# holomorphic function associated with continuous function

If $f(z)$ is continuous   on a (finite) contour $\gamma$ of the complex plane, then the contour integral

 $\displaystyle g(z)\;=:\;\int_{\gamma}\frac{f(t)}{t\!-\!z}\,dt,$ (1)

defines a function$z\mapsto g(z)$  which is holomorphic in any domain $D$ not containing points of $\gamma$.  Moreover, the derivative  has the expression

 $\displaystyle g^{\prime}(z)\;=\;\int_{\gamma}\!\frac{f(t)}{(t\!-\!z)^{2}}\,dt.$ (2)

Proof.  The right hand side of (2) is defined since its integrand is continuous.  On has to show that it equals

 $\lim_{\Delta z\to 0}\frac{g(z\!+\!\Delta z)\!-\!g(z)}{\Delta z}.$

Let  $z_{1}=:z\!+\!\Delta z\notin\gamma$,  $\Delta z\neq 0$.  We may write first

 $\frac{g(z_{1})-g(z)}{z_{1}\!-\!z}\;=\;\frac{1}{\Delta z}\int_{\gamma}f(t)\left% [\frac{1}{t\!-\!z_{1}}-\frac{1}{t\!-\!z}\right]\,dt\;=\;\int_{\gamma}\frac{f(t% )}{(t\!-\!z_{1})(t\!-\!z)}\,dt,$

whence

 $E\;=:\;\frac{g(z_{1})-g(z)}{z_{1}\!-\!z}-\int_{\gamma}\frac{f(t)}{(t\!-\!z)^{2% }}\;=\;\Delta z\cdot\!\int_{\gamma}\frac{f(t)}{(t\!-\!z_{1})(t\!-\!z)^{2}}\,dt.$

Because $f$ is continuous in the compact set $\gamma$, there is a positive constant $M$ such that

 $|f(t)|\;<\;M\quad\forall\;t\in\gamma.$

As well, we have a positive constant $d$ such that

 $|t\!-\!z|\;\geqq\;d\quad\forall\;t\in\gamma.$

When we choose  $|\Delta z|<\frac{d}{2}$,  it follows that

 $|t\!-\!z_{1}|\;=\;|(t\!-\!z)-\Delta z|\;\geqq\;|t\!-\!z|-|\Delta z|\;>\;d\!-\!% \frac{d}{2}\;=\;\frac{d}{2}.$

Consequently,

 $\left|\frac{f(t)}{(t\!-\!z_{1})(t\!-\!z)^{2}}\right|\;=\;\frac{|f(t)|}{|t\!-\!% z_{1}||t\!-\!z|^{2}}\;<\;\frac{M}{\frac{d}{2}\cdot d^{2}}\;=\;\frac{2M}{d^{3}}$

and, by the estimating theorem of contour integral,

 $|E|\;=\;|\Delta z|\cdot\left|\int_{\gamma}\frac{f(t)}{(t\!-\!z_{1})(t\!-\!z)^{% 2}}\,dt\right|\;<\;|\Delta z|\cdot\frac{2M}{d^{3}}\cdot k,$

where $k$ is the length of the contour.  The last expression tends to zero as  $\Delta z\to 0$.  This settles the proof.

Remark 1.  By induction  , one can prove the following generalisation of (2):

 $\displaystyle g^{(n)}(z)\;=\;n!\!\int_{\gamma}\!\frac{f(t)}{(t\!-\!z)^{n+1}}\,% dt\qquad(n\;=\;0,\,1,\,2,\,\ldots)$ (3)

Remark 2.  The contour $\gamma$ may be .  If it especially is a circle, then (1) defines a holomorphic function inside $\gamma$ and another outside it.

Title holomorphic function associated with continuous function HolomorphicFunctionAssociatedWithContinuousFunction 2013-03-22 19:14:29 2013-03-22 19:14:29 pahio (2872) pahio (2872) 11 pahio (2872) Theorem msc 30E20 msc 30D20 DifferentiationUnderIntegralSign CauchyIntegralFormula