Prove the function $Q: \Bbb R^3\to C[0,1],\;\langle a,b,c\rangle \mapsto ax^2+bx+c$ is continuous The following map $Q : \Bbb R^3 \to C[0, 1]$: if $(a, b, c)$ is any point in $\Bbb R^3$, then $Q(a, b,c) = f$ is the quadratic polynomial in $C[0, 1]$ given by the formula：  
$$Q(a, b, c)(x) = f(x) = ax^2 + bx + c$$ 
for all $x ∈ [0, 1]$. 
Assume that $\Bbb R^3$ has the Euclidean metric and $C[0,1]$ has the metric
$$d(f,g) = \sup\{|f(x)-g(x)|: x \in [0,1]\}$$
Prove that $Q$ is continuous. 
 A: Let $(a_0,b_0,c_0) \in \mathbb{R}^3$ and $\epsilon > 0$. Choose $(a,b,c) \in \mathbb{R}^3$ such that $\vert (a,b,c) - (a_0,b_0,c_0) \vert < \epsilon/3$ and note that
\begin{equation*}
\begin{aligned}
&\mathrel{\phantom{=}} \sup_{x \in [0,1]}\vert (a - a_0)x^2 + (b - b_0)x + c - c_0\vert \\
&\leq \sup_{x \in [0,1]}\vert a - a_0\vert x^2 + \sup_{x \in [0,1]}\vert b - b_0 \vert x + \sup_{x \in [0,1]} \vert c - c_0 \vert \\
&= \vert a - a_0 \vert + \vert b - b_0 \vert + \vert c - c_0 \vert.
\end{aligned}
\end{equation*}
A: Take $(a_n,b_n,c_n)\rightarrow (a,b,c)$ which is equivalent to $a_n\rightarrow a$, $b_n\rightarrow b$, and $c_n\rightarrow c$
Then
$d(Q(a_n,b_n,c_n),Q(a,b,c))=\sup\{|(a_n-a)x^2+(b_n-b)x+(c_n-c)|\}$
$\le\sup\{|a_n-a|\cdot|x^2|+|b_n-b|\cdot|x|+|c_n-c)|\}$
$\le\sup\{|a_n-a|+|b_n-b|+|c_n-c)|\}\rightarrow 0$
as $|x|\le 1$ and $|x^2|\le 1$
So $Q$ is continuous.
A: If $x\in [0,1]$ we have $\vert (Q(a, b, c)-Q(a_0,b_0,c_0))(x)\vert$  = 
$\vert (a-a_0)x^2 + (b-b_0)x+(c-c_0)\vert\le \vert a-a_0\vert +\vert b-b_0\vert+\vert c-c_0\vert$. 
Since this is true for all $x\in [0,1]$ it follows that 
$\sup_{x\in [0,1]}\vert Q(a,b,c)(x)-Q(a_0,b_0,c_0)(x)\vert \le \vert a-a_0\vert +\vert b-b_0\vert+\vert c-c_0\vert$. 
Now let $a\to a_0;\ b\to b_0$ and $c\to c_0$
