How to show that space of all infinitely differentiable functions is separable? My idea is to construct approximation polynomial with rational coefficients. How to construct by using Weierstrass Approximation Theorem? Thanks!
[Math] Show that space of all infinitely differentiable functions is separable
real-analysis
Related Solutions
This follows from the density of $\mathbb{Q}$ in $\mathbb{R}$. Let $p(x) = \sum_{k=0}^n a_k x^k$ be any polynomial with real coefficients and fix an $\epsilon > 0$. Note that for any $q(x) = \sum_{k=0}^n q_k x^k$ and any $x \in [a,b]$ $$ |p(x) - q(x)| \leq \sum_{k=0}^n |a_k - q_k| |x|^k \leq \sum_{k=0}^n |a_k - q_k| M^k $$ where $M = \max(|a|,|b|)$. So you can show that $\|p-q\|_\infty < \epsilon$ by, for each $k \in\{0,\ldots,n\}$, choosing a $q_k \in \mathbb{Q}$ such that $$ |a_k - q_k| < \frac{\epsilon}{(n+1)M^k}. $$
The way it's stated in the passage you've selected, no: the Weierstrass Approximation Theorem doesn't say anything about the coefficients of the approximating polynomials. However, we've just said quite a bit! In fact, it's not difficult to combine Weierstrass's theorem with 1. to conclude that $\mathbb{Q}[x]$ -- the set of polynomials with rational coefficients -- is dense in $C[a,b]$. Do you see how?
In fact, this is exactly what the author is saying when he points out that $C[a,b]$ is separable! I'm not sure how generally separability is defined in Carother, but a metric space $M$ is separable if it contains a subset $S$ such that
- $S$ is countable
- $S$ is dense in $M$ (i.e., every element of $M$ is the limit of a sequence $\{s_n\}$ of elements $s_n$ in $S$) under the metric within $M$
(Typically such a subset $S$ is called a countably dense subspace of $M$, naturally!) Here $M = C[a,b]$ with norm $\|\cdot\|_\infty$ (or metric $d(f_1,f_2) = \|f_1-f_2\|_\infty$), and $S = \mathbb{Q}[x]$. If you can show that $\mathbb{Q}[x]$ is uniformly dense in $C[a,b]$ as I recommended above, then you've also shown that $C[a,b]$ is separable. (Side note: How do you know that $\mathbb{Q}[x]$ is countable? If you don't immediately see an argument, now would be a good time to prove that this is true.)
Edit: Here's how to conclude explicitly that $\mathbb{Q}[x]$ is dense in $C[a,b]$ given what we now know.
Let $f \in C[a,b]$. Assume $\{p_k\}$ is a sequence of polynomials converging uniformly to $f$ on the interval $[a,b]$. Accordingly, if we fix an $\epsilon > 0$, we know that there exists $K_\epsilon \in \mathbb{N}$ such that for all $k \geq K_\epsilon$ $$ \|f - p_k\|_\infty < \epsilon/2. $$ At the same time, by 1. we know that for each $k$ there exists a sequence $\{q_{k,i}\}$ of polynomials with coefficients in $\mathbb{Q}$ such that $q_{k,i} \to p_k$ uniformly on $[a,b]$ as $i \to \infty$. Hence for each $k$, there exists $I_{k,\epsilon} \in \mathbb{N}$ such that and for all $i \geq I_{k,\epsilon}$ $$ \|p_k - q_{k,i}\|_\infty < \epsilon/2. $$ Set $k = K_{\epsilon}$, set $i = I_{k,\epsilon}$, and then conclude that there exists a polynomial $q := q_{k,i}$ with rational coefficients such that $$ \|f - q\|_\infty = \|(f-p_k)+(p_k-q)\|_\infty \leq \|f-p_k\|_\infty + \|p_k - q\|_\infty < \epsilon/2 + \epsilon/2 = \epsilon. $$
(Expanding on a comment by John Ma.)
The family is uniformly bounded: $|f_n|\le M_0$. It is also equicontinuous, thanks to $|f'_n|\le M_1$: $$ |f_n(x)-f_n(y)|\le M_1|x-y| $$ Therefore, there is a subsequence $\{f_{n_k}\}$ converging to a continuous function $f$.
Next step, consider $\{f_{n_k}'\}$ which is also a uniformly bounded equicontinuous family. Extract a further subsequence $f'_{n_{k_l}}$ which converges uniformly. Since $f_{n_{k_l}}\to f$ and $f'_{n_{k_l}}$ converge uniformly, it follows that $f$ is differentiable and $f'_{n_{k_l}}\to f'$.
This process of choosing subsequences continues indefinitely, demonstrating that $f$ is infinitely differentiable. It remains to apply the diagonal argument: let $F_n$ be the $n$th term of the $n$th subsequence. Then $F_n$ converge, with all its derivatives, to the corresponding derivatives of $f$.
Best Answer
The proof is outlined in Wikipedia, Stone-Weierstraß-Theorem for the $C([a,b],\mathbb{R})$ case. As every $C^\infty $ function is continuous this is a subspace, and as the polynomial with rational coefficients are $C^\infty$ you still can take the same countable set which is dense.