Sampling Theorem and Dirac Comb I am reading "The Scientist and Engineer's Guide to Digital Signal Processing" and trying to understand Figure 3.5 below which is about the sampling theorem and aliasing.

I do not understand the picture of the frequency domain. The signal can be represented as
$$f(t)=\frac{1}{2\pi}\int_{-\infty}^{\infty}F(x)e^{ixt}dx$$
where $F(x)$ is the Fourier transform of $f$. This represents $f$ with the "frequencies" $e^{ixt}$ and these can have $x$ negative. How then am I suppose to understand Figure 3-5 which shows the frequency domain for only greater than zero and moreover real and not complex.

Also why does the frequency domain of the impulse train have the original frequency spectrum appear flipped half the time? My current understanding is that the repetition comes from something like
$$\mathcal{F^{-1}} \{f(t)\Delta(t)\} = F(x)*\Delta(x)$$
where $\Delta$ is the dirac comb and probably some kind of constant like $2\pi$ is needed in there somewhere. How does the flipping half the time in the frequency domain arise from $F(x)*\Delta(x)$. The image I think for convolution is that $\Delta$ is flipped and as it is moved to the right by $s$ the "area" of $F(x)\Delta(s-x)$ is the value of the convolution at $s$. This would seem to replicate the original spectrum multiple times but without the flipping half the time as shown in Figure 3.5.
Let $X(f)$ denote the Fourier transform of $x(t)$ where
$$\begin{align*}
X(f) &amp;= \int_{-\infty}^{\infty} x(t) \exp(-j2\pi ft) \mathrm dt\\
x(t) &amp;= \int_{-\infty}^{\infty} X(f) \exp(+j2\pi ft) \mathrm df
\end{align*}$$
which I will denote via $x(t) \leftrightarrow X(f)$.
The following transform pairs will be needed in what follows.
$$\begin{align*}
\delta(t) &amp;\leftrightarrow 1\\
\delta(t-t_0) &amp;\leftrightarrow \exp(-j2\pi f \,t_0)\\
\sum_{n=-\infty}^{\infty}\delta(t-nT) &amp;\leftrightarrow
\frac{1}{T}\sum_{k=-\infty}^{\infty}\delta\left(f-\frac{k}{T}\right)\\
\end{align*}$$

Given a signal $x(t)$, its sampled pulse train (at intervals of $T$ seconds) is 
$$x(t) \sum_{n=-\infty}^{\infty}\delta(t-nT) 
= \sum_{n=-\infty}^{\infty} x(t)\delta(t-nT) 
= \sum_{n=-\infty}^{\infty} x(nT)\delta(t-nT).$$
Since multiplication in the time domain corresponds to convolution in the
frequency domain, we have
$$\begin{align*}
x(t)\sum_{n=-\infty}^{\infty}\delta(t-nT) &amp;\leftrightarrow
X(f)\circledast\frac{1}{T}\sum_{k=-\infty}^{\infty}\delta\left(f-\frac{k}{T}\right)\\
&amp;\leftrightarrow \frac{1}{T}\sum_{k=-\infty}^{\infty}
X(f)\circledast\delta\left(f-\frac{k}{T}\right)\\
&amp;\leftrightarrow \frac{1}{T}\sum_{k=-\infty}^{\infty}
\int_{-\infty}^{\infty} X(f-w)\delta\left(w-\frac{k}{T}\right) \mathrm dw\\
x(t)\sum_{n=-\infty}^{\infty}\delta(t-nT)
&amp;\leftrightarrow \frac{1}{T}\sum_{k=-\infty}^{\infty}
X\left(f-\frac{k}{T}\right) = \hat{X}(f)
\end{align*}$$
Thus, the Fourier transform of the impulse train formed by
sampling $x(t)$ at $T$ second intervals is $\hat{X}(f)$ which
is obtained by repeating $X(f)$ along the $f$ axis at intervals of 
$T^{-1}$ Hz and summing the result.
Furthermore, $\hat{X}(f)$ is a periodic function of the frequency
variable $f$ with period $T^{-1}$ Hz.  That is, for all $f$,
$$\hat{X}\left(f + \frac{1}{T}\right) = \hat{X}(f).$$  Note that
all of this holds regardless of what $X(f)$ is: $X(f)$ could be
nonzero for all $f$, and the result would still be valid.

Now suppose that $X(f)$ is zero if $f &lt; a$ or $f &gt; b$.  Then,
$\hat{X}(f)$, obtained by repeating $X(f)$ periodically along
the $f$ axis,  is nonzero only in the intervals
$$\ldots, \left[a-\frac{2}{T}, b-\frac{2}{T}\right], 
\left[a-\frac{1}{T}, b-\frac{1}{T}\right],
[a, b] 
\left[a+\frac{1}{T}, b+\frac{1}{T}\right], 
\left[a+\frac{2}{T}, b+\frac{2}{T}\right], 
\ldots
$$
and so if 
$$b -\frac{1}{T} &lt; a \Rightarrow b-a &lt; \frac{1}{T},$$
that is, the support $b-a$ of $X(f)$ is smaller than 
the repetition interval $T^{-1}$ Hz, then the repetitions
of $X(f)$ do not overlap.  Indeed as the OP's figures show, 
for a real-valued signal whose spectrum extends from $a = -f_0$
to $b = f_0$ (support of $X(f)$ is of length $2f_0$), sampling 
$x(t)$ at intervals of $T = (3f_0)^{-1}$ (and thus repeating $X(f)$ at
intervals of $3f_0$ Hz on the frequency axis)
leads to no overlap while sampling $x(t)$ at intervals of
$T = (1.5f_0)^{-1}$ leads to overlap of the repetitions
of $X(f)$.

Finally, the OP asks about negative frequencies and the
interpretation of the pictures which show positive frequencies
only.  For a real-valued signal, $X(f)$ has conjugate symmetry,
meaning that $X(-f) = X^*(f)$, and so specifying $X(f)$ for
positive values of $f$ suffices.  In any case, the pictures
he is looking at are of $|X(f)|$ and $|\hat{X}(f)|$ which
are even functions of $f$, and so showing only the positive
axis saves space, though it does make the pictures look a 
bit lopsided since only half the lobe is shown at low frequencies.
For complex-valued signals, $X(f)$ does not have conjugate
symmetry and $|X(f)|$ need not be an even function of $f$
and so the whole axis would need to be shown.  But, the general
development above is still applicable, and we still need 
to sample at rate exceeding $b-a$ Hz, and it helps to keep in
mind that in this general case, each sample is actually two real
numbers, not one, since we are sampling a complex-valued signal.