Energy in baseband and bandpass signal
Let $x(t)$ denote a real-valued finite-energy signal of energy $E_x$, that is,

$$\int_{-\infty}^\infty |x(t)|^2\,\mathrm dt = E_x &lt; \infty \tag{1}$$ 

and let $X(f)$ denote the Fourier transform of $x(t)$.
Parseval's theorem assures
us that

$$\int_{-\infty}^\infty |x(t)|^2\,\mathrm dt = \int_{-\infty}^\infty |X(f)|^2\,\mathrm df \tag{2}$$

and thus the energy $E_x$ can also be computed via the right side of $(2)$.

Now consider the DSB-AM (Double-Side-Band Amplitude-Modulated) 
signal $x(t)\cos(2\pi f_c t)$. Its energy is

$$\begin{align}
\int_{-\infty}^\infty |x(t)\cos(2\pi f_c t)|^2\,\mathrm dt
&amp;= \int_{-\infty}^\infty |x(t)|^2|\cos(2\pi f_c t)|^2\,\mathrm dt\\
&amp;= \frac 12 \int_{-\infty}^\infty |x(t)|^2[1 + \cos(2\pi 2f_c t)]\,\mathrm dt\\
&amp;= \frac 12 E_x + \int_{-\infty}^\infty |x(t)|^2\cos(2\pi 2f_c t)\,\mathrm dt\tag{3}
\end{align}$$
Now, if the value of the integral on the right side of $(3)$ is exactly zero,
then the modulated signal $x(t)\cos(2\pi f_c t)$
has exactly half the energy of the unmodulated signal $x(t)$.
In the time domain, it can be plausibly argued that if the integral is
not identically zero, it is at least very small when $x(t)$ is a
low-pass signal and $f_c$ is very large compared to the highest frequency
in $x(t)$.  Denote by $T = (2f_c)^{-1}$ the period of the double-frequency
sinusoid in $(3)$ and consider that
$$\int_{-\infty}^\infty |x(t)|^2\cos(2\pi 2f_c t)\,\mathrm dt
= \sum_{n=-\infty}^\infty \int_{nT}^{(n+1)T}|x(t)|^2\cos(2\pi 2f_c t)\,\mathrm dt.
\tag{4}$$
Now, $x(t)$ is a slowly-varying signal that does not change very much
during $[nT, (n+1)T]$ and so
$$\int_{nT}^{(n+1)T}|x(t)|^2\cos(2\pi 2f_c t)\,\mathrm dt
\approx |x(nT)|^2 \int_{nT}^{(n+1)T}\cos(2\pi 2f_c t)\,\mathrm dt \approx 0.$$
Note also that 
$\displaystyle \int \cos(2\pi 2f_c t)\,\mathrm dt 
= -\frac{1}{4\pi f_c}\sin(2\pi 2f_c t)$ so that any contribution from
$\int_{nT}^{(n+1)T}|x(t)|^2\cos(2\pi 2f_c t)\,\mathrm dt$, which we have
already argued is going to be small is further tamped down towards $0$ by
that $\frac{1}{4\pi f_c} &lt;&lt; 1$ factor.

Calculations such as these are more easily handled in the frequency
domain. We have the Fourier transform pair
$$x(t)\cos(2\pi f_c t) \leftrightarrow 
\left.\left.\frac 12\right[X(f+f_c)+X(f-f_c)\right]$$
so that Parseval's theorem gives us that
$$\begin{align}
\int_{-\infty}^\infty |x(t)\cos(2\pi f_c t)|^2\,\mathrm dt
&amp;= \frac 14 \int_{-\infty}^\infty |X(f+f_c)+X(f-f_c)|^2\,\mathrm df\\
&amp;= \frac 14 \left[\int_{-\infty}^\infty |X(f+f_c)|^2\,\mathrm df
+ \int_{-\infty}^\infty |X(f-f_c)|^2\,\mathrm df
\right.\\
&amp;= \qquad \left. + \int_{-\infty}^\infty X(f+f_c)X^*(f-f_c)
+X^*(f+f_c)X(f-f_c)\,\mathrm df\right]\\
&amp;= \frac 12 E_x + \frac 14 \left[\int_{-\infty}^\infty X(f+f_c)X^*(f-f_c)
+X^*(f+f_c)X(f-f_c)\,\mathrm df\right].\tag{5}
\end{align}$$
We now argue (as in the comments) that if the support
of $X(f)$ is $[-W,W]$ and $f_c &gt; W$, then the supports of
$X(f+f_c)$ and $X(f-f_c)$ are $[-f_c-W, -f_c+W]$ and
$[f_c-W, f_c+W]$ respectively, and since $-f_c+W &lt; 0 &lt; f_c-W$,
the supports do not overlap so that the integrand in $(5)$
has value $0$ for all $f$. Thus, $f_c &gt; W$ is an explicit sufficient
condition for the energy in $x(t)\cos(2\pi f_c t)$ to be half of
the energy in $x(t)$.  The modulation process "loses" half the
energy because $\cos(2\pi f_c t)$ is a carrier signal of power $\frac 12$;
if we had used a unit-power signal such as $\sqrt{2}\cos(2\pi f_c t)$,
there would be no "loss".