Is W Closed in the Space C[-Pi, Pi]?

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In summary, W is a subspace of C[-Pi,Pi] consisting of all finite linear combinations of 1, cos(nx), and sin(nx). It is closed in C[-Pi,Pi] as shown by the convergence of the series and the lemma. The argument for the closure of W may not be fully understood, but it is a reasonable conclusion. Additionally, the sequence of functions can be taken as the Fourier approximations to the heaviside function.
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dikmikkel
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Homework Statement


W is a subset of C[-Pi,Pi] consisting of all finite linear combinations:
1,cos(nx),sin(nx)
i) Show that W is a subspace of C[-Pi,Pi]
ii) Is W closed in C[-Pi,Pi]. Hint from Fourier analysis: For x in [-Pi,Pi]:
[itex]|x^2-(\dfrac{\pi^2}{3}+4\sum\limits_{n=1}^N\dfrac{(-1)^{n}\cos(nx)}{n^2})|\leq 4\sum\limits_{n=N+1}^{\infty} \dfrac{1}{n^2}[/itex]

Homework Equations


C[-Pi,Pi] could be equipped with a norm
Lemma: W is closed <=> For any convergent sequence [itex]\{ v_k \}_{k=1}^\infty[/itex] of elements in W the V = limit (vk) for k->infinity also belongs to W.

The Attempt at a Solution


I have shown that W is a subspace by realising that linearcombinations of sines cosines and 1's are also in C[-Pi,Pi].
ii)
I concluded as sum(1/n^2) is convergent and the left hand side is less than or equal(also convergent) to for x in [-Pi,Pi], then V is closed in C[-Pi,Pi]
 
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1) looks reasonable

2) I'm not too sure I understand your argument or the hint...

As a thought exercise how about taking your sequence of functions as the Fourier approximations to the heaviside function
 

FAQ: Is W Closed in the Space C[-Pi, Pi]?

What is a closed subset of a vectorspace?

A closed subset of a vectorspace is a set of vectors that is closed under vector addition and scalar multiplication. This means that if you take any two vectors from the subset and add them together, the resulting vector will also be in the subset. Similarly, if you multiply any vector in the subset by a scalar, the resulting vector will still be in the subset.

How is a closed subset different from an open subset?

A closed subset is different from an open subset in that it includes all of its boundary points, while an open subset does not. This means that a closed subset is considered "complete" and has no holes or gaps, while an open subset may have missing points or gaps between its elements.

What is the significance of closed subsets in linear algebra?

Closed subsets are important in linear algebra because they help define the structure and properties of a vectorspace. They allow us to perform operations on subsets of vectors and still maintain the same properties as the original vectorspace, such as closure under vector addition and scalar multiplication.

How can you identify a closed subset of a vectorspace?

To identify a closed subset of a vectorspace, you can check if the set satisfies the closure properties. This means that if you add any two vectors from the subset together, the resulting vector should also be in the subset. Similarly, if you multiply any vector in the subset by a scalar, the resulting vector should still be in the subset.

Can a closed subset of a vectorspace be infinite?

Yes, a closed subset of a vectorspace can be infinite. As long as the subset satisfies the closure properties, it can contain an infinite number of vectors. This is because the closure properties ensure that any operation performed on the subset will still result in a vector within the subset.

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