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I have a sequence of functions $f_n:[0,1] \rightarrow \mathbb{R}^n$ such that $f_n$ is uniformly bounded, i.e. $\|f_n\|\leq M$ with $M$ independent of $n$.

Is it true that that the sequence $(f_n)$ converges pointwise to some function $f$, $\lim f_n(x)=f(x)$ for all $x \in [0,1]$ ?

cmk
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Lucas27
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2 Answers2

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It is not true. For example, let the range just be $\mathbb{R}$, and take $f_n(x)=(-1)^n.$ This is a uniformly bounded sequence, but it does not converge pointwise. You can extend this to $\mathbb{R}^m$ in the obvious way.

cmk
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  • True, does this also work if we extract a subsequence of functions? – Lucas27 Aug 24 '19 at 18:39
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    I believe that $f_n(x)=\sin nx$ will give you a negative answer. – cmk Aug 24 '19 at 18:44
  • @cmk I belive as well that the example with the sin is a counterexample, but I have a hard time proving it. Do you have an idea? – Severin Schraven Aug 24 '19 at 19:18
  • @SeverinSchraven see https://math.stackexchange.com/questions/545781/looking-for-proof-that-this-uniformly-bounded-sequence-of-functions-has-no-point, https://math.stackexchange.com/questions/2633927/how-to-prove-sinnx-has-no-pointwise-convergent-subsequence-without-prior-knowl/2657440, https://math.stackexchange.com/questions/1380286/pointwise-almost-everywhere-convergent-subsequence-of-sin-nx?noredirect=1&lq=1 – cmk Aug 25 '19 at 17:53
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    Thanks! The first one is such a beautiful proof. – Severin Schraven Aug 25 '19 at 19:49
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If you want that no subsequence converges, you might define $f_n(x)=a_n$ where $x=a_0.a_1 a_2 a_3\dots$ is the decimal expansion (with the usual restrictions that renders the expansion unique). Pick your favourite subsequence $(f_{n_k})_{k\geq 1}$. Then we consider $x=a_0.a_1\dots$ such that $a_{n_k}=1+(-1)^k$ and $a_j=0$ otherwise. then we have $f_{n_k}(x)=1+(-1)^k$ which does not converge.

Severin Schraven
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