- Morera’s theorem states that if f is continuous in \(\mathbb { C } .\) and \(\textstyle \int _ { T } f ( z ) d z = 0\) for all triangles T , then f is holomorphic in C. Naturally, we may ask if the conclusion still holds if we replace triangles by other sets.
- Suppose that f is continuous on \(\mathbb { C } .\) and
\[ \int_ {C} f (z) d z = 0\tag{16} \]
for every circle C. Prove that f is holomorphic.
- More generally, let Γ be any toy contour, and \(\mathcal { F }\) the collection of all trans lates and dilates of Γ. Show that if f is continuous on \(\mathbb { C } .\) and
\[ \int_ {\gamma} f (z) d z = 0 \quad \text { for all } \gamma \in \mathcal {F} \]
then \(f\) is holomorphic. In particular, Morera’s theorem holds under the weaker assumption that \(\begin{array} { r } { \int _ { T } f ( z ) d z = 0 } \end{array}\) for all equilateral triangles.
[Hint: As a first step, assume that f is twice real diferentiable, and write \(f ( z ) =\) \(f ( z _ { 0 } ) + a ( z - z _ { 0 } ) + b ( \overline { { z - z _ { 0 } } } ) + O ( | z - z _ { 0 } | ^ { 2 } )\) for z near \(z _ { \mathrm { 0 } }\) . Integrating this expansion over small circles around \(z_0\) yields \(\partial f / \partial { \overline { { z } } } = b = 0\) at \(z _ { 0 }\) . Alternatively, suppose only that \(f\) is diferentiable and apply Green’s theorem to conclude that the real and imaginary parts of f satisfy the Cauchy-Riemann equations.
In general, let \(\varphi ( w ) = \varphi ( x , y )\) (when \(w = x + i y )\) denote a smooth function with \(0 \leq \varphi ( w ) \leq 1\) , and \(\begin{array} { r } { \int _ { \mathbb { R } ^ { 2 } } \varphi ( w ) d V ( w ) = 1 } \end{array}\) , where \(d V ( w ) = d x d y .\) and \(\scriptstyle \int\) denotes the usual integral of a function of two variables in \(\mathbb { R } ^ { 2 }\) . For each \(\epsilon > 0\) , let \(\varphi _ { \epsilon } ( z ) =\) \(\epsilon ^ { - 2 } \varphi ( \epsilon ^ { - 1 } z )\) , as well as
\[ f _ {\epsilon} (z) = \int_ {\mathbb {R} ^ {2}} f (z - w) \varphi_ {\epsilon} (w) d V (w), \]
where the integral denotes the usual integral of functions of two variables, with \(d V ( w )\) the area element of \(\mathbb { R } ^ { 2 }\) . Then \(f _ { \epsilon }\) is smooth, satisfies condition (16), and \(f _ { \epsilon } \to f\) uniformly on any compact subset of C.]
Solution
1. If \(f\) is \(C^2\) and \(\int_C f\,dz=0\) for every circle \(C\), then \(f\) is holomorphic.
Near \(z_0\) write \(f(z)=f(z_0)+a(z-z_0)+b\,\overline{(z-z_0)}+O(|z-z_0|^2)\) with \(b=\partial f/\partial\overline z(z_0)\). On \(\abs{z-z_0}=r\) the constant and \(a(z-z_0)\) integrate to \(0\), the error term integrates to \(O(r^3)\), and \(\int_{|z-z_0|=r}\overline{(z-z_0)}\,dz=2\pi i r^2\). Hence \(0=2\pi i b r^2+O(r^3)\), so \(b=0\). As \(z_0\) is arbitrary, \(\partial f/\partial\overline z=0\) and \(f\) is holomorphic.
2. (a) If \(f\) is continuous and \(\int_C f\,dz=0\) for every circle \(C\), then \(f\) is holomorphic.
With \(\varphi_\epsilon\) as in the hint, \(f_\epsilon(z)=\int_{\mathbb R^2}f(z-w)\varphi_\epsilon(w)\,dV(w)\) is smooth, and \(f_\epsilon\to f\) uniformly on compact sets. For a circle \(C\), Fubini’s theorem gives \(\int_C f_\epsilon\,dz=\int_{\mathbb R^2}\bigl(\int_{C-w}f\,dz\bigr)\varphi_\epsilon(w)\,dV(w)=0\), since \(C-w\) is again a circle. By step 1 each \(f_\epsilon\) is holomorphic, and a locally uniform limit of holomorphic functions is holomorphic.
3. (b) If \(\int_\gamma f\,dz=0\) for all translates and dilates \(\gamma\) of a toy contour \(\Gamma\), then \(f\) is holomorphic.
Since \(\mathcal F\) is closed under translation, the Fubini argument of step 2 shows that each \(f_\epsilon\) also satisfies \(\int_\gamma f_\epsilon\,dz=0\) for all \(\gamma\in\mathcal F\); so, as in step 2, it suffices to treat smooth \(f\). For smooth \(f\), Green’s theorem on the interior \(U\) of \(\Gamma\), which has positive area \(\abs U\), gives for \(\gamma=z_0+\delta\Gamma\) \[0=\int_\gamma f\,dz=\pm2i\iint_{z_0+\delta U}\frac{\partial f}{\partial\overline z}\,dA=\pm2i\,\delta^2\iint_U\frac{\partial f}{\partial\overline z}(z_0+\delta w)\,dA(w),\] the sign depending on the orientation of \(\Gamma\). Dividing by \(\delta^2\) and letting \(\delta\to0\) gives \(\abs U\,\frac{\partial f}{\partial\overline z}(z_0)=0\). Hence \(\partial f/\partial\overline z=0\) everywhere and \(f\) is holomorphic.
4. Morera’s theorem holds when \(\int_T f\,dz=0\) is assumed only for equilateral triangles \(T\).
Take \(\Gamma\) to be an equilateral triangle; then \(\mathcal F\) consists of the equilateral triangles with the orientation of \(\Gamma\), and step 3 applies.