DTFT and a Downsampled Sinc Function I found the answers to this question and this question to be extremely helpful in understanding the derivation of the downsampling or decimation property of the DTFT.  Thank you!  I am now struggling to come up with a good example that shows clearly the role of the frequency-shifted term(s).
To summarize, let $y[n]$ be a discrete-time signal formed by downsampling $x[n]$ by an integer factor $M$,
$$ y[n] = x[Mn]. $$
Then the discrete-time Fourier transforms of $y[n]$ and $x[n]$ are related through
$$ Y \left( e^{j \omega} \right)
= \frac{1}{M} \sum_{k=0}^{M-1} X \left( e^{j (\omega - 2 \pi k)/M} \right). $$
The simplest case is for $M=2$,
$$ Y \left( e^{j \omega} \right)
= \frac{1}{2} \, X \left( e^{j \omega/2} \right) 
+ \frac{1}{2} \, X \left( e^{j (\omega/2 - \pi)} \right) . $$
I want to develop an example that clearly shows the role of the second term.  I have already derived the result for a causal exponential signal, but for that example the role of the second term is not at all obvious from a plot of the spectrum.  Therefore, let $x[n]$ be a sinc signal,
$$ x[n] = \frac{\sin(\omega_0 n)}{\pi n} 
\qquad \Rightarrow \qquad X \left( e^{j \omega} \right) = 
    \begin{cases}
        1, &amp; |\omega| &lt; \omega_0 \\
        0, &amp; \omega_0 &lt; |\omega| &lt; \pi.
    \end{cases}
$$
I am suppressing the periodic repetition of the rectangular spectrum here for simplicity.
Assume for the moment that $\omega_0 &lt; \pi/4$.
The computation of the DTFT of $y[n]$ by the downsampling property gives
$$
\begin{array}{rclcl}
    Y_1 \left( e^{j \omega} \right)
    &amp; = &amp; \dfrac{1}{2} \, X \left( e^{j \omega/2} \right) &amp; = &amp;
    \begin{cases}
        \frac{1}{2}, &amp; |\omega| &lt; 2 \omega_0 \\
        0, &amp; 2 \omega_0 &lt; |\omega| &lt; \pi.
    \end{cases} \\
    \\
    Y_2 \left( e^{j \omega} \right)
    &amp; = &amp; \dfrac{1}{2} \, X \left( e^{j (\omega - 2 \pi)/2} \right) &amp; = &amp;
    \begin{cases}
        \frac{1}{2}, &amp; \pi - 2 \omega_0 &lt; |\omega| &lt; \pi\\
        0, &amp; |\omega| &lt; \pi - 2 \omega_0.
    \end{cases}\\
    \\
    Y \left( e^{j \omega} \right)
    &amp; = &amp; Y_1 \left( e^{j \omega} \right) + Y_2 \left( e^{j \omega} \right)
    &amp; = &amp; 
    \begin{cases}
        \frac{1}{2}, &amp; |\omega| &lt; 2 \omega_0 \\
        \frac{1}{2}, &amp; \pi - 2 \omega_0 &lt; |\omega| &lt; \pi\\
        0, &amp; 2 \omega_0 &lt; |\omega| &lt; \pi - 2 \omega_0.
    \end{cases}\\
\end{array}
$$
This expression tells us that the DTFT of $y[n]$ consists of two rectangular spectra, one centered around $\omega = 0$, and the other around $\omega = \pm \pi$.  This would seem to be exactly what I am looking for.  However, if we return to the original expression for $y[n]$,
$$ y[n] = x[2n] = \frac{\sin(2 \omega_0 n)}{2 \pi n} . $$
Therefore, the DTFT of $y[n]$ should be
$$
    Y \left( e^{j \omega} \right) =
    \begin{cases}
        \frac{1}{2}, &amp; |\omega| &lt; 2 \omega_0 \\
        0, &amp; 2 \omega_0 &lt; |\omega| &lt; \pi,
    \end{cases}
$$
which contains only the low-frequency rectangular component.  How can I resolve this apparent contradiction?
You got the term $X \left( e^{j (\omega - 2 \pi)/2} \right)$ wrong. It is centered at $2\pi$ and it is non-zero in the interval $(2\pi-2\omega_0,2\pi+2\omega_0)$. So the results obtained in the frequency domain and in the time domain, respectively, are identical.
Note that the term $X \left( e^{j \omega/2} \right)$ is $4\pi$-periodic, so the term $X \left( e^{j (\omega - 2 \pi)/2} \right)$ makes sure that the spectrum of the downsampled signal is $2\pi$-periodic. Clearly, there is no aliasing as long as $\omega_0&lt;\pi/2$ is satsified.