Exchange of limit of complex integral

Question

Suppose $f=u+iv$ a complex function of complex variable, continuous on an open set $A$, $u,v \in C^1(A)$.

I have to demonstrate that $\lim_{h \rightarrow 0} \int _0^1f(z+th)dt= \int _0^1\lim_{h \rightarrow 0} f(z+th)dt$, where $z\in A,h \in \mathbb{C}$. How can I do this, and how much can this result be generalized (in term of smoothness of $f$) ?

PS. I'm not really into measure theory.

Answer 1

No need of uniform continuity nor uniform convergence. Continuity of $f$ at the point $z$ is sufficient:

Let $\epsilon>0.$ There exists $\delta>0$ such that $$\forall k\in\Bbb C\quad|k|<\delta\Rightarrow|f(z+k)-f(z)|<\epsilon.$$ For such a $\delta,$ we immediately deduce: $$\forall h\in\Bbb C\quad|h|<\delta\implies\left|\int_0^1\left(f(z+th)-f(z)\right)dt\right|\le\epsilon.$$

Answer 2

It is enough to show that for any sequence $h_n\to 0$ we have $\lim_{n\to\infty}\int_0^1f(z+th_n)dt=\int_0^1\lim_{n\to\infty}f(z+th_n)dt$. So let $h_n$ be a sequence that converges to $0$. From here we can build a sequence of functions $f_n:[0,1]\to\mathbb{C}$ by $f_n(t)=f(z+th_n)$. We'll show that it converges uniformly to the constant function $f(z)$. Take any $\epsilon>0$. $f$ is continuous at the point $z$ so there exists a $\delta>0$ such that $|z-w|<\delta$ implies $|f(z)-f(w)|<\epsilon$. So now take a number $n_0\in\mathbb{N}$ such that $\forall n\geq n_0$ we have $|h_n|<\delta$. There exists such an index $n_0$ because $h_n\to 0$. So now take any $n\geq n_0$ and any $t\in[0,1]$. Then $|(z+th_n)-z|<\delta$ and hence $|f_n(t)-f(z)|=|f(z+th_n)-f(z)|<\epsilon$. That is for all $n\geq n_0$ and for all $t\in[0,1]$. That means the sequence converges uniformly. And it is known that in that conditions we can exchange complex integral with limit. Formally, we can define a curve $\gamma:[0,1]\to\mathbb{C}$ by $\gamma(t)=t$. Then:

$\lim_{n\to\infty}\int_0^1f(z+th_n)dt=\lim_{n\to\infty}\int_0^1 f_n(t)dt=\lim_{n\to\infty}\int_\gamma f_n(w)dw=$

$=\int_\gamma\lim_{n\to\infty}f_n(w)dw=\int_0^1 \lim_{n\to\infty} f_n(t)=\int_0^1\lim_{n\to\infty}f(z+th_n)dt$

Answer 3

Consider the function $g:[0,1]\times \{z:|z| \leq 1\} \to \mathbb C$ defined by $g(t,h)=f(z+th)$. This function is continuous on a compact set hence uniformly continuous. Hence, given $\epsilon >0$ there exists $\delta >0$ such that $|g(t,h)-g(t',h')|<\epsilon$ whenever $\|(t,h)-(t',h')\| <\delta$. This holds in particular if $|h-h'| < \delta$ and $t'=t$. Now $|\int [f(z+th) -f(z)] \, dt| \leq \epsilon$ whenever $|h| <\delta$. can you complete the proof now?