Double exponential distribution is not an exponential family

Define a one-parameter exponential family as a family of densities of the form $$f_\theta(x)=\exp(\eta(\theta)T(x) + \xi(\theta))h(x)$$ where $T(x)$ and $h(x)$ are Borel functions, $\theta\in\Theta\subset\mathbb R$ and $\eta$ and $\xi$ are real-valued functions defined on $\Theta$.

Double exponential distribution is a distribution having the density $$p_\theta(x)= \frac{1}{2}\exp(-|x - \theta|)$$ for $\theta\in\mathbb R$.

I am looking for a simple proof of the theorem in the title. I found a proof in the book of Shao "Mathematical Statistics. Exercises and Solutions." but it uses a more general definition of exponential families and doesn't show why the classes are not compatible. What is the special feature of $p_\theta(x)$ that makes the representation as exponential family impossible?

Assume that $\Theta$ has at least three distinct points $\theta_1$, $\theta_2,$ and $\theta_3$. Suppose that $$\exp(\eta(\theta)T(x)+\xi(\theta)) h(x)={1\over 2}\exp(-|x-\theta|).$$ Since $h(x)$ never takes the value zero, we can write it as $h(x)={1\over 2}\exp(w(x))$, and deduce that, for all $x\in\mathbb{R}$ and $\theta\in\Theta$, $$\eta(\theta)T(x)+\xi(\theta) +w(x)= -|x-\theta|.$$

Substitute $\theta_1, \theta_2$ and subtract the two equations to get $$[\eta(\theta_1)-\eta(\theta_2)]\ T(x)+\xi(\theta_1)- \xi(\theta_2) = |x-\theta_2|-|x-\theta_1|.$$ Since the right hand side is not a constant function of $x$, we find that $\eta(\theta_1)\neq\eta(\theta_2)$ and hence that $T$ is differentiable in $x$, except possibly at $\theta_1$ and $\theta_2$. The same argument using the pairs $\{\theta_1 ,\theta_3\}$ and $\{\theta_2 ,\theta_3\}$ shows that $T$ is, in fact, differentiable everywhere.

We conclude that $|x-\theta_2|-|x-\theta_1|$ is everywhere differentiable in $x$, which is a contradiction.

Meanwhile I figured out another proof. But the one of Byron is clearly more elegant.

Following the idea of Shao we consider the quotients $\frac{p_\theta(x)}{p_{-\theta}(x)}= \frac{f_\theta(x)}{f_{-\theta}(x)}$. This allows us to get rid of $h(x)$ and yields \begin{equation} |x+\theta| - |x-\theta| = \left( \eta(\theta) - \eta(-\theta) \right)T(x) - \left( \xi(\theta) - \xi(-\theta) \right) \end{equation} Since $\eta(\theta) - \eta(-\theta)$ must be non-zero for some $\theta$ we define \begin{equation} A = \eta(\theta) - \eta(-\theta) \end{equation} \begin{equation} B =\xi(\theta) - \xi(-\theta) \end{equation} and get $ |x+\theta| - |x-\theta| = A T(x) + B $. Moreover $|x+\theta| - |x-\theta|$ is an antisymmetric function and therefore $A T(x) + B = -(A T(-x) + B)$. This yields $T(x) + \frac{B}{A} = - T(-x) - \frac{B}{A}$ and implies that, $T(x)$ shifted by $\frac{B}{A}$ is antisymmetric.

On the other hand, if we consider $\frac{p_\theta(x)}{p_{0}(x)}= \frac{f_\theta(x)}{f_{0}(x)}$ we get \begin{equation} |x| - |x-\theta| = \left( \eta(\theta) - \eta(0) \right)T(x) - \left( \xi(\theta) - \xi(0) \right) \end{equation} For positive $\theta$ this function is constant in $x$ for $x<0$. But $T(x)$ is also antisymmetric. This implies $T(x) \equiv T$ is constant for all $x\in\mathbb R$. A contradiction.