How to find a sequence of differentiable functions${f_n}$ with limit $0$ for which \{f'_n\} diverges?
Divergent sequence of derivatives
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real-analysis
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1Think of rapidly oscillating sinusoidal functions that have small amplitudes. – 2012-03-10
2 Answers
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How about $ f_n (x) := \frac{\sin (n^2 x) }{ n }$
Then the pointwise limit is \lim_{n \to \infty} \left | f_n(x_0) \right | = \lim_{n \to \infty} \frac{\left | \sin (n^2 ) \right |}{\left | n \right |} \leq \lim_{n \to \infty} \frac{1}{|n |} = 0
But $ f_n^\prime (x) = \frac{n^2 \cos (n^2 x)}{ n } = n \cos (n^2 x)$
Whose pointwise limit diverges at the points $x_0 = 2 \pi k$ since
\lim_{n \to \infty} n \cos (n^2 x_0) = \lim_{n \to \infty} n
Hope this helps. I thought I'd post this anyway even though I'm using the same function as AD in their answer. But I was typing this when their answer appeared and I see no reason to delete mine.
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0@DidierPiau Yes. Thanks Didier. – 2012-03-10
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Put $f_n(x)=\frac{\sin(nx)}{\sqrt{n}}$
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0@AD.: You are welcome :-) – 2012-03-10