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I If A is an algebra, then its uniform closure is an algebra.

  1. Dec 26, 2017 #1

    Math_QED

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    Let me give some context.

    Let X be a compact metric space and ##C(X)## be the set of all continuous functions ##X \to \mathbb{R}##, equipped with the uniform norm, i.e. the norm defined by ##\Vert f \Vert = \sup_{x \in X} |f(x)|##

    Note that this is well defined by compactness. Then, for a subset ##A \subset C(X)##, we define the uniform closure as the set ##\overline{A}## with respect to the uniform norm.

    Now, in a proof I'm going through, it is claimed that if ##A## is an algebra (subvectorspace + closed under pointwise multiplication), then ##\overline{A}## is an algebra too.

    I decided to prove this, and did the following:

    Let ##f,g \in \overline{A}##. There are sequences of functions such that ##f_n \to f, g_n \to g## for the uniform norm. But convergence for the uniform norm is the same thing as uniform convergence of sequences of functions.

    So, let ##\epsilon > 0##. Choose ##n_0## such that for all ##n \geq n_0, x \in X##, we have both

    ##|f_n(x) - f(x)| < \epsilon## and ##|g_n(x) - g(x)| < \epsilon##

    Then, for ##n \geq n_0, x \in X##, we have:

    ##|f_n(x)g_n(x) - f(x)g(x)| \leq |f_n(x)||g_n(x)-g(x)| + |f_n(x) -f(x)||g(x)| \leq \Vert f_n \Vert \epsilon + \Vert g \Vert\epsilon < M \epsilon## for some number M (because all the functions are bounded, as they are continuous.

    This shows that ##f_ng_n \to fg##, and hence ##fg\in \overline{A}## as ##f_ng_n \in A## because it is an algebra.

    Analoguous, we can prove that linear combinations remain in the set. Is this a correct proof?
     
    Last edited: Dec 27, 2017
  2. jcsd
  3. Dec 27, 2017 #2

    fresh_42

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    Except for a lost ##\varepsilon ##, yes. You basically prove ##\lim (\alpha f_n + \beta g_n) = \alpha \lim f_n + \beta \lim g_n## and ##\lim (f_n \cdot g_n) = \lim f_n \cdot \lim g_n## so the algebraic structure extends from ##A## to ##\overline{A}##.
     
    Last edited: Dec 27, 2017
  4. Dec 27, 2017 #3

    Math_QED

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    Except for a lost epsilon?

    Edit: Nevermind, found it (also edited it out). This was a typo.

    Thanks a lot! I actually had a question about this proof, but while typing it out the question resolved itself. As the question was already typed, I decided that I would post it. It is good to get feedback sometimes :)
     
  5. Dec 27, 2017 #4

    fresh_42

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    That's one of the best tricks: If you want to understand something, explain it to others!
     
  6. Dec 27, 2017 #5

    Math_QED

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    Indeed, or just formulating the question in a formal way! I have had many times that even this was enough to made me realise what I was missing.
     
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