Assume that $P(x)$ be the set of all polynomials with real coefficients. $P(x)$ is an infinite-dimensional real vector space. Consider $P(x)^*$ as dual of $P(x)$ and for $f^*\in P(x)^*$ and $g\in P(x)$ denote by $\langle g, f^*\rangle= f^*(g)$. By the $\mathbb{R}$-linear topology on $P(x)$ is meant the least topology of $P(x)$ with respect to which each element of $P(x)^*$ is continuous.
In my research, I need to have some information about neighborhood $U$ of $0\in P(x)$.
Indeed I study linear transformation $T:P(x)\to P(x)$ by $T(f(x))= \int_{0}^x f(t)dt$.
Assume that $f\in U$,
Is there $n\in \mathbb{N}$ such that $T^n(f)\notin U$? where $T^2(f)= T\circ T(f)$ , $T^3(f)=T\circ T\circ T(f)$, ...
We will find a neighborhood $U$ of $0$ such that for any nonzero $f\in P(x)$, $T^n(f)$ leaves $U$. Of course, if $f=0$, then $T^n(f)=T^n(0)=0\in U$ for all $n\in\mathbb{N}$.
For $f^*\in P(x)^*$ and $\varepsilon>0$, let $$U(f^*)=\{f\in P(x):|\langle f,f^*\rangle|<1\}.$$ Every neighborhood $U$ of $0$ in $P(x)$ has a subset of the form $$\bigcap_{i=1}^n U(f^*_i)$$ for some $f^*_1,\ldots, f^*_n\in P(x)^*$. So searching for $U$, it is sufficient to check sets of this form. A good starting point is to look for a set of the form $U(f^*)$ for a single $f^*$, and indeed this is what we will find.
For $n=0,1,2,\ldots$, let $b_n(x)=\frac{x^n}{n!}$. We note that this sequence forms a basis for $P(x)$. That means each $f\in P(x)$ can be uniquely represented as $f=\sum_{n=0}^\infty a_nb_n$, where only finitely many of the $a_n$ are non-zero. In order to know what a linear functional $f^*\in P(x)^*$ does on $P(x)$, it is necessary and sufficient to know the values $\langle b_n,f^*\rangle$ for $n=0,1,\ldots$. For convenience, for each $f^*\in P(x)^*$, let $f^*(n)=\langle b_n,f^*\rangle$. In order to define a $f^*\in P(x)^*$, we only need to give the values $f^*(n)$.
We also note that $Tb_n=b_{n+1}$ for all $n=0,1,\ldots$. By induction $T^mb_n=b_{n+m}$. So $$\langle T^mb_n,f^*\rangle= \langle b_{n+m},f^*\rangle=f^*(n+m).$$ So we just need these to grow very rapidly in order to obtain a neighborhood $U$ which such that for each $f\in P(x)$, $T^m f$ eventually leaves $U$.
Let $f^*(n)=n!$. That is, $$\langle \sum_{n=0}^d a_nb_n,f^*\rangle = \sum_{n=0}^d a_n n!.$$ This defines a linear functional on $P(x)$. Let $$U=\{f\in P(x):|\langle f,f^*\rangle|<1\},$$ which is a neighborhood of $0$.
Fix $f\neq 0$ and write $f=\sum_{n=0}^d a_nb_n$ with $a_d\neq 0$, so $f$ has degree $d$. In other words, the $a_n$ are such that $f(x)=\sum_{n=0}^d \frac{a_nx^n}{n!}$. Of course, these are the coefficients for the Maclaurin series of $f$ at zero, which we know satisfy $a_n=f^{(n)}(0)$, where $f^{(n)}$ is the $n^{th}$ derivative of $f$.
Note that $$\langle T^mf,f^*\rangle = \Bigl\langle \sum_{n=0}^d a_nT^mb_n,f^*\Bigr\rangle = \Bigl\langle \sum_{n=0}^d a_nb_{n+m},f^*\Bigr\rangle = \sum_{n=0}^d a_n (n+m)!.$$
We will show that $$\lim_{m\to \infty}\frac{\langle T^mf,f^*\rangle}{a_d(d+m)!}=1.\tag{$E$}$$ If $a_d>0$, $\lim_m a_d(d+m)!=+\infty$, and $\lim_m \langle T^mf,f^*\rangle =+\infty$ by $(E)$. If $a_d<0$, $\lim_m a_d(d+m)!=-\infty$, and $\lim_m \langle T^m f,f^*\rangle = -\infty$ by $(E)$. In either case, $\lim_m |\langle T^mf,f^*\rangle|=\infty>1$, so for sufficiently large $m$, $T^mf\notin U$.
We prove $(E)$. \begin{align*} \frac{\langle f,f^*\rangle}{a_d(d+m)!} & = \frac{\sum_{n=0}^d a_n(n+m)!}{a_d(d+m)!} = \sum_{n=0}^d \frac{a_n}{a_d}\cdot \frac{(n+m)!}{(d+m)!} \\ & = \sum_{n=0}^d \frac{a_n}{a_d}\cdot \frac{1}{(d+m)(d-1+m)\ldots (n+2+m)(n+1+m)} \\ & = \frac{a_d}{a_d}\cdot 1 + \frac{a_{d-1}}{a_d}\cdot \frac{1}{d+m} + \frac{a_{d-2}}{a_d}\cdot \frac{1}{(d-1+m)(d+m)} +\ldots \\ & +\frac{a_0}{a_d}\cdot \frac{1}{(d+m)(d-1+m)\ldots (m+1)} \underset{m\to\infty}{\to} 1+0+0+\ldots+0=1.\end{align*}