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I think the problem is simple, but I am somehow unable to get it. Let $L_{k}^{1}([0,1])$ be the space of all $f\in C^{k-1}([0,1])$ such that $f^{(k-1)}$ is absolutely continuous on $[0,1]$ (and hence $f^{(k)}$ exists a.e. and is in $L^{1}([0,1])$. Then $||f|| = \sum_{0}^{k} \int_{0}^{1}|f^{(j)}(x)|dx$ is a norm on $L_{k}^{1}([0,1])$ that makes $L_{k}^{1}([0,1])$ into a Banach space. Note: $f^{(j)}$ is $j^{th}$ derivative of $f$.

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Reposting your question which had -4 votes in nearly the same fashion is foolish, again you show no effort to show us what you have tried. You simply prepended your original question with `this problem is easy, but I am somehow unable to get it'. Show us why you aren't able to get it, where you get stuck, instead of just coming here and asking for the solution. –  sxd Jan 25 '12 at 4:48
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Having set something very like this on an analysis qualifying exam recently, I think the author should show more evidence of what he or she has tried, which part is causing problems, and so on –  user16299 Jan 25 '12 at 4:52
    
Clarification: not that recently. My point is that this kind of question is designed to test a student's understanding of several things in combination, and so anyone asking for help with it should first demonstrate that they have broken it down into smaller problems –  user16299 Jan 25 '12 at 5:19
    
@Dimitri I guess the OP has deleted the previous question. Just for the record, can you post a link to it (in case you are able to locate it, of course)? –  Srivatsan Jan 25 '12 at 5:27
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@Srivatsan math.stackexchange.com/questions/102212/… –  sxd Jan 25 '12 at 5:39

1 Answer 1

Of course, the hard part is showing completeness. You may find it helpful to prove the following lemma.

Suppose $f_n$ is a sequence of absolutely continuous functions such that $f_n \to f$ and $f_n' \to g$ in $L^1$-norm. Then $f$ is absolutely continuous and $f' = g$.

Use this in conjunction with the fact that $L^1$ itself is complete. This should get you the $k=1$ case. Then you can get the general case by using induction.

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thanks...got it –  user24367 Jan 25 '12 at 6:49

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