= Solution
A <simple hypothesis> specifies one probability distribution. Let the <null hypothesis> and alternative have <probability density functions> $f_0,f_1$ with respect to a common measure. A possibly randomized <statistical test> is a measurable function $\varphi(x)\in[0,1]$, its conditional probability of rejecting the <null hypothesis>. Its <size of a statistical test> is $\mathbb E_0\varphi$, and its <statistical power> at the alternative is $\mathbb E_1\varphi$. A <most powerful test> at level $\alpha$ maximizes power among tests with size at most $\alpha$.
The <Neyman-Pearson lemma> says to choose $k\geq0$ and a test $\varphi_*$ of size $\alpha$ with $\varphi_*=1$ where $f_1>kf_0$ and $\varphi_*=0$ where $f_1<kf_0$, choosing values on the equality set to reach the specified size. Equivalently, threshold the <likelihood ratio>, treating positive alternative density with zero null density as an infinite ratio. Randomization on a threshold atom ensures exact size when needed. If $k=0$, all remaining null-only mass can be used to reach size without changing power.
For any test $\varphi$ of size at most $\alpha$, the signs of the two factors give
$$
(\varphi_*-\varphi)(f_1-kf_0)\geq0
$$
pointwise. Integrating proves
$$
\mathbb E_1\varphi_*-\mathbb E_1\varphi
\geq k(\mathbb E_0\varphi_*-\mathbb E_0\varphi)\geq0.
$$
This is the required optimality proof. If $k>0$ and another test is equally powerful, both inequalities must be equalities. It must have size $\alpha$ and agree with $\varphi_*$ wherever $f_1\ne kf_0$, apart from null sets for the relevant densities. Thus freedom is confined to the equality set and sets null under both hypotheses; at $k=0$, additional size need not improve power. This also explains the nonuniqueness below.
For the unit-scale <Laplace distribution>, put $\delta=\theta_1-\theta_0>0$. Its <likelihood ratio> is
$$
L(x)=\frac{f(x\mid\theta_1)}{f(x\mid\theta_0)}=
\begin{cases}
e^{-\delta},&x\leq\theta_0,\\
e^{2x-\theta_0-\theta_1},&\theta_0<x<\theta_1,\\
e^\delta,&x\geq\theta_1.
\end{cases}
$$
It is nondecreasing, so an upper-tail <critical region> has the <Neyman-Pearson lemma> form, even if the cutoff lies inside a constant-ratio tail. Solving $\mathbb P_{\theta_0}(X>c_\alpha)=\alpha$ gives the <most powerful test for a Laplace location shift>,
$$
\boxed{\text{reject }H_0\text{ when }X>c_\alpha,\qquad
c_\alpha=\begin{cases}
\theta_0-\log(2\alpha),&0<\alpha\leq1/2,\\
\theta_0+\log(2(1-\alpha)),&1/2<\alpha<1.
\end{cases}}
$$
Continuity of the distribution makes this nonrandomized test exactly size $\alpha$. It need not be the unique <most powerful test>: within a flat extreme <likelihood ratio> region, other selections having the same null probability are equally good.
Its <statistical power> is
$$
\beta(\theta_1)=\begin{cases}
\frac12e^{\theta_1-c_\alpha},&c_\alpha\geq\theta_1,\\
1-\frac12e^{c_\alpha-\theta_1},&c_\alpha<\theta_1.
\end{cases}
$$
Here $\beta$ denotes power, not the probability of a type-II error. For $\alpha=0.05$, $c_\alpha=\theta_0+\log10$. Thus the power is $0.05e^\delta$ for $\delta\leq\log10$, and $1-5e^{-\delta}$ otherwise. The first branch is at most $1/2$. On the second branch, power at least $0.95$ is equivalent to $5e^{-\delta}\leq0.05$. Therefore \b[the required pairs are exactly $\theta_1-\theta_0\geq\log100$], with no restriction on the common location.
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