# Pointwise convergence resisting averaging

Can you give an example of a sequence of continuous functions $f_n:[0,1]\to [0,1]$, such that $f_n\to 0$ pointwise and there is no subsequence $(f_{n_k})$ for which $\frac 1 m\sum_{k=1}^{m}f_{n_k}$ tends to zero uniformly?

I think it's the same as asking whether the Banach space of continuous real valued functions has the "weak Banach-Saks property", but I was unable to find out the answer.

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What have you tried so far? –  Jesse Madnick Feb 9 '12 at 17:16
@JesseMadnick: not much. I was hoping someone would just know the answer... –  KotelKanim Feb 9 '12 at 17:56
Did you try some triangular wave which increases frequency but such that $f_{n + 1}$ has the same peaks as $f_n$ but more of them? –  Jonas Teuwen Feb 9 '12 at 18:27
[What have you tried so far?] not much. I was hoping someone would just know the answer... –  Did Feb 9 '12 at 20:03
It was somewhat rhetoric, @Didier. (if you are replying on my comment) –  Jonas Teuwen Feb 9 '12 at 21:20

That $C[0,1]$ fails the weak Banach Saks property appears as problem 17 in chapter VII in Joseph Diestel's Sequences and Series in Banach Spaces.

The problem gives an outline:

1) For $k$ a fixed positive integer, construct a nonnegative sequence $(g_n^k)_n$ in $B(C[0,1])$ satisfying:

• $g_n^k(t)=0$ if $t\notin ((k-1)/k, k/(k+1))$.
• $(g_n^k)$ converges pointwise to 0 on $[0,1]$.
• If $n_1<n_2<\cdots<n_k$, then there is an $a\in[0,1]$ with $g_{n_1}^k(a)=g_{n_2}^k(a)=\cdots=g_{n_k}^k(a)=1$.

2) Define $f_n=g_n^1+g_n^2+\cdots+g_n^n$. Show that:

• $(f_n)$ is weakly null in $C[0,1]$ (which is equivalent to saying $(f_n)$ converges pointwise to 0 and is bounded).
• If $n_1<n_2<\cdots<n_m<n_{m+1}<\cdots< n_{2m}$, then $(f_{n_1}+\cdots f_{n_{2m}})(t)\ge {1\over 2}$ for all $t\in[0,1]$ (I believe Diestel has a typo here, it should be $\ge m/2$ for some $t\in[0,1]$).

This was first proved by J. Schreier in Ein Gegenbeispiel zur Theorie der schwachen Konvergence, Studia Math. 2 (1930), 58–62. H. P. Rosenthal's article in volume 2 of Handbook of the Geometry of Banach Spaces (proposition 4.21, page 1585) may also be helpful.

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The question did say $f_n : [0,1] \to [0,1]$ so they are bounded in $C[0,1]$. (I missed this at first also.) –  Nate Eldredge Feb 10 '12 at 1:18
@Nate Ah! I should have seen that... –  David Mitra Feb 10 '12 at 1:22
How is $(\frac12, n)$ in $[0, 1]$ for $n \geqslant 1$? –  Jonas Teuwen Feb 10 '12 at 8:44
@Jonas I initially misread the question, and thought the OP wanted only pointwise convergence. –  David Mitra Feb 10 '12 at 13:30