A non continuous linear map $A:\Bbb{R}[X]\rightarrow \Bbb{R}$ such that $A(P)=P(1).$ 
I have a linear map $A:\Bbb{R}[X]\mapsto \Bbb{R}$ such that $A(P)=P(1)$, for the $p-$norm : $\Vert P\Vert=\bigl(\sum_{i=1}^n\vert a_i\vert^p\bigr)^{1/p}$ where $p\in[1,+\infty].$

For the cas where $p=1$ it's continuous (triangle inequality), anyway for the case $p>1$ I would like to prove that is not continuous.
For that I can use the sequential characterization of continuity, if $p=\infty$, the following polynomial $P(X)=1+X+X^2+\cdots+X^n$ works, because $\Vert P\Vert_{\infty}=1$ but $\vert P(1)\vert=1+1+1+\cdots+1=n+1\rightarrow+\infty.$
Anyway, I am stuck for $1<p<\infty$,I get always the same result...
Any idea ?
EDIT : in fact my polynomial works for any $p$.
 A: Hint Define a sequence of polynomials by
$$
P_n(x)= \frac{1}{\log(\log(n))}\sum_{k=0}^n \frac 1{k+1}x^k
$$
A: For a given $p\gt1$, consider
$$
a_{n,k}=\left\{\begin{array}{cl}
\dfrac1{n^{1/p}}&\text{if }1\le k\le n\\
0&\text{if }k\gt n
\end{array}\right.
$$
Then
$$
\begin{align}
\left(\sum_{k=1}^\infty a_{n,k}^p\right)^{1/p}
&=\left(\sum_{k=1}^n\frac1n\right)^{1/p}\\[6pt]
&=1
\end{align}
$$
while
$$
\begin{align}
\sum_{k=1}^\infty a_{n,k}
&=\sum_{k=1}^n\frac1{n^{1/p}}\\[6pt]
&=n^{1-\frac1p}
\end{align}
$$
A: Partial answer, for $p>2.$ Let $P_n(x)=\sum_{j=1}^n j x^j .$  We have $$\|P_n\|^p=\sum_{j=1}^n j^p<\sum_{j=1} ^n\int_j^{j+1}y^p dy=\int_1^{n+1}y^p dy=$$ $$=\frac{(n+1)^{1+p}-1}{1+p}<\frac {(n+1)^{1+p} }{1+p}.$$  $$\text {So }\; \|P_n\|<\frac {(1+n)^{(1+1/p)}}{(1+p)^{1/p}}\;\text { But }\; P_n(1)=(n^2-n)/2.$$
Let $Q_n=P_n/n^{1+2/p}.\;$ Then $\lim_{n\to \infty}\|Q_n\|=0$ but $\lim_{n\to \infty}Q_n(1)=\infty,$ so $A$ is discontinuous at $0.$
I expect a variation of  this approach should work for $1<p\leq 2.$
