The Stone-Weierstrass theorem states that for any continuous function $f$ on $[a,b]$ there exists a polynomial function $p$ such that $\|f-p\|<\varepsilon$ for any $\varepsilon>0$. Where $\|\cdot\|$ is the sup norm.
If I replace "a polynomial function" with "a polynomial function with rational coefficient" from the above statement, will the statement still hold true?
For sure it is true. For two polynomials $P_n,P_n^\prime$of the same degree $n$ with coefficients $p_0, \dots p_n$ and $p_0^\prime, \dots , p_n^\prime$ respectively, you have $$\Vert P_n -P_n^\prime\Vert_\infty =\sup\limits_{t \in [a,b]} \vert P_n(t)-P_n^\prime(t)\vert \le c \left(\sup\limits_{1\le i\le n}\vert p_i -p_i^\prime\vert\right)$$
with $c=\sup\limits_{t \in [a,b]} \sum_{k=0}^n \vert t \vert^k$.
Hence if you have a real polynomial $P_n$ at a distance less than $\epsilon/2$ of $f$, you can find a polynomial of same degree with rational coefficients $P_n^\prime$ at a distance less then $\epsilon/2$ of $P_n$. And you can conclude with triangular inequality.
Yes.
Observe:
if
$p(x) \in \Bbb R[x], \tag 1$
then given any $\epsilon$ there exists a polynomial
$q(x) \in \Bbb Q[x] \tag 2$
such that
$\vert p(x) - q(x) \vert < \epsilon, \; x \in [a, b]; \tag 3$
for, writing
$p(x) = \displaystyle \sum_0^n p_i x^i, \; p_i \in \Bbb R, \tag 4$
$q(x) = \displaystyle \sum_0^n q_i x^i, \; q_i \in \Bbb Q, \tag 5$
then, assuming $a \ne b$,
$\vert p(x) - q(x) \vert = \vert \displaystyle \sum_0^n p_i x^i - \sum_0^n q_i x^i \vert \le \sum_0^n \vert p_i - q_i \vert \vert x \vert^i \le \sum_0^n \vert p_i - q_i \vert (\max(\vert a \vert, \vert b \vert))^i; \tag 6$
given $p_i$, $0 \le i \le n$, we may choose $q_i$ such that
$\vert p_i - q_i \vert < \dfrac{\epsilon}{(n + 1)(\max_{0 \le i \le n}(\max(\vert a \vert, \vert b \vert))^i)}; \tag 7$
then
$\vert p(x) - q(x) \vert \le \displaystyle \sum_0^n \vert p_i - q_i \vert (\max(\vert a \vert, \vert b \vert))^i$ $< \displaystyle \sum_0^n \dfrac{\epsilon(\max(\vert a \vert, \vert b \vert))^i)}{(n + 1)(\max_{0 \le i \le n}(\max(\vert a \vert, \vert b \vert))^i)}$ $\le \dfrac{\epsilon(\max(\vert a \vert, \vert b \vert))^i)}{(\max_{0 \le i \le n}(\max(\vert a \vert, \vert b \vert))^i)} \le \epsilon. \tag 8$
It follows that for every $\epsilon > 0$ and any $p(x)$ as in (1), there exists $q(x)$ as in (2) with $\vert p(x) - q(x) \vert < \epsilon$ on $[a, b]$. Now replacing $\epsilon$ with $\epsilon / 2$ and choosing $p(x) \in \Bbb R[x]$ such that
$\vert f(x) - p(x) \vert < \dfrac{\epsilon}{2}, \; x \in [a, b], \tag 9$
we may find some $q(x) \in \Bbb Q[x]$ such that
$\vert p(x) - q(x) \vert < \dfrac{\epsilon}{2}, \tag{10}$
and combining (9) and (10) we have
$\vert f(x) - q(x) \vert = \vert f(x) - p(x) + p(x) - q(x) \vert$ $\le \vert f(x) - p(x) \vert + \vert p(x) - q(x) \vert < 2 \dfrac{\epsilon}{2} = \epsilon, \; x \in [a, b], \tag{11}$
or
$\Vert f(x) - q(x) \Vert < \epsilon. \tag{12}$
Yes this is true. We know that we can approximate $f$ arbitrarily closely by a polynomial $p = \sum_k a_k x^k$ with real coefficients. Let $q = \sum_k c_kx^k$ be a polynomial with rational coefficients. Then for any $x \in [a,b]$ $$ |p(x)-q(x)| \leq \sum_{k=1}^n |a_k-c_k||x^k| \leq \sum_{k=1}^n |a_k-c_k| b^k $$ By density of rationals, we can make $|a_k-c_k| < \frac{\epsilon}{nb^k}$.
Yes. We know there is some $p$ of degree $n$ with $||f-p|| < \epsilon$. You can show that the map $f: \mathbb{R}^{n+1} \rightarrow \mathbb{R}$ that sends $(a_0, \ldots, a_n)$ to $||f - p||$ where $p(x) = \sum_{i=0}^n a_i x^i$ is continuous, so $f^{-1}( (0, \epsilon))$ is a nonempty open set; since rationals are dense in $\mathbb{R}^{n+1}$, there must be a rational point in $f^{-1}((0, \epsilon))$.
All holomorphic (or complex differentiable) functions (which include continuous functions) have a laurent series, which is a sum of the form: $$f(z)=\sum_{k=-\infty}^\infty a_k(z-z_0)^k$$Which is an infinite sum of polynomials. You can find a proof of this here. Since we have a sum of polynomials, then we could look at a few terms to get a rough approximation of the function.
