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{\bf Question}

State the maximum Modulus Principle for a non-constant function
$f(z)$ that is analytic within a simple closed contour $\gamma$
and continuous on $\gamma$.

Show that if $f(z)\not=0$ for all $z$ within and on $\gamma$ then
the minimum value of $|f(z)|$ cannot be achieved in the interior
of $\gamma$.

By considering the function $e^{f(z)}$ show that no non-constant
harmonic function can achieve its maximum or minimum values in the
interior of $\gamma$.

Hence find the maximum and minimum values of $x^3-3xy^2$ in the
set

$\{(x,y)|x^2+y^2\leq1\}$ and find where the bounds are attained.


\vspace{0.25in}

{\bf Answer}

Statement of max mod.  Proof of min mod applies max mod to
$\frac{1}{f(z)}$ - bookwork

${}$

Let $u$ be harmonic.  Find a harmonic conjugate $v$ so that
$f(z)=u+iv$ is analytic.

Now $|e^{f(z)}|=e^u$ so $e^{f(z)}\not=0$

Thus $|e^{f(z)}|$ achieves its max and min on $\gamma$.

Now $e^u$ is an increasing function of $u$, so $u$ achieves its
max and min on $\gamma$.

Let $u(x,y)=x^3-3xy^2$, then $u_{xx}=6x$ and $u_{yy}=-6x$, so $u$
is harmonic.  So the unit disc $u$ achieves max and min on
$\gamma$, i.e. where $y^2=1-x^2$.

So $u(x,1-x^2)=x^3-3x(1-x^2)=4x^3-3x=h(x) \hspace{0.3in} -1\leq
x\leq1$

$h'(x)=12x^2-3=0$ where $4x^2=1$ i.e. $x=\pm\frac{1}{2}$

$h''(x)=24x \begin{array}{ll}>0 {\rm\ at\ } x=\frac{1}{2} & {\rm\
-\ a\ local\ minimum}\\ <0 {\rm\ at\ } x=-\frac{1}{2} & {\rm\ -\
a\ local\ maximum}\end{array}$

$h(\frac{1}{2})=-1 \hspace{0.3in} h(-\frac{1}{2})=1$

At the end points $h(-1)=-1 \hspace{0.3in} h(1)=1$

so $u$ achieves max values at
$\ds\left(-\frac{1}{2},\pm\frac{\sqrt3}{2}\right)$ and $(1,0)$

and $u$ achieves min values at
$\ds\left(\frac{1}{2},\pm\frac{\sqrt3}{2}\right)$ and $(-1,0)$.

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