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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 $L^2(0,\infty;H^1_0(\Omega))$.