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What is the norm of a function?

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    I know that a norm of a vector, v in a vector space is ||v||= \sqrt{(<v,v>)} , i.e the square root of the dot product of v with itself.

    But what is the norm of a function in a function space C[a,b]? Am I right in assuming that if f is in C[a,b] then:

    ||f||= \sqrt{(<f,f>)} =  \sqrt{ \int_a^b f(x)f(x) dx} ?

    Thanks for any replies!
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    (Original post by Lewk)
    I know that a norm of a vector, v in a vector space is ||v||= \sqrt{(<v,v>)} , i.e the square root of the dot product of v with itself.

    But what is the norm of a function in a function space C[a,b]? Am I right in assuming that if f is in C[a,b] then:

    ||f||= \sqrt{(<f,f>)} =  \sqrt{ \int_a^b f(x)f(x) dx} ?

    Thanks for any replies!
    Without further details, I'd go with that. However, sometimes it'll involve weight functions or conjugates, but that would be stated in the question (or lecture notes ).
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    A norm is a function f of an element of a vector space with a number of properties:
    1)f(v) = 0 if and only if v = 0 and f(v) > 0,
    2)f(kv) = |k|f(v),
    3)The triangle inequality.

    Any function on a vector space of functions with these three properties is a norm. Possibly the most common is the maximum norm, which simply returns the largest value the function achieves.

    You have assumed that your vector space is equipped with an inner product (your <> function) which makes it a Euclidean space. A vector space does not necessarily have such a product, and equally the product you have suggested is not the only inner product that there can be. The one you have suggested induces the norm you have given which does fit into the classification of 'norm', but is not commonly used.
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    This is similar to your last question. A priority, a vector space doesn't have a norm - a norm is an extra structure and as before, a vector space may be endowed with many different norms that give your vector space different structures as a normed vector space.

    If you are dealing with an inner product space (i.e. a vector space together with a fixed choice of inner product) then the canonical choice of norm is the one induced by the inner product. That is \mid v \mid := \sqrt{\langle v, v \rangle} where the angled brackets denote the inner product.

    Therefore, in your situation the norm will depend on the context. Since you were considering the continuous functions on an interval as an inner product space, the norm will presumably be the one as defined above. Note again that this is completely dependent on the choice of inner product that you are using.
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    (Original post by Bobifier)
    You have assumed that your vector space is equipped with an inner product (your <> function) which makes it a Euclidean space. A vector space does not necessarily have such a product, and equally the product you have suggested is not the only inner product that there can be. The one you have suggested induces the norm you have given which does fit into the classification of 'norm', but is not commonly used.
    The norm that he was using is also a very common norm.

    Also note that not every norm is induced by an inner product.
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    This is a very good question, and leads to some interesting mathematical theory. In short, the answer is that there is no single way of defining a norm.

    The norm you've described is indeed a norm, and one that is commonly used. I'll give two good reasons why it is of interest. First, in physical situations, this norm might describe the energy of whatever the function represents (if you've done Fourier transforms, you might've seen something like this). The sum or integral of squares of something crops up a lot in applied mathematics. Secondly, it's directly analogous to the Euclidean norm usually used for the finite-dimensional vector spaces you're used to thinking of - instead of summing over the squares of the components of the vector, we integrate over the squares of values of the function.

    However, we could use a completely different norm - the uniform norm. This is defined as the supremum (the "smallest upper bound") of the absolute value of the function. For example, the uniform norm of the function \sin x on the interval [0, 2\pi] is 1, because that's the greatest value | \sin x | attains. This is important in analysis, where a sequence of functions which converges when we use this norm is called "uniformly convergent", which leads to things having nice properties.

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Updated: May 18, 2012
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