Take for example the function $u(x)=sin(\pi x)$, which is in $H^1_0(0,1)$, and satisfies  
$$
u_t  -u^{\prime\prime}=f,\quad u_0=sin(\pi x), \mbox{ on } L^2(0,T;H^{-1}(\Omega))
$$
with $f=\pi^2\sin(\pi x)$. The $L^2(0,T;H^1_0(\Omega))$ norm of $u$ is
$$
\|u\|_{L^2(0,T;H^1_0(\Omega))}=\int_0^T \int_0^1 \pi^2 \cos^2(\pi x)dxdt= \frac{T \pi^2}{2}.
$$ 
It isn't in $L^2(0,\infty;H^1_0(\Omega))$.

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This answer corresponded to a previous version of the question. The key point that you skip in your question is indeed if the norm stays bounded!
If your norm is bounded independently of $n$, then you can show directly (by verifying the definition) that the $v$ you have constructed is indeed in $L^2(0,\infty;V)$.