Show that $T$ is not a homeomorphism

Question

Let $X$ be the space of all polynomials in one variable over $\mathbb R$ .If $p=a_0+a_1x+\cdot\cdot\cdot\cdot +a_nx^n$

Define :$||p||=|a_0|+|a_1|+\cdot\cdot\cdot\cdot+|a_n|$.

Prove that $T:X\to X$ by $Tp=a_0+a_1x+\dfrac{a_2}{2}x^2+\cdot\cdot\cdot+\dfrac{a_n}{n}x^n$ is continuous but not a homeomorphism.

Solution:

let $p_n$ be a sequence in $X$ such that $p_n\to p$. To show $Tp_n\to Tp$.

let $p_n=a_o^{(n)}+a_1^{(n)}x+\cdot\cdot\cdot+a_n^{(n)}x^n$ .Then $Tp_n=a_o^{(n)}+a_1^{(n)}x+\cdot\cdot\cdot+\dfrac{a_n^{(n)}}{n}x^n$. Suppose $p_n\to p $ where $p=a_o+a_1x+\cdot\cdot\cdot+a_nx^n$.

Then $||p_n-p||\to 0 (n\to \infty)\implies |a_0^{(n)}-a_0|+|a_1^{(n)}-a_1|+\cdot\cdot\cdot\cdot+|a_n^{(n)}-a_n|\to 0$

Since sum of non-negative quantities tends to zero so $|a_i^{(n)}-a_i|\to 0\forall i$ and hence $Tp_n\to Tp$.

$T$ is bijective.But how to show it is not a homeomorphism.

Any help will be useful.I think it will be enough to show that $T$ is not an open map.

Answer 1

$T$ has inverse $S(p) = a_0 + a_1x + 2a_2x^2 + \dots +na_nx^n$. If $T$ were a homeomorphism, $S$ would be continuous. So we construct a sequence that $S$ behaves poorly on:

Consider $a_n = x^n/n$. Then $a_n \rightarrow 0$, since $\lVert a_n \rVert = 1/n \rightarrow 0$. What happens to $S(a_n)$?