1}^{+\infty}\left\lVert f_{n_i}\right\rVert_p^{-1/2}$$ and by the first bullet, $$\sum_{i=j+1}^{+\infty}\left\lVert f_{n_i}\right\rVert_p^{-1/2}\leqslant \left\lVert f_{n_j}\right\rVert_p^{-1}2^{-j+1}$$ hence $$\left\lvert \int f_{n_j}g\right\rvert\geqslant\left\lVert f_{n_j}\right\rVert_p^{1/2}-\frac 1j-2^{-j+1},$$ which contradicts weak convergence.