Bounded sequence of functions has subsequence convergent a.e.?

Does bounded sequence of functions have subsequence convergent a.e.?

There is a equivalent question :

Does weak convergent sequence of functions have subsequence convergent a.e.?

These two are equivalent because of Banach-Steinhaus theorem. Before I ask here, I thought Bolzano-Weierstrass theorem which is every bounded sequence has convergent subsequence. And my question comes up from 'this theorem can be applied to sequence of functions?'. If the space of functions should be special (like reflexive or compact or housdorff), please mention about it.

I didn't study functional analysis but attended courses : Partial differential equation and real analysis for graduate student.

• First point: if you speak about "weak convergence", you should specify what is the TVS you are considering, in order to define the related notion of "weak topology". It seems here that you are considering the space of continuous functions $C(K)$ over some compact space $K$, with the sup norm. Mar 24 '18 at 11:37
• Second point: the weak convergence of a sequence $(f_n)$ in $C(K)$ implies, but is NOT equivalent to the fact that the sequence of functions $(f_n)$ is bounded: see Weal Convergence of Continuous Functions. In any case, you are excused because you have never studied Functional Analysis! Mar 24 '18 at 11:38
• Third Point: however, it is true that for any Banach space $X$, the weak convergence of sequence $(x_n)$ can be characterized by using ALSO the boundedness condition, that is $||x_n|| \leq M$ for all $n$: this is an immediate corollary of Banach-Steinhaus theorem. See e.g. Kantorovich and Akilov, Functional Analysis, Chapter VIII, Theorem 2, p. 216. Mar 24 '18 at 12:02

Not necessarily. For each $x\in [0,1)$ let $B(x)=(x_n)_{n\in \mathbb N}$ be the sequence of digits for the representation of $x$ in base $2,$ with $\{n:x_n=0\}$ being infinite. For $n\in \mathbb N$ let $f_n(x)=x_n.$ Let $g:\mathbb N\to \mathbb N$ be strictly increasing. A convergent binary sequence is eventually constant . So if $(f_{g(n)}(x))_n=(x_{g(n)})_n$ converges then $$x\in \cup_{m\in \mathbb N}(A_m\cup B_m)$$ $$\text {where } A_m=\{x: \forall n\geq m\;(x_{g(n)}=0)\}$$ $$\text { and } B_m=\{x:\forall n\geq m\;(x_{g(n)}=1)\}.$$ We show that $\mu(A_m)=\mu(B_m)=0$ for each $m,$ where $\mu$ is Lebesgue measure, so the set of $x$ for which $(f_{g(n)}(x))_n$ converges is Lebesgue-null.

(i). Fix $m$. Let $S(n)=\{g(m+j):0\leq j\leq n\}.$ Let $2^{S(n)}$ denote the set of all functions from $S(n)$ to $\{0,1\}.$

(ii). For $h\in 2^{S(n)}$ let $$R(h)=\{x\in [0,1): \forall i\in S(n)\;(x_i=h(i)\}$$ and let $V(h)=\sum_{i\in S(n)}2^{-i}h(i).$

Let $h_0\in 2^{S(n)}$ where $h_0(j)=0$ for all $j\in S(n).$

(iii). For each $h\in 2^{S(n)}$ we have $R(h)=\{x+V(h): x\in R(h_0)\},$ so $$\mu (R(h))=\mu (R(h_0)).$$ Now if $h,h'$ are distinct members of $2^{S(n)}$ then $R(h)\cap R(h')=\emptyset.$ Also $\cup \{R(h):h\in 2^{S(n)}\}=[0,1).$ And the number of members of $2^{S(n)}$ is $2^n.$

(iv).Putting all the parts of (iii) together we have $$1=\mu ([0,1))=\sum_{h\in 2^{S(n)}}\mu (R(h))=2^n\mu (R(h_0)).$$ Since $A_m\subset R(h_0)$ we have $1\geq 2^n \mu (A_m)$ for every $n.$ Therfore $\mu (A_m)=0.$

The proof that $\mu (B_m)=0$ is similar.

Note: I neglected to show that $R(h_0)$ is a Lebesgue-measurable set. I will leave this as an exercise because it's 8 A.M. and I've been up all night. Happy New Year.

• Have a happy new year to you, too. I'm reading line by line now. But it does not look that intuitive. Is this example well-known? Jan 1 '17 at 14:22
• I dk whether it is well-known. I'm an amateur. $\mu (A_m)=0$ is likely well-known. If we regard $\mu$ as a probability measure, the odds that $x_i=0$ for all $i$ in a finite $S\subset \{g(n)\}_n$ should be $2^{-|S|}.$ Jan 2 '17 at 5:20

There is no subsequence of $\sin (nx)$ that converges a.e. In fact, every subsequence $\sin (n_kx)$ diverges a.e. For a proof of this, see Pointwise almost everywhere convergent subsequence of $\{\sin (nx)\}$