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You wrote in fact the equation $\dot u+A u=0$ in a suitable weak formulation which makes sense provided $V$ can be embedded in $V'$. Now what you call the null set of the test function does not appear in the equation: you multiply both sides of your equation by any continuous function $\phi$ and you integrate with respect to $t$, getting an integral version of your equation.
@Danqing You are right, I have withdrawn the middle part of my answer. The commutation of the essential supremum with a supremum may pose some problems.
