# Why my endomorphisme between Polynomial fonction is not continuous?

1. Jul 28, 2015

### Calabi

Hello let be $$E = \mathbb{R}[X]$$ with the norme $$||P|| = sup_{t \in \mathbb{R}}e^{-|t|}|P(t)|$$. Let be $$A \in E$$. How to show that $$\Psi_{A} : P \rightarrow AP$$ is not continue please?

Thank you in advance and have a nice afternoon.

2. Jul 28, 2015

### Calabi

For A differente of the nul. And of a 0 degrre polynom(a constant.); otherwhise it's continu in both case.

3. Aug 1, 2015

4. Aug 1, 2015

### disregardthat

I haven't found a solution, but let me make clear that I have understood this correctly. You are defining a norm on $\mathbb{R}[x]$ by defining $||P(x)|| = \sup_{t \in \mathbb{R}} e^{-|t|} |P(t)|$. This defines a metric on $\mathbb{R}[x]$. Then you are defining multiplication maps $\psi_{A(x)} : \mathbb{R}[x] \to \mathbb{R}[x]$ for polynomials $A(x) \in \mathbb{R}[x]$ by putting $\psi_{A(x)}(P(x)) = A(x)P(x)$. For constant polynomials $A(x) = a_0$, these functions are clearly continuous. And the question is whether constant polynomials are the only polynomials $A(x)$ for which the corresponding multiplication map is continuous?

My attempt was trying to find a sequence of polynomials $P_r(x)$ of degree $r$ such that $||P_r(x)|| \leq 1$ for each $r$, and then try to show that $\psi_{A(x)}$ is not continuous at $0$. The idea is to find an appropriate sequence of such $P_r(x)$'s such that $||P_r(x)A(x)||$ grows without bound as $r$ gets larger for any nonconstant polynomial $A(x)$. In this case, choose $\epsilon = 1$, and assume a $\delta > 0$ exists such that $||P(x)|| < \delta \Rightarrow ||P(x)A(x)|| < 1$. Then $||\frac{\delta}{2} P_r(x)|| \leq \frac{\delta}{2} < \delta$, but $||\frac{\delta}{2} P_r(x)A(x)||$ grows without bound, which shows that $\psi_{A(x)}$ is not continuous at $0$.

For example, in the case of $A(x) = x$, define $P_r(x) = \frac{e^r}{r^r}x^r$. Then $||P_r(x)|| = \frac{e^r}{r^r} \sup_{t \in \mathbb{R}} e^{-|t|} |t^r|$. To find this maximum value, we can clearly assume that $t \geq 0$ by symmetry. So to find the maximum value, we differentiate $e^{-t}t^r$ and get $e^{-t}(rt^{r-1}-t^r)$. This expression is $0$ when $t = r$, in which case the maximum value is $\frac{e^r}{r^r} e^{-r}r^r = 1$ (of course $t = 0$ is a minimum value). On the other hand, $||P_r(x)A(x)|| = \frac{e^r}{r^r} \sup_{t \in \mathbb{R}} e^{-|t|} |t^{r+1}|$. Similarly, we differentiate and get the equation $e^{-t}((r+1)t^{r}-t^{r+1}) = 0$. Here $t = r+1$. But then the maximum value is $\frac{e^r}{r^r} e^{-(r+1)} (r+1)^{r+1} = e^{-1}(1+ \frac{1}{r})^r (r+1) \geq e^{-1}(r+1)$. But $\lim_{r \to \infty} e^{-1}(r+1) = \infty$. Thus the condition I described above is satisfied, and this proves that $\phi_{x}$ is not continuous at $0$.

I do not know a proof for general nonconstant polynomials $A(x)$, but the above example might be adapted.

Last edited: Aug 1, 2015
5. Aug 1, 2015

### Calabi

In the case of $$A = X$$ I find it's not continue. But could I, with that show it's not continu in general please?

6. Aug 1, 2015

### disregardthat

Slight generalization: we will show that $A(x) = x^n$ is not continuous for each $n \geq 1$. We use the $P_r(x)$ defined as above. Then $||A(x)P_r(x)|| = \frac{e^r}{r^r} \sup_{t \in \mathbb{R}} e^{-|t|} |t^{r+n}|$. To find this, we differentiate and get $e^{-t}((r+n)t^{r+n-1}-t^{r+n}) = 0$. So $t = r+n$. This yields the maximum value $\frac{e^r}{r^r} e^{-(r+n)} (r+n)^{r+n} = e^{-n}(1+ \frac{n}{r})^r (r+n)^n \geq e^{-n} (r+n)^n$. But $\lim_{r \to \infty} e^{-n} (r+n)^n= \infty$, so $\phi_{x^n}$ is not continuous.

Do you have any ideas?

7. Aug 1, 2015

### disregardthat

Okay, for a general $A = a_nx^n + a_{n-1}x^{n-1}+... + a_0$, we have

$||AP_r(x)|| = ||a_nx^nP_r(x) +P_r(x)(a_{n-1}x^{n-1}+... + a_0)|| \geq |a_n| ||x^nP_r(x)|| - \sum_{k=0}^{n-1}|a_k| ||P_r(x)x^k|| = |a_n| e^{-n}(1+ \frac{n}{r})^r (r+n)^n - \sum_{k=0}^{n-1} |a_k| e^{-k}(1+ \frac{k}{r})^r (r+k)^k$

Now we use that

1) $(1+ \frac{k}{r})^r \leq e^k$ for all $r$ (it is an increasing sequence converging to $e^k$).
2) $(1+ \frac{n}{r})^r \geq 1$ for all $r$.
3) $(r+k)^k \leq (2r)^k = 2^kr^k$ for sufficiently large $r$.
4) $(r+n)^n \geq r^n$ for all $r$.

So

$|a_n| e^{-n}(1+ \frac{n}{r})^r (r+n)^n - \sum_{k=0}^{n-1} |a_k| e^{-k}(1+ \frac{k}{r})^r (r+k)^k \geq |a_n| e^{-n} r^n - \sum_{k=0}^{n-1} |a_k| 2^k r^k$

for sufficiently large $r$. But this is a polynomial $B(r)$ of degree $n$ with positive coefficient $|a_n| e^{-n}$ in front of $r^n$, so $\lim_{r \to \infty} B(r) = \infty$. By the same argument as before, we conclude that $\phi_A$ is not continuous.

Last edited: Aug 1, 2015