Unitary representation of the 2-D Poincare Group Define
$$U_1(\lambda, a)f^+ := (f_{\lambda, a})^+,$$
where $f_{\lambda, a}(x) = f(\Lambda_{\lambda}^{-1}(x-a))$ with
$$\Lambda_{\lambda} = \begin{pmatrix} \cosh(\lambda) & \sinh(\lambda) \\ \sinh(\lambda) & \cosh(\lambda) \end{pmatrix}.$$
Show that $U_1$ is a Unitary representation of the 2-D Poincare Group.
I understand that the definition of a unitary representation means that we have to show that essentially $U_1(f_1f_2) = U_1f_1\cdot U_2f_2$ and that these $U_1$ are unitary operators, but I'm not sure how to compute this.
EDIT:
Let the Hilbert space be $L^2 ( \mathbb{R}, d\theta)$, then for $f \in C^{\infty}_0 (\mathbb{R}^2)$ (meaning a smooth function with compact support dependent on 2 variables) we then define a new function $f^{+}(\theta) : = \tilde{f}(p(\theta))$ where $\tilde{f}$ is the Fourier transform of $f$ and $p(\theta) = m\begin{pmatrix} \cosh(\theta) \\ \sinh(\theta) \end{pmatrix}$ with $m>0$ a constant. With $f_{\lambda, a}$ defined as before, you can then define $(f_{\lambda,a})^+ (\theta)$ which then leads to the question of the operator $U$.
 A: My comment was getting too long to simply remain a comment so I typed out a full answer, I hope this is more clear.
Abstractly the Poincare group is simply the set 
of ordered pairs $(\Lambda, a )$
where $\Lambda$ is a 2-dimensional Lorentz transformation and $a \in \mathbb{R}^2$,
along with the multiplication rule $(\Lambda, a) \cdot (\bar{\Lambda}, b) = (\Lambda \cdot \bar{\Lambda}, \Lambda b +  a)$.
This is simply how multiplication for elements of the Poincare group are defined. You could consider some other multiplication rule, but then you won't necessarily be talking about the Poincare group. 
As I will demonstrate, this is simply the usual definition of the Poincare group as defined in physics, just in a bit more abstract terms. Let $x^\mu$ be a set of coordinates. Consider the coordinate transformation of $x^\mu$ defined by $x'^{\mu} = {\bar{\Lambda}}^\mu_{\; \nu} x^\nu  +b^\mu$ where $
\bar{\Lambda}$ is a Lorentz transformation and $b \in \mathbb{R}^2$ is a vector. Now consider a second transformation $x''^{\mu}= {\Lambda}^\mu_{\; \nu} x'^\nu  +a^\mu$ again where $
\Lambda$ is a Lorentz transformation and $a \in \mathbb{R}^2$ is a vector. Considering the composition of those two transformation gives us
$$
x''^{\mu} = \Lambda^\mu_{\; \nu} (\bar{\Lambda}^\nu_{\; \alpha} x^\alpha  +b^\nu) +a^\mu = {(\Lambda \bar{\Lambda})}^{\mu}_{\; \nu} x^\nu + \Lambda^{\mu}_{\; \nu} b^\nu + a^\mu.
$$
So indeed this corresponds to the Poincare transformation with Lorentz transformation $\Lambda \bar{\Lambda}$ and translation by $\Lambda b+ a$.

Now to prove a map $U:G \to \text{GL}(V)$ with $G$ a group and $V$ a vector space, defines a Unitary representation you need to prove $U$ is a group homomorphism and $U(g)$ is unitary for every $g \in G$. In other words $$
U(\Lambda, a) U(\bar{\Lambda},b) = U(\Lambda \bar{\Lambda}, \Lambda b +a).
$$
This isn't too hard to check, it's just a little bit of manipulation of the matrices. 
To prove that $U$ is unitary will probably be a bit harder. Let $\langle \cdot, \cdot \rangle$ be the inner product of $L^2(\mathbb{R}, d\theta)$. Then you have to prove that for every transformation $(\Lambda, a)$ we have
$$
U(\lambda, a)^\dagger = U(-\lambda, - \Lambda_{-\lambda} a)
$$
because the inverse of $(\lambda, a)$ is $(-\lambda, - \Lambda_{-\lambda} a)$.
Equivalently you can prove that for $f,g \in C^\infty_0(\mathbb{R}^2)$ we have
$$
\langle f_{\lambda,a}^+, g^+ \rangle= \langle f^+, U(\lambda, a)^\dagger g^+ \rangle  = \langle f^+,U(-\lambda, - \Lambda_{-\lambda} a)g^+ \rangle = \langle f^+, g_{-\lambda, - \Lambda_{-\lambda} a}^+ \rangle.
$$
To explicitly verify this you will probably have to perform some change of variable in the integral.
