i have no idea what the "reconstruction fidelity term" is or what it's about.
Hermitian symmetry is a term usually applied to some form the Fourier Transform of a signal that is purely real.
for continuous-time, continuous-frequency Fourier Transform:
$$ X(f) = \int\limits_{-\infty}^{\infty} x(t) \, e^{-j 2 \pi f t} \ dt $$
if $x(t)$ is purely real (that is $\Im\{x(t)\}=0$ for all real $t$), then we know that there is this symmetry about $f=0$:
$$ X(-f) = X(f)^* = \operatorname{conj}\{X(f)\} \qquad \forall f \in \mathbb{R} $$
or
$$\begin{align}
\Re\{X(-f)\} &= \Re\{X(f)\} \\
\Im\{X(-f)\} &= -\Im\{X(f)\} \\
|X(-f)| &= |X(f)| \\
\arg\{X(-f)\} &= -\arg\{X(f)\} \\
\end{align}$$
Similarly, for the Discrete Fourier Transform:
$$ X[k] = \sum\limits_{n=0}^{N-1} x[n] \, e^{-j 2 \pi \frac{nk}{N}} $$
if $x[n]$ is purely real (that is $\Im\{x[n]\}=0$ for integer $n \in [0..N-1]$), then we know that there is this symmetry about $k=\tfrac{N}{2}$:
$$ X[N-k] = X[k]^* = \operatorname{conj}\{X[k]\} \qquad 1 \le k \le N-1 $$
or
$$\begin{align}
\Re\{X[N-k]\} &= \Re\{X[k]\} \\
\Im\{X[N-k]\} &= -\Im\{X[k]\} \\
|X[N-k]| &= |X[k]| \\
\arg\{X[N-k]\} &= -\arg\{X[k]\} \\
\end{align}$$
Now in my religion regarding the DFT, i insist that
$$\begin{align}
x[n+N] &= x[n] \qquad & \forall n \in \mathbb{Z} \\
X[k+N] &= X[k] \qquad & \forall k \in \mathbb{Z} \\
\end{align}$$
this is always the case (the DFT maps a discrete and periodic sequence of period $N$ in one domain to another discrete and periodic sequence of period $N$ in the reciprocal domain). then the Hermitian symmetry takes a simpler form:
if $x[n]$ is purely real (that is $\Im\{x[n]\}=0$ for all integer $n$), then we know that there is this symmetry about $k=0$:
$$ X[-k] = X[k]^* = \operatorname{conj}\{X[k]\} \qquad \forall k \in \mathbb{Z} $$
or
$$\begin{align}
\Re\{X[-k]\} &= \Re\{X[k]\} \\
\Im\{X[-k]\} &= -\Im\{X[k]\} \\
|X[-k]| &= |X[k]| \\
\arg\{X[-k]\} &= -\arg\{X[k]\} \\
\end{align}$$
likewise you can identify Hermitian symmetry for the DTFT and relate it to the real-ness of the input to the DTFT. (do i have to do that one, too?)