-3
$\begingroup$

Is this proposition established?

Suppose that $0<\nu<1$, $x\in[0,1]$ and absolutely converge power series $$p(x)=\sum_{n=0}^\infty a_nx^n,$$ $$P(x)=\sum_{n=0}^\infty \frac{\Gamma(n+1)}{\Gamma(n+1+\nu)}a_nx^{n+\nu}.$$ Suppose that $p'(x),P'(x)$ don't exist. For any $a\in[0,1)$ and any sufficiently small $\delta>0$, there exists a certain $C>0$ such that $$\sup_{x,y\in[a,a+\delta]}|P(x)-P(y)|\geq C\sup_{x,y\in[a,a+\delta]}|p(x)-p(y)|\delta^{\nu}. $$

Improve this question
$\endgroup$
2
  • $\begingroup$ What is $C$ supposed to depend on? As stated, $C$ may depend on $(a_n)$, $a$ and $\delta$, in which case such $C$ obviously exists. $\endgroup$
    – Mateusz Kwaśnicki
    Commented Feb 14, 2020 at 15:11
  • 2
    $\begingroup$ It does not makes sense to assume that $p'(x)$ does not exist. If $p(x)$ converges in $(0,1)$, then the function defined by it is infinitely many times differentiable in $(0,1)$. In fact $p(x)$ extends holomorphically to the unit disk $\{z\in\mathbb{C}:|z|<1\}$. This is one of the basic theorems in complex analysis. See en.wikipedia.org/wiki/… See also my last comment under my response. $\endgroup$
    – GH from MO
    Commented Feb 17, 2020 at 23:18

1 Answer 1

Reset to default
1
$\begingroup$

If $P'(a)=0\neq p'(a)$, then there is no such constant. Indeed, in this situation, we have for sufficiently small $\delta$, \begin{align*} \sup_{x,y\in[a,a+\delta]}|P(x)-P(y)|&\ \ll_a\ \delta^2\\ \sup_{x,y\in[a,a+\delta]}|p(x)-p(y)|\delta^{\nu}&\ \gg_a\ \delta^{1+\nu}. \end{align*} These bounds follow readily from the Taylor series expansion of $P(x)$ and $p(x)$ around $a$. In particular, the ratio of the left hand sides tends to zero under $\delta\to 0+$, hence it is not bounded away from zero.

Improve this answer
$\endgroup$
4
  • $\begingroup$ What if P'(x) and p'(x) do not exist? $\endgroup$
    – Burnie
    Commented Feb 14, 2020 at 15:07
  • 1
    $\begingroup$ @Burnie: They always exist for $x \ne 0$, do they not? $\endgroup$
    – Mateusz Kwaśnicki
    Commented Feb 14, 2020 at 15:12
  • $\begingroup$ the convergence of Σan does not indicate a convergence of Σn*an. @Mateusz Kwaśnicki $\endgroup$
    – Burnie
    Commented Feb 14, 2020 at 15:14
  • 1
    $\begingroup$ @Burnie: Any power series is differentiable in the interior of its set of convergence. Hence $P'(a)$ and $p'(a)$ exist. More generally, if $u_n$ are holomorphic functions on an open set $M\subset\mathbb{C}$, and the function series $\sum u_n$ converges locally uniformly on $M$, then the series defines a holomorphic function on $M$, and its derivative equals $\sum u_n'$ on $M$, which itself converges locally uniformly on $M$. $\endgroup$
    – GH from MO
    Commented Feb 14, 2020 at 19:19

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .