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Let $\{f_k(x)\}_{k=1}^\infty$ be a sequence of differentiable functions defined on $[0,1]$. Suppose $f_k(0)=0$ and $|f_k'(x)|\le M$ for all $x\in[0,1]$ and for all $k\ge1$. Prove that after passing to a subsequence, $f_k$ converges uniformly on $[0,1]$ as $k\to\infty$.

In the above question, what is meant by "after passing to a subsequence"? My interpretation is that I am to show that there exists a subsequence which converges uniformly on $[0,1]$ as $k\to\infty$. Is this correct?

Note that I am not asking $why$ we pass to a subsequence. I am asking $what$ passing to a subsequence means.

zbrads2
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  • Possible duplicate of Why is "passing to a subsequence" necessary? – Sam Skywalker Jun 10 '19 at 23:21
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    @SamSkywalker The question you linked appears to ask why it is necessary to pass to a subsequence in some particular problem. I'm asking what the phrase itself means. – zbrads2 Jun 10 '19 at 23:27
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    Your interpretation is correct; often, people will merely state that they "pass to a suitable subsequence" and then rename their sequence as that subsequence. This is so that they can avoid double subscripts i.e. $f_{n_k}(x)$. It's a slight abuse of notation, but it's not uncommon. – Ben Jun 10 '19 at 23:28
  • the phrase means: Pick a suitable subsequence of your sequence (for example, a subsequence that converges to a limit), and perform the remaining analysis using this selected subsequence. – rubikscube09 Jun 10 '19 at 23:50

2 Answers2

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The term "passing to a subsequence" means considering a subsequence instead of the original sequence. However, to avoid making the notation more cumbersome, the subsequence keeps the same name as the original sequence.

For example, consider a sequence $\{a_n\}$, and the subsequence $a_{n_1}, a_{n_2},\ldots$. We could express this as $\{a_{n_i}\}$, but this would be awkward. Since the specific indices of the subsequence don't matter (in some cases), we simply rename the subsequence to have the name of the original sequence.

For the specific problem asked, you are asked to prove not that $\{f_k\}$ converges uniformly, but that there exists some $k_1, k_2,\ldots \subseteq \mathbb{N}$ such that $\{f_{k_i}\}$ converges uniformly.

vadim123
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Since my reputation is at this moment below 50, I cannot add a comment to reply to @zsbrads2, so I am answering in here. Mostly, I would like to point you all towards the comments in Why is "passing to a subsequence" necessary?, where the question is almost answered.

By passing to a subsequence we get rid of the elements that are least convenient. The pertinent subsequence will depend on the specific problem, as well as the techniques involved.

By the way, the statement you want to prove, as it is formulated, is false. Consider the sequence where $f_{2n} = \frac{M}{2}x^2$ and $f_{2n+1} = {M}x$. All of them have their first derivatives bounded by $M$ and are $0$ at $0$, so they satisfy the hypotheses. However, there is no uniform convergence, since the sequence keeps oscillating between these two functions ad infinitum.

Sam Skywalker
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    The statement to be proved is correct. Passing to a subsequence is precisely what's needed to avoid the oscillation problem that you noticed. In your example, the subsequence $(f_{2n})$ converges uniformly, as required. – Andreas Blass Jun 10 '19 at 23:53