# Equivalent statement for uniform convergence

Let $f_n:D\to\mathbb{C}$ be a sequence of holomorphic functions for which $|f_n(s)|^2$ converges uniformly, i.e. for $\varepsilon>0$ there always exists $N>0$ such that for all $n,m>N$ we have $||f_n(s)|^2-|f_m(s)|^2|<\varepsilon$ is true.

Does it follow that $f_n(s)$ is also uniformly convergent?

Assuming $||f_n(s)|+|f_m(s)||\neq 0$, then from my working so far I know that $$||f_n(s)|-|f_m(s)||<\frac{\varepsilon}{||f_n(s)|+|f_m(s)||}<\varepsilon,$$ but then I'm unsure if I can proceed...

• Are you sure of your hypothesis ? Try $f_n(x)=(-1)^n$ for all $x$. – Kelenner Jul 12 '17 at 19:02
• note that $|-1| = |1|$. You are making no assumptions on whether the $f_n$ are continuous...consider a sequence $f_n$ which is $=1$ or $= -1$ on a set depending on $n$, so that the sequence does not converge. You will still have $||f_n|^2 - |f_m|^2 | = 0$ – Thomas Jul 12 '17 at 19:02
• @Antinous do you mean that you assume uniform convergence of $|f_n|$ and pointwise convergence of $f_n$? – haemi Jul 12 '17 at 19:07
• @haemi Apologies for the confusion. If it is assumed true that $||f_n|^2-|f_m|^2|<\varepsilon$ for a sequence of functions $|f_n|^2$, then I am wondering if I can conclude from this that $f_n$ also converges uniformly. – Pixel Jul 12 '17 at 19:10
• @Antinous Gotcha :) I'll think about it... – haemi Jul 12 '17 at 19:33

No, it doesn't: Set $S:=[0,1]$ and $f_n(s):=\exp(i\pi\cdot s^n)$. Then $|f_n|^2$ is equal to $1$ everywhere on $S$, hence $|f_n|^2$ converges uniformly. $f_n$ converges pointwise to the function $$f(s):=1, s\in[0,1); f(1):=-1.$$ Since all $f_n$ are continuous but $f$ isn't, the convergence cannot be uniform.