# A proposition about power series

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 $$\sup_{x,y\in[a,a+\delta]}|P(x)-P(y)|\geq C\sup_{x,y\in[a,a+\delta]}|p(x)-p(y)|\delta^{\nu}.$$

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• 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. – Mateusz Kwaśnicki Feb 14 at 15:11
• 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. – GH from MO 3 hours ago

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.
• @Burnie: They always exist for $x \ne 0$, do they not? – Mateusz Kwaśnicki Feb 14 at 15:12
• @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 $M$. – GH from MO Feb 14 at 19:19