Holomorphic functions

[Camillo]

Show that no holomorphic function \( \displaystyle {f{{\left({z}\right)}}} \) exists such that \( \displaystyle {R}{e}{\left({f{{\left({z}\right)}}}\right)}={3}{{x}}^{{2}}+{{y}}^{{2}} \); find 2 functions continuous on \( \displaystyle \mathbb{C} \) of which \( \displaystyle {3}{{x}}^{{2}}+{{y}}^{{2}} \) is the real part.

[gugo82]

For the sake of semplicity, let \(u(x,y):=3x^2+y^2\).
There are at least four simple ways to prove \(u(x,y)\) is not the real part of an analytic function, and I list them in what follows.

Testo nascosto, fai click qui per vederlo

The function \(u(x,y)\) cannot be the real part of an analytic function because:

i) it is not harmonic, for \(\Delta u(x,y)=8\neq 0\);

ii) it does not possess any harmonic conjugate function \(v(x,y)\), for the system of PDEs:
\[\begin{cases} v_x=-2y \\ v_y=6x\end{cases}\]
has no solution. In fact, if \(v(x,y)\) were a solution, then it would be of the type \(v(x,y)=6xy+\gamma (x)\); but this is a contradiction, because then the derivative of \(\gamma\) should depend also on the \(y\) variable;

iii) it does not satisfy the MVP on circles, for:
\[u(0,0)=0< \frac{1}{2\pi}\int_{|z|=1} u\ \text{d} s \; ;\]
iv) neither it satisfies the MVP on disks, since:
\[u(0,0)=0<\frac{1}{\pi} \int_{|z|\leq 1} u(x,y)\ \text{d} x\text{d} y\; .\]

Moreover, here's two functions having \(u(x,y)\) as real part which are continuous but not analytic:

Testo nascosto, fai click qui per vederlo

\[f(z):=3(\text{Re} z)^2+(\text{Im} z)^2 \quad \text{and} \quad g(z):=f(\overline{z})\; .\]