\begin{claim}
If $x(n)$ is a real-valued signal of an even function, then its fourier transform is even and real-valued. In other words, 

\begin{nicebox}
$$x \in \left[ \Z \to \R \right]  \mbox{ AND } x(n) = x(-n) \mbox{ } \forall n$$
$$\Rightarrow$$
$$X(-\omega) = X(\omega) \mbox{ AND } X \in \left[ \R \to \R \right]$$
\end{nicebox}

\end{claim}

To show that the frequency response is odd, we show that $X(\omega) = X(-\omega)$. To show that the frequency response is real-valued, we need to show that the conjugate of the frequency response is equal to the frequency response, namely $X(\omega) = X^*(\omega)$

\begin{proof}

First show that the frequency response is even by noting some properties of the DTFT:
\begin{align*}
x(n) &\ftp X(\omega) \\
x(-n) &\ftp X(-\omega) \\
\end{align*}

Putting it all together,

\begin{align*}
x(n) &= x(-n) \\
\ftrans{x(n)} &= \ftrans{x(-n)} \\
X(\omega) &= X(-\omega)
\end{align*}

Now to show that the frequency response is real, we simply note that $X(\omega) = X^*(-\omega)$ because $x$ is real-valued, then 
\begin{align*}
X(-\omega) &= X(\omega) = X^*(-\omega) = X^*(\omega) \\
X(\omega) &= X^*(\omega) \\
\Rightarrow X(\omega) &\in \R \mbox{ } \forall \omega
\end{align*}

\end{proof}

Now we make make the general claim that the fourier transform of a real-valued even signal is equal to the real part of the signal's frequency response:

\begin{align*}
\ftrans{\frac{x(n) + x(-n)}{2}} &= \Real{\frac{X(\omega)+X(-\omega)}{2}} \\
\ftrans{x_e(n)} &= \Real{\frac{1}{2}X(\omega)+\frac{1}{2}X(-\omega)} \\
\ftrans{x_e(n)} &= \Real{\frac{1}{2}X(\omega)+\frac{1}{2}X(\omega)} \quad \mbox{since } X(\omega) = X(-\omega) \\
\ftrans{x_e(n)} &= \Real{X(\omega)} \\
\end{align*}

\begin{nicebox}
$$x \in \left[ \Z \to \R \right] $$
$$\Rightarrow$$
$$x_e(n) \ftp \Real{X(\omega)} $$
\end{nicebox}


Signals of Even Functions