# Prove operator $T$ is onto

Consider the Hilbert spaces $X := H^{1}(\Omega)\times H^{1}(\Omega)$ and $Y:=L^2(\Omega)\times L^2(\Omega)$, where $\Omega =\ ]{-}\pi, \pi[$, and \begin{eqnarray*} \langle(u,v), (z,w)\rangle_X & = & \langle u,z\rangle_{H^1(\Omega)} + \langle v,w\rangle_{H^1(\Omega)},\\ & =& \int_{\Omega}\left\{uz + u^{\prime}z^{\prime} + vw + v^{\prime}w^{\prime}\right\},\quad \forall\ (u,v),(z,w)\in X,\\[0.2cm] \langle(u,v), (z,w)\rangle_Y & = & \langle u,z\rangle_{L^2(\Omega)} + \langle v,w\rangle_{L^2(\Omega)},\\ & = & \int_{\Omega}\left\{uz + vw\right\},\quad\quad\quad\quad\quad\quad\quad \forall\ (u,v),(z,w)\in Y. \end{eqnarray*} Now, define the operator $$\begin{array}{rcccl} T & : & X & \longrightarrow & Y,\\ & & (u,v) & \longmapsto & T(u,v)\ :=\ (u^{\prime}, v^{\prime}). \end{array}$$ Is very easy to check that $T\in\mathcal{L}(X,Y)$, but the problem (for me) is: How can I prove that $T$ is onto, i.e, $\mbox{Im}(T) = Y$ ?

By the way, I read that $A:H^m(\Omega)\rightarrow L^2(\Omega)$, defined by $A(u) = u^{(m)}$ is onto, but I really don't know why, and I think that may be usefull in order to prove that $T$ is onto.
(1) It's sufficient to show that $T_1 u = u'$ is surjective as a map from $H^1(\Omega)$ to $L^2(\Omega)$; can you see why? (2) The fundamental theorem of calculus tells you what the "inverse" of $T_1$ should be. – Nate Eldredge Jun 25 '13 at 1:22