If $E< V(x) $ everywhere, then the TISE implies

$$ 0 ~\leq~ \langle \psi | T | \psi \rangle ~=~ \langle \psi | (E-V) | \psi \rangle~<~ 0, $$

which is impossible. Here $H=T+V$ is the Hamiltonian operator.

If $E< V(x) $ everywhere, and if we assume that the kinetic energy operator $T=\frac{p^{\dagger}p}{2m}$ is a (semi)positive operator, then the TISE implies

$$ 0 ~\leq~ \langle \psi | T | \psi \rangle ~=~ \langle \psi | (E-V) | \psi \rangle~<~ 0, $$

which is impossible. Here $H=T+V$ is the Hamiltonian operator.