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let $f_n$ be a sequence of functions in $C[0,1]$ and they are differentiable continuously in $(0,1)$. Also $|f_n(x)|\leq 1$ and $|f_n{'}(x)|\leq 1$ forall $x\in [0,1]$ and for each $n$.

Since $[0,1]$ is compact so $C[0,1]$ is compact and so every sequence has a convergent subsequence so $f_n$ will also have so.But is $f_n$ convergent?

Are my arguements correct?Also any hints to do my problem.I am a bit weak in analysis

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  • $\begingroup$ Convergent in what sense? Pointwise? Uniformly? In integral norm? $\endgroup$ – Gyu Eun Lee Nov 22 '14 at 11:07
  • $\begingroup$ under supremum norm $\endgroup$ – Learnmore Nov 22 '14 at 11:09
  • $\begingroup$ $C[0,1]$ is not compact. $\endgroup$ – user99914 Nov 22 '14 at 11:12
  • $\begingroup$ Please give the correct arguements wherever wrong $\endgroup$ – Learnmore Nov 22 '14 at 11:19
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For one thing, $C[0,1]$ is far from compact. If it were compact in supremum norm it would necessarily be bounded, which is manifestly false.

Since you have that your functions are continuously differentiable, your best strategy is to use the mean value theorem. We need to try and control $$ |f_n(x) - f_m(x)| $$ uniformly in $x$. Using the triangle inequality, $$ |f_n(x) - f_m(x)| \leq |f_n(x) - f_n(y)| + |f_n(y) - f_m(y)| + |f_m(y) - f_m(x)|. $$ You can control the first and third terms using the mean value theorem, and the middle term after that by pointwise convergence.

This is already too big of a hint. I leave the rest to you.

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  • $\begingroup$ But does that conclude $f_n$ is convergent or $f_n $ has a convergent subsequence? $\endgroup$ – Learnmore Nov 22 '14 at 11:30
  • $\begingroup$ I've already said this is too much of a hint. I don't claim anything about the convergence of $f$, careful analysis of this inequality should tell you what is or is not true and hint at counterexamples if there is no convergence. You're the one who needs to take over here. $\endgroup$ – Gyu Eun Lee Nov 22 '14 at 11:32
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$f_n$ will have a subsequence uniformly convergent, but may not converge. Example: $f_n(x)=(-1)^n$.

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  • $\begingroup$ Thank you sir for your useful answer $\endgroup$ – Learnmore Nov 23 '14 at 3:22

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