Eignvalues of Laplacian operator and Sobolev spaces Let $\Omega$ an open bounded in $\mathbb{R}^n$ and $I$ un interval in $\mathbb{R}$, let $t_0 \in I$, $f\in H^1_0(\Omega)$, $g\in L^2(\Omega)$. Let $(\lambda_n)_{n\in \mathbb{N}}$ the eigen values od the Laplacian operator $\Delta$, and $(e_n)_n$ the vector associates to the eigen values.
We denote 
$$u=\sum_{n=1}^{+\infty} [(f,e_n) \cos[(t-t_0) \sqrt{\lambda_n}] + \dfrac{(g,e_n)}{\sqrt{\lambda_n}}\sin[(t-t_0)\sqrt{\lambda_n}]]e_n$$
How we can prouve that $u \in C^0(I,H^1_0(\Omega))$ and $\dfrac{\partial u}{\partial t} \in C^0(I,L^2(\Omega))$?
II
I undestand the answer of this question. I have an seconde question please. 
How we prouve that $$\Box u = 0 \quad\mbox{in} \quad \mathcal{D}'(I^{0} \times \Omega)$$
where $\Box$ is wave operator, and $I^{0}$ is the interior od $u$.
Thank's for the help beaucause i lost.
 A: I'll assume $\Omega$ has a smooth boundary (or something similar.) The Laplacian has a natural domain $\mathcal{D}(\Delta)=H^{2}(\Omega)\cap H^{1}_{0}(\Omega)$ because
$$
        -\Delta : \mathcal{D}(\Delta) \subset L^{2}(\Omega)\rightarrow L^{2}(\Omega)
$$
is selfadjoint and non-negative on its domain. The domain of the positive square root of $-\Delta$ is $H_{0}^{1}(\Omega)$. Furthermore, $-\Delta$ has a complete orthonormal set of eigenfuntions $\{ e_{n}\}_{n=0}^{\infty}$ with corresponding eigenvalues $\{ \lambda_{n} \}_{n=0}^{\infty} \subset (0,\infty)$. By the spectral theorem,
$$
   f \in \mathcal{D}(-\Delta) \iff \sum_{n=0}^{\infty}|\lambda_{n}(f,e_{n})|^{2} < \infty,\;\;\;
   f \in \mathcal{D}(\sqrt{-\Delta}) \iff \sum_{n=0}^{\infty}|\sqrt{\lambda_{n}}(f,e_{n})|^{2} < \infty.
$$
Using Friedrichs extensions, one finds that $\mathcal{D}(\sqrt{-\Delta})=H_{0}^{1}(\Omega)$. So the above characterization of $\mathcal{D}(\sqrt{-\Delta})$ becomes a characterization of $H_{0}^{1}$.
Because $\sqrt{-\Delta}$ is selfadjoint, then $f(t)=e^{it\sqrt{-\Delta}}f_{0}$ is a continuous function from $[0,\infty)$ to $L^{2}(\Omega)$ which is given by Functional Calculus:
$$
         f(t) = e^{it\sqrt{-\Delta}}f_{0}=\sum_{n=0}^{\infty} e^{i\sqrt{\lambda_{n}}t}(f_{0},e_{n})e_{n}.
$$
If $f_{0} \in \mathcal{D}(\sqrt{-\Delta})=H_{0}^{1}(\Omega)$, then $f$ is a continuous function from $[0,\infty)$ to $H_{0}^{1}(\Omega)$, and $f'(t)=\sqrt{-\Delta}f(t)=e^{it\sqrt{-\Delta}}(\sqrt{-\Delta}f_{0})$ is a continuous function from $[0,\infty)$ to $L^{2}(\Omega)$. Your situation is a mild variation of this.
Your function $u$: (Discussion added after your questions.) So you have defined $u=v+w$ where
$$
    v(t) = \sum_{n=0}^{\infty}(f,e_{n})\cos((t-t_{0})\sqrt{\lambda_{n}})e_{n},
                 \;\;\; f \in H_{0}^{1}(\Omega)\\
    w(t) = \sum_{n=0}^{\infty}\frac{(g,e_{n})}{\sqrt{\lambda_{n}}}\sin((t-t_{0})\sqrt{\lambda_{n}})e_{n},\;\;\; g \in L^{2}(\Omega).
$$
Both $v$ and $u$ are in $L^{2}(\Omega)$ for all $t\in\mathbb{R}$ because the Fourier coefficents with respect to the complete orthonormal set $\{ e_{n}\}_{n=0}^{\infty}$ in each series are square integrable:
$$
        \|v(t)\|^{2}_{L^{2}}=\sum_{n=0}^{\infty}|(f,e_{n})|^{2}\cos^{2}((t-t_{0})\sqrt{\lambda_{n}})
                \le \sum_{n} |(f,e_{n})|^{2} = \|f\|^{2},\\
      \|w(t)\|_{L^{2}}^{2}=\sum_{n=0}^{\infty}\frac{|(g,e_{n})|^{2}}{\lambda_{n}}\sin^{2}((t-t_{0})\sqrt{\lambda_{n}}) \le \sum_{n=0}^{\infty}\frac{1}{\lambda_{1}}|(g,e_{n})|^{2}
               = \frac{1}{\lambda_{1}}\|g\|^{2}.
$$
Because $f \in H^{1}_{0}$, then the equivalent condition $\sum_{n=0}^{\infty}|\sqrt{\lambda_{n}}(f,e_{n})|^{2} < \infty$ holds.
This forces $v(t) \in H^{1}_{0}$ for all $t$ because
$$
    \sum_{n=0}^{\infty}|\sqrt{\lambda_{n}}(v(t),e_{n})|^{2}=
      \sum_{n=0}^{\infty}|\sqrt{\lambda_{n}}(f,e_{n})|^{2}\sin^{2}((t-t_{0})\sqrt{\lambda_{n}})\le \sum_{n=0}^{\infty}|\sqrt{\lambda_{n}}(f,e_{n})|^{2} < \infty.
$$
Similarly, $w(t) \in H^{1}_{0}$ for all $t$ because $g\in L^{2}$ and you're dividing the Fourier coefficients of $g$ by $\sqrt{\lambda_{n}}$, which puts the result in $H^{1}_{0}$. So everything is properly defined and $u(t) \in H^{1}_{0}$ for all $t$.
As I pointed out before, $h \in H^{1}_{0}$ iff $h\in\mathcal{D}(\sqrt{-\Delta})$, and, in that case,
$$
               \|h\|^{2}_{H^{1}_{0}}=\|h\|^{2}_{L^{2}}+\|\nabla h\|^{2}_{L^{2}}
      = \|h\|^{2}_{L^{2}}+\|\sqrt{-\Delta}h\|^{2}_{L^{2}}=\sum_{n=0}^{\infty}(1+\lambda_{n})|(f,e_{n})|^{2}.
$$
In fact, for $h \in L^{2}$, the sum on the right is finite iff $h \in H^{1}_{0}$. To shorten the expressions, let $s_{n}(t)=\sin((t-t_{0})\sqrt{\lambda_{n}})$ and $c_{n}(t)=\cos((t-t_{0})\sqrt{\lambda_{n}})$. What you've got is
$$
\begin{align}
      \|u(t)-u(t')\|_{H^{1}_{0}}
   & \le\left[\sum_{n=0}^{\infty}(1+\lambda_{n})|(f,e_{n})|^{2}|c_{n}(t)-c_{n}(t')|^{2}\right]^{1/2} \\
   & +\left[\sum_{n=0}^{\infty}\frac{1+\lambda_{n}}{\lambda_{n}}|(g,e_{n})|^{2}|s_{n}(t)-s_{n}(t')|^{2}\right]^{1/2}.
\end{align}
$$
The sums are absolutely and uniformly convergent with respect to $t$, $t'$ because $f \in H^{1}_{0}$ and $g\in L^{2}$. It follows that the above is continuous in $t,t'$. So $u : \mathbb{R}\rightarrow H^{1}_{0}$ is a continuous vector function. And when you take a derivative of $u$, you end up with a continuous function $u' : \mathbb{R}\rightarrow L^{2}$. The reduced regularlity is because of the added $\sqrt{\lambda_{n}}$ which appears in the derived sums. I'll let you work out the details.
Your New Modifications: The new mention of $\Box$ requires that you be able to take two time derivatives. This can be done if $f\in \mathcal{D}(-\Delta)$ and $g \in \mathcal{D}(\sqrt{-\Delta})=H_{0}^{1}(\Omega)$. If $f$, $g$ are as stated, then taking two time derivatives gives the same result as applying $-\Delta$ (both multiply the n-th Fourier coefficient by $\lambda_{n}$.) The expression $\Box$ is now interpreted in terms of a derivative of a vector function in $t$ and in terms of an operator $-\Delta$, which is more subtle than classical derivatives. So $\frac{d^{2}}{dt^{2}}u(t) = \Delta u(t)$ in this operator language, where $u : [t_{0},\infty)\rightarrow L^{2}(\Omega)$ is a vector function whose value lies in the domain of $-\Delta$ for all $t > 0$ so that $-\Delta u(t)$ makes sense, provided $f \in \mathcal{D}(-\Delta)$ and $g\in H_{0}^{1}$.
A: I'm adding this about a weak solution. Editing is becoming slow, so I'm adding another post.
I'm assuming that $f \in \mathcal{D}(-\Delta)$ and that $g \in \mathcal{D}(\sqrt{-\Delta})=H^{1}_{0}(\Omega)$. That way, $u(t)$ as you have defined is a continuously differentiable vector function of $t$ from $\mathbb{R}$ to $\mathcal{D}(-\Delta)$. And,
$$
\begin{align}
    (-\Delta)u(t) & =(-\Delta)
     \sum_{n=1}^{\infty}\left[(f,e_{n})\cos((t-t_{0})\sqrt{\lambda_{n}})
 + \frac{(g,e_{n})}{\sqrt{\lambda_{n}}}\sin((t-t_{0})\sqrt{\lambda_{n}})\right]e_{n} \\
   & =
     \sum_{n=1}^{\infty}\left[(f,e_{n})\cos((t-t_{0})\sqrt{\lambda_{n}})
 + \frac{(g,e_{n})}{\sqrt{\lambda_{n}}}\sin((t-t_{0})\sqrt{\lambda_{n}})\right]\lambda_{n}e_{n} \\
   & =
     -\frac{d^{2}}{dt^{2}}\sum_{n=1}^{\infty}\left[(f,e_{n})\cos((t-t_{0})\sqrt{\lambda_{n}})
 + \frac{(g,e_{n})}{\sqrt{\lambda_{n}}}\sin((t-t_{0})\sqrt{\lambda_{n}})\right]e_{n}\;.
\end{align}
$$
All of the sums involved are convergent in $L^{2}$ because of the assumptions on $f$, $g$. Suppose that $\psi \in C^{\infty}_{c}(I\times \Omega)$. Then $t\mapsto \psi(x,t)$ can be considered to be an infinitely differentiable function of $t$ with values in $L^{2}(\Omega)$. Using the inner-product $(\cdot,\cdot)$ on $L^{2}(\Omega)$, the function $t\mapsto (u(t),\psi(\cdot,t))$ is a twice continuously function of $t$ which is compactly supported in $I$. Furthermore,
$$
\begin{align}
  (\frac{d^{2}}{dt^{2}}u(t),\psi)
      &=\frac{d}{dt}(\frac{d}{dt}u(t),\psi)
        -(\frac{d}{dt}u(t),\frac{\partial}{\partial t}\psi) \\
  (u(t),\frac{\partial^{2}}{\partial t^{2}}\psi)
      &=\frac{d}{dt}(u(t),\frac{\partial}{\partial t}\psi)
        -(\frac{d}{dt}u(t),\frac{\partial}{\partial t}\psi)
\end{align}
$$
Integrating over $I$ then gives
$$
     \int_{I}(\frac{d^{2}}{dt^{2}}u,\psi)\,dt=-\int_{I}(\frac{d}{dt}u,\frac{\partial}{\partial t}\psi)\,dt = \int_{I}(u,\frac{\partial^{2}}{\partial t^{2}}\psi)\,dt.
$$
For fixed $t$, one can consider $\psi(t,\cdot)$ to be a vector in $\mathcal{D}(-\Delta)$, and, because $-\Delta$ is selfadjoint on $L^{2}(\Omega)$,
$$
        (\frac{d^{2}}{dt^{2}}u(t),\psi)=(\Delta u(t),\psi)=(u(t),\Delta\psi). 
$$
Combining these last two equations gives
$$
        \int_{I}(u,\Delta\psi)\,dt = \int_{I}(u,\frac{\partial^{2}}{\partial t^{2}}\psi)\,dt
$$
In other words,
$$
          (u,\Box \psi)_{L^{2}(I\times\Omega)}=0,\;\;\; \psi \in C_{c}^{\infty}(I\times\Omega).
$$
Therefore $T_{u} \in \mathcal{D}'(I^{0}\times\Omega)$ defined by $T_{u}(\psi)=(u,\psi)_{L^{2}(I\times\Omega)}$ satisfies the weak equation $\Box T_{u}=0$.
You've added more to your question. Assuming $F=F(t)$ is in $C^{0}(I,L^{2}(\Omega))$, you must now solve for $u(t) \in L^{2}(\Omega)$ such that
$$
           \Box u = F,\;\;\; u(0)=u'(0)=0.
$$
This $u$ can be found by assuming the separation of variables solution $u(t)=\sum_{n}a_{n}(t)e_{n}$. Formally, the equation reduces to
$$
   \sum_{n}(a_{n}''+\lambda_{n}a_{n})e_{n} = \sum_{n}(F,e_{n})e_{n},\\
        a_{n}(0)=a_{n}'(0)=0.
$$
So the coefficients $a_{n}$ satisfy and ODE
$$
             a_{n}''(t)+\lambda_{n}a_{n}(t) = (F,e_{n}),\;\; a_{n}(0)=a_{n}'(0)=0.
$$
The function $(F,e_{n})$ is an inner-product in $L^{2}(\Omega)$, and $(F,e_{n})$ is a continuous function of $t$ because $t\mapsto F(t)\in L^{2}(\Omega)$ is continuous. The solution I find by variation of parameters is the same as yours:
$$
\begin{align}
        a_{n} & =-\frac{1}{\sqrt{\lambda_{n}}}\cos(\sqrt{\lambda_{n}}t)\int_{0}^{t}\sin(\sqrt{\lambda_{n}}s)(F,e_{n})|_{s}\,ds \\
      & +\frac{1}{\sqrt{\lambda_{n}}}\sin(\sqrt{\lambda_{n}}t)\int_{0}^{t}\cos(\sqrt{\lambda_{n}}s)(F,e_{n})|_{s}\,ds \\
    & = \frac{1}{\sqrt{\lambda_{n}}}\int_{0}^{t}\sin(\sqrt{\lambda_{n}}(t-s))(F,e_{n})|_{s}\,ds.
\end{align}
$$
The coefficients $a_{n}(t)$ are twice continuously differentiable and satisfy
$$
    |\sqrt{\lambda_{n}}a_{n}|^{2} \le \int_{0}^{t}|\sin(\cdots)|^{2}\,ds\int_{0}^{t}|(F,e_{n})|^{2}\,ds.
$$
So, as a crude estimate, $|\sin(\cdots)|\le 1$ and
$$
     \sum_{n}|\sqrt{\lambda_{n}}a_{n}|^{2} \le t\int_{0}^{t}\sum_{n}|(F,e_{n})|^{2}\,ds
   = t \int_{0}^{t}\|F(s)\|^{2}\,ds
$$
Therefore, $u(t)=\sum_{n}a_{n}e_{n}$ is in $H^{1}_{0}$ for each $t \ge 0$. It's not too hard show that $t \mapsto u(t)\in H_{0}^{1}$ is continuous. Taking a derivative with respect to $t$ of $u$ leads to sum where the $1/\sqrt{\lambda_{n}}$ disappears, and $t\mapsto u'(t)\in L^{2}(\Omega)$ is continuous.
Note: The previous arguments used to show that you have a weak solution no longer work because the assumptions you have given are not enough to guarantee that $t\mapsto u(t)$ is a twice continuously differentiable vector function. However, it is still the case that $\Box T_{u} = F$ in the weak sense, which adds an interesting and important twist to the problem. The argument is still simple, just a little different. If $\psi \in C_{c}^{\infty}(I^{0}\times \Omega)$, then $\Box(a_{n}(t)e_{n}(x))=(F,e_{n})_{L^{2}(\Omega)}e_{n}$ gives
$$
\begin{align}
        (\Box T_{u})(\psi) = \int_{I^{0}\times\Omega}u\Box \psi\,dx\,dt 
  & = (u,\Box\psi)_{L^{2}(I\times\Omega)} \\
  & =(\sum_{n}a_{n}(t)e_{n}(x),\Box\psi(t,x))_{L^{2}(I\times\Omega)} \\
  & = \sum_{n}(a_{n}e_{n},\Box\psi)_{L^{2}(I\times\Omega)} \\
  & = \sum_{n}(\Box(a_{n}e_{n}),\psi)_{L^{2}(I\times\Omega)} \\
  & = \sum_{n}((F,e_{n})_{L^{2}(\Omega)}e_{n},\psi)_{L^{2}(I\times\Omega)}\\
  & = (\sum_{n}(F,e_{n})_{L^{2}(\Omega)}e_{n},\psi)_{L^{2}(I\times\Omega)}\\
  & = (F,\psi)_{L^{2}(I\times\Omega)}.
\end{align}
$$
Of course, this assumes you know that the eigenfunctions $e_{n}(x)$ are smooth (which they are) and vanish at the boundary of the smooth region $\Omega$ (which they do.)
Justification of interchange of sum and inner-product: You have edited the question again, to ask about the justification for going from line (2) to (3) of the multi-line equation above. All you need to do in order to establish that equality is to show that $\sum_{n} a_{n}(t)e_{n}(x)$ converges in $L^{2}(I\times\Omega)$. First notice that $(a_{n}e_{n},a_{k}e_{k})_{L^{2}(I\times\Omega)}=0$ for $n\ne k$, which follows from the orthogonality of $\{ e_{n} \}$ in $L^{2}(\Omega)$. So, look at
$$
    \|a_{n}e_{n}\|^{2}_{L^{2}(I\times\Omega)}=\|a_{n}\|^{2}_{L^{2}(I)}\|e_{n}\|^{2}_{L^{2}(\Omega)}=\|a_{n}\|^{2}_{L^{2}(I)}
$$
To get what you want, all you need to show is that $\sum_{n}\|a_{n}\|^{2}_{L^{2}(I)} < \infty$. From the crude estimate given above,
$$
     \sqrt{\lambda_{1}}\sum_{n}|a_{n}(t)|^{2} \le \sum_{n}|\sqrt{\lambda_{n}}a_{n}(t)|^{2} \le 
    t \int_{0}^{t}\|F(s)\|_{L^{2}(\Omega)}^{2}\,ds.
$$
The integral over $I$ of the right side is finite; so that gives you all you need to get convergence in $L^{2}(I\times\Omega)$ of the orthogonal series $u = \sum_{n}a_{n}(t)e_{n}(x)$.
