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

[member 587159]

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?

 

[fresh_42]

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:

$|f_n(x)g_n(x) - f(x)g(x)| \leq |f_n(x)||g_n(x)-g(x)| + |f_n(x) -f(x)||g(x)| \leq \Vert f_n \Vert \epsilon + \Vert g \Vert < M \epsilon$ 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?

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 $\overline{A}$.

 

[member 587159]

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 $\overline{A}$.

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 :)

 

[fresh_42]

I actually had a question about this proof, but while typing it out I saw it.

That's one of the best tricks: If you want to understand something, explain it to others!

 

[member 587159]

That's one of the best tricks: If you want to understand something, explain it to others!

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