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Let $f(x,s)=\sum_{n=0}^\infty a_n(s)x^n$ where $|a_n(s)|\le1$ is a bounded function theory. Suppose for every $|x|<1$, $f(x,s)$ is Holder-$\alpha$ for $s$-variable, i.e. $|f(x,s_1)-f(x,s_2)|\le C|s_1-s_2|^\alpha$. Then can we have $a_n(s)$ is Holder-$\alpha$ for every $n$?

Equivalently, can we have, for every $z\in\mathbb C$, $|z|<1$, $f(z,s)$ is of Holder-$\alpha$ for $s$?

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  • $\begingroup$ You mean for $x\in(-1,1)$? Also, is it OK to assume that the estimate is uniform for $|x|\le r$ when $r<1$, or you want it exactly as written? $\endgroup$
    – fedja
    Commented Jun 27, 2019 at 4:19
  • $\begingroup$ @fedja Yes the radius does not really matters, all I need is the local convergence $\endgroup$
    – Liding Yao
    Commented Jun 27, 2019 at 4:20
  • $\begingroup$ Anyway, the answer is "No". All you can get is that the function in the complex domain is $\beta$-Holder with any $\beta<\alpha$ in a sufficiently small disk depending of $\beta$. $\endgroup$
    – fedja
    Commented Jun 27, 2019 at 4:42
  • $\begingroup$ @fedja But what is the example $\endgroup$
    – Liding Yao
    Commented Jun 27, 2019 at 4:43
  • $\begingroup$ Consider $f(z)=\sum_{m\ge 1}b_me^{-m}e^{-imz}\min(1,e^m s)$ with $b_m=m^{-2}$, say and $s\in[0,1]$. It is bounded in the strip $|\Im z|\le 1$, Lipschitz in $s$ uniformly on the real line but only $\frac 12$-Holder in $s$ (at $s=0$) on $\mathbb R+\frac i2$. $\endgroup$
    – fedja
    Commented Jun 27, 2019 at 4:57

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