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Why my endomorphisme between Polynomial fonction is not continuous?

  1. Jul 28, 2015 #1
    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:oldbiggrin:.
     
  2. jcsd
  3. Jul 28, 2015 #2
    For A differente of the nul. And of a 0 degrre polynom(a constant.); otherwhise it's continu in both case.
     
  4. Aug 1, 2015 #3
    Can you helpe me please?
     
  5. Aug 1, 2015 #4

    disregardthat

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    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
  6. Aug 1, 2015 #5
    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?
     
  7. Aug 1, 2015 #6

    disregardthat

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    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?
     
  8. Aug 1, 2015 #7

    disregardthat

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    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
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