A operator is unitary if and only if it is a surjective isometry I'm trying to prove the following result.
Let U be an operator of a Hilbert space H, then $U$ is an unitary operator $\iff$ $U$ is an isometry and $R_u = H$ ($U$ is onto and isometry)
I tried to use the fact that $U$ is a linear operator and the fact that $U^*U=I$ defines an isometry but I couldn't proceed.
 A: Let $U$ be a unitary, then $UU^*=U^*U=1$.
$U^*U=1$ implies that $U$ is an isometry. Also $UU^*=1 $ implies that $U$ is onto, because ($UU^*H=H$).
Conversely, Suppose $U$ is an onto isometry, so $U^*U=1$. It's just necessary to show that $UU^* = 1$. If $\eta\in H$, then there is $\xi\in H$ such that $U\xi=\eta$ ($U $ is onto). Also $U$ is an isometry, so $ \xi=U^*U\xi = U^*\eta$ .Now clearly $UU^*\eta = U\xi =\eta$ for $\eta \in H$. Therefore $UU^*=1$.
A: Below I will prove the equivalence in a little more general settings.
The equivalence to be proven:

If  $\mathcal{H}_1$ and $\mathcal{H}_2$ are Hilbert spaces over $\mathbb{C}$ with inner products $\langle \cdot,\cdot \rangle_1$ and $\langle \cdot,\cdot \rangle_2$, respectively, and $U:\mathcal{H}_1\rightarrow\mathcal{H}_2$ is a bounded linear operator between them, then $U$ is unitary $\Longleftrightarrow$ $U$ is surjective and an isometry.

To avoid any misunderstanding, below is the definition of a unitary operator:

Given the above settings, a unitary operator from $\mathcal{H}_1$ to $\mathcal{H}_2$ is an invertible linear operator $U$ that preserves inner products: $\langle Ux,Uy\rangle_2=\langle x,y\rangle_1$ for $\forall x,y\in\mathcal{H}_1$.

Note that this definition is equivalent to the normal form that $UU^*=U^*U=I$.

*

*Proof of $\Longrightarrow$:

Since the definition of unitary operator requires that it be bijective, so $U$ is surjective. Isometry can be established by equalities $\|Ux\|_2^2=\langle Ux,Ux\rangle_2=\langle x,x \rangle_1=\|x\|_1^2$ for $\forall x\in \mathcal{H}_1$, which are just applications of the definition.

*

*Proof of $\Longleftarrow$:

Before we proceed, let's rewrite the polarization identity as follows:

Re$\langle x,y\rangle=\frac{1}{4}(\|x+y\|^2-\|x-y\|^2),$
Im$\langle x,y\rangle=\frac{1}{4}(\|x+iy\|^2-\|x-iy\|^2)$.

Supposing any $x,y\in\mathcal{H}_1$ satisfying $Ux=Uy$, we have $0=\|Ux-Uy\|_2=\|U(x-y)\|_2=\|x-y\|_1$ due to isometry. So $x=y$ by positive definiteness of norm. This proves injectivity of $U$. Together with the surjectivity, we know $U$ is bijective and therefore invertible. Boundedness is clear from isometry.
To show $\langle Ux,Uy\rangle_2=\langle x,y\rangle_1$, we need to show that this equality is true for both the real and imaginary part (note that the inner product is a complex number). But this is straightforward in what follows:
$\begin{eqnarray}
\mathop{\rm Re}\langle Ux,Uy\rangle_2&=&\frac{1}{4}(\|Ux+Uy\|_2^2-\|Ux-Uy\|_2^2)\\
&=&\frac{1}{4}(\|U(x+y)\|_2^2-\|U(x-y)\|_2^2)\\
&=&\frac{1}{4}(\|x+y\|_1^2-\|x-y\|_1^2)\\
&=&\mathop{\rm Re}\langle x,y\rangle_1\enspace,
\end{eqnarray}$
$\begin{eqnarray}
\mathop{\rm Im}\langle Ux,Uy\rangle_2&=&\frac{1}{4}(\|Ux+iUy\|_2^2-\|Ux-iUy\|_2^2)\\
&=&\frac{1}{4}(\|U(x+iy)\|_2^2-\|U(x-iy)\|_2^2)\\
&=&\frac{1}{4}(\|x+iy\|_1^2-\|x-iy\|_1^2)\\
&=&\mathop{\rm Im}\langle x,y\rangle_1\enspace.
\end{eqnarray}$
