$T$ is bijective and homeomorphism. Suppose $X$ be the set of all polynomial with real coefficients in one variable with norm $$\|p(x)\|=|a_0|+|a_1|+\dots+|a_n|$$ where $p(x)=a_0+a_1x+\dots+a_nx^n$ which induces a metric $$d(p,q)=\|p-q\|$$
could anyone help me to find out whether the following statements are true/false
1.$X$ is a complete metric space.


*$T:X\to X$ defined by $T(p(x))=a_0+a_1x+{a_2x^2\over 2}+\dots+{a_x^n\over n}$ is continuous

*$T$ is bijective and homeomorphism.
Thanks for helping.
 A: *

*$X$ is not a complete metric space.  Observe the sequence


$$
p_n(x) \;\; =\;\; \sum_{k=0}^n \frac{x^k}{k!}.
$$
Each term is a finite sum of polynomials but in their limit $p_n(x) \to e^x$.  This remains true independent of the norm put on $X$.


*Yes, $T$ is continuous. Let $\epsilon > 0$ and let $p(x) = a_0 + a_1x + \ldots + a_nx^n$ and $q(x) = b_0 + b_1x + \ldots + b_mx^m$ where 


$$
||p - q || \;\; =\;\; |a_0 - b_0| + |a_1 - b_1| + \ldots + |a_n - b_n| + |b_{n+1}| + \ldots + |b_m| \;\; < \;\; \epsilon.
$$
It then follows that 
$$
||T(p-q)|| \;\; =\;\; |a_0 - b_0| + |a_1 - b_1| + \frac{|a_2 - b_2|}{2} + \ldots + \frac{|a_n - b_n|}{n} + \ldots + \frac{|b_m|}{m} \;\; < \;\; \epsilon.
$$


*$T$ is certainly bijective (check this), and it is also a homeomorphism.  The inverse map $T^{-1}$ is given by 


$$
T^{-1}(p) \;\; =\;\; a_0 + a_1x + 2a_2x^2 + \ldots + na_nx^n.
$$
Let $\epsilon > 0$, and let $m$ be the max degree between the polynomials $p$ and $q$ as given above, and adjust their coefficients so that $||p-q|| < \frac{\epsilon}{m}$.  We then have 
\begin{eqnarray*}
||T^{-1}(p-q)|| & = & |a_0-b_0| + |a_1 - b_1| + 2|a_2 - b_2| + \ldots + n|a_n - b_n| + \ldots + m|b_m| \\
& < & m (|a_0 - b_0| + |a_1 - b_1| + |a_2 - b_2| + \ldots + |a_n - b_n| + \ldots + |b_m| ) \\
& < & \epsilon. 
\end{eqnarray*}
