Let me change your notations. You deal with a 1D quasilinear system with size $N=4$: the standard Cauchy problem is
$$
\frac{\partial u}{\partial t}+A(t,x,u)\frac{\partial u}{\partial x}= f(t,x),\quad u(t=0,x)=u_0(x),
$$
where $t\in \mathbb R$ (time variable) as well as $x$ (this is a 1D problem), $u$ is valued in $\mathbb R^N$, $A$ is a real-valued $N\times N$ matrix. You may also assume that $A$ depends smoothly (or even analytically) of its arguments.
(1) Your first case: all eigenvalues of $A_0:=A(0,x_0, u_0(x_0))$ are real and distinct, this is indeed the strictly hyperbolic case. In this case, you can guarantee local existence, uniqueness and continuous dependence on the data, i.e. local well-posedness. Of course you cannot expect global existence in general because of the nonlinearity (think about the scalar Burgers).
(2) Let me skip some of your cases and go directly to the case where $A_0$ has a non-real eigenvalue (and thus a pair of non-real eigenvalues). Big trouble ahead: even if the matrix $A$ is analytic, in which case, Cauchy-Kovalevskaya theorem is providing a local (unique) analytic solution, that solution is very unstable in the Hadamard sense. It means that even though $v_0-u_0$ is very small in a very strong topology, such as the $C^\infty$ topology or the $H^s$ topology for a very large $s$, you will not be able to control $u(t)-v(t)$ in a quite weak topology such as $L^2$ (all this is local of course). You will find precise statements in a paper by Métivier, Remarks on the well-posedness of the nonlinear Cauchy
problem, with MR number MR2127041.
(3) When all eigenvalues are real, some with multiplicity larger than 2, then instability could or could not occur, depending on other structural factors such as semi-simplicity of $A_0$. Generally speaking, multiple roots will trigger difficulties.
(4) A very important class of theorems, with the name of Lax-Mizohata theorems is establishing a weak converse to (1): if the problem is well-posed (e.g. meaning that you have some Sobolev norm control of $u(t)$ by some Sobolev norm of $u(0)$), then it implies that the system is
weakly hyperbolic (case (3)$\cup$(1)). So if you expect your system to be well-behaved in the sense of Hadamard, no choice, the roots must be real-valued, possibly with multiplicity.
Do not think that all physically relevant problem of that type are hyperbolic: to quote just one example, Van der Waals classical system is
$$
\partial_t u+\partial_x v=0\quad \partial_t v+\partial_x q(u)=0.
$$
When $q'(u)>0$, you are in a hyperbolic region, but when $q'(u)<0$, you have a non-real eigenvalue.
The revised question, where you just want uniformly bounded Holder norm on the approximating polynomials, the answer is yes.
In fact, you have more.
Theorem 1 If $f\in C^{m,\alpha}_M([0,1])$ is such that the function $\tilde{f}(x) = \begin{cases} f(x) & x\in [0,1]\\ 0 & x\not\in [0,1]\end{cases}$ is in $C^{m,\alpha}(\mathbb{R})$, then there exists a polynomial approximation sequence in $C^{m,\alpha}_M([0,1])$.
This theorem is based on a proof of the Stone-Weierstrass Theorem using a polynomial approximation to identity. See e.g. Theorem 11.7.1 of Lebl's textbook (volume 2).
A nice property of the convolution argument is that:
If $\phi$ is a non-negative function with total integral 1, then $\|\phi*f\|_{\infty} \leq \|f\|_\infty$.
Since $(\phi*g)^{(j)} = \phi* (g^{(j)})$, for any function $g$, applying this to $\tilde{f}$ you see that the convolution by the polynomial approximation to identity means that the polynomials constructed have $C^m$ norms bounded by that of $\tilde{f}$.
Similarly, letting $g = \tilde{f}^{(m)}$, observe for each fixed $t\neq 0$, $x\mapsto \frac{1}{|t|^\alpha}({g}(x+t) - {g}(x))$ is a $C^{0,\alpha}$ function, we have that
$$ \frac{1}{|t|^\alpha} \| \phi*{g}(x+t) - \phi*{g}(x)\|_{L^\infty_x} \leq \frac{1}{|t|^\alpha} \|{g}(x+t) - {g}(x)\|_{L^\infty_x} $$
Taking the sup of both sides over $t\neq 0$ this shows that the Holder semi-norm of $\phi*\tilde{f}$ is bounded by that of $\tilde{f}$. And this proves the Theorem.
Now, given $f\in C^{m,\alpha}_M([0,1])$, there exists a polynomial $q$ such that $f-q$ satisfies the hypotheses of Theorem 1. The polynomial $q$ is determined entirely by $f(0), f'(0), \ldots, f^{(m)}(0), f(1), f'(1), \ldots, f^{(m)}(1)$.) You can select $q$ to be of degree $2m+1$, in which case the $q$ is uniquely determined by those values. The relation between those values and the coefficients of $q$ are given by a linear operator, depending only on $m$. This means that there exists $\tilde{M}$ that uniformly bounds the coefficients of $q$, where $\tilde{M}$ depends only on $M$ and on $m$ (since by assumption $f(0), \ldots f^{(m)}(1)$ are all bounded by $M$). (Note that in the context of Theorem 1, this unique $q$ is the 0 polynomial, and hence $\tilde{M} = 0$.) The mapping from $\mathbb{R}^{2m+2}\times \mathbb{R} \to \mathbb{R}$ where the first factor represents the coefficients of $q$ and the last factor represents $x$ mapping $(q,x)\mapsto q(x)$ is continuous (in fact a polynomial), and hence achieves a maximum on compact regions. Therefore knowing that the coefficients are bounded by $\tilde{M}$ and $x\in [0,1]$ means that there exists $\hat{M}$ depending only on $M$ and $m$ that bounds the $C^{m+1}$ and hence $C^{m,\alpha}$ norm of $q$. Finally set $M' = M + 2 \hat{M}$.
So you conclude:
Theorem 2 Given $m, \alpha,M$, there exists $M' = M'(m,M,\alpha)$ such that whenever $f\in C^{m,\alpha}_M([0,1])$, you can approximate $f$ by polynomials $p_j$ in the uniform norm, with $p_j$ selected from $C^{m,\alpha}_{M'}([0,1])$.
Best Answer
No, take any $p$ such that $p(x)=|x|^\alpha$ in some neighbourhood of the origin.