Let me give some context.(adsbygoogle = window.adsbygoogle || []).push({});

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?

**Physics Forums | Science Articles, Homework Help, Discussion**

Join Physics Forums Today!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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

Tags:

Have something to add?

Draft saved
Draft deleted

**Physics Forums | Science Articles, Homework Help, Discussion**