Proving C^k[a,b] is Not Closed in C^0[a,b]

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Homework Statement



Is C^{k}[a,b] closed in C^{0}[a,b]?

The Attempt at a Solution



C^{k}[a,b] is obviously a subset of C^{0}[a,b].

My gut feeling says no. I thought the best way would be to construct a function f_{n}(x) which converges to f(x) and where f_{n}(x) is in C^{k}[a,b] but f(x) is not.

I thought maybe f_{n}(x)=x^{k+1}sin(\frac{1}{nx}) would do it since it's not k+1 differentiable at 0. But then f(x)=0 which can be differentiated infinitely (since each derivative is 0).
 
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You are definitely on right track. Try writing your sequence as a series
f_n = \sum_{i=0}^n a_n \sin(b_n x) and then choose an and bn so that limit of fn exists but f'n diverges.
... Or just google "Weierstrass function", if you're lazy :)
 
If you're asking such questions, then you should always say which metric you're working with.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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