= Solution
Consider a nonreal <phase velocity> $c$, so $U-c$ never vanishes. Substitute $\phi=(U-c)\psi$ into the <Rayleigh equation for inviscid shear flow>. Differentiating cancels the two terms involving $U''$, leaving
$$
(U-c)\psi''+2U'\psi'-k^2(U-c)\psi=0,
\qquad
\boxed{\bigl[(U-c)^2\psi'\bigr]'-k^2(U-c)^2\psi=0.}
$$
For $k>0$, the wall condition on normal velocity gives $\phi=0$, hence $\psi=0$, at both walls. Multiply by $\overline\psi$ and use <integration by parts> to obtain the <weighted identity for Howard's semicircle theorem>:
$$
\int_{-y_0}^{y_0}(U-c)^2W\,dy=0,\qquad
W=|\psi'|^2+k^2|\psi|^2\ge0.
$$
For a nonzero <normal mode>, $N=\int W\,dy>0$. The imaginary and real parts of this identity give, respectively,
$$
-2c_i\int(U-c_r)W\,dy=0,\qquad
\int[(U-c_r)^2-c_i^2]W\,dy=0.
$$
Since $c_i\ne0$, these imply
$$
\boxed{\int UW\,dy=c_rN,\qquad
\int U^2W\,dy=(c_r^2+c_i^2)N.}
$$
The first equation says that $c_r$ is a positive <weighted mean> of $U$, and consequently $U_{\min}\le c_r\le U_{\max}$. The division by $c_i$ is essential: the weighted-mean assertion concerns nonreal modes. A neutral mode may have a <critical level> where $U=c$ and the substitution for $\psi$ is singular.
Pointwise, $(U-U_{\max})(U-U_{\min})\le0$. Integrating against the nonnegative weight $W$ and substituting both identities yields
$$
[c_r^2+c_i^2-(U_{\max}+U_{\min})c_r+U_{\max}U_{\min}]N\le0.
$$
Completing the square proves <Howard's semicircle theorem>:
$$
\boxed{\left(c_r-\frac{U_{\max}+U_{\min}}2\right)^2+c_i^2
\le\left(\frac{U_{\max}-U_{\min}}2\right)^2.}
$$
Thus all nonreal <phase velocities> lie in the closed disk with the real interval $[U_{\min},U_{\max}]$ as diameter. For the convention $e^{ik(x-ct)}$ and $k>0$, exponentially growing <normal modes> occupy its upper semicircle, while decaying modes occupy its lower semicircle.
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