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A Theorem is 'obvious' when one does not see an immediate obstruction (for instance a counter-example). Of course it may be true or false, depending on how you are lucky or not. An obvious true theorem whose proof is notoriously difficult is the existence of non-trivial solutions to linear PDEs $P(i\nabla_x)u=f$ for constant coefficients operators (Malgrange-Ehrenpreis theorem). I don't mean elliptic, hyperbolic, parabolic PDEs, or PDEs of principal type. No, just PDEs. It is not only true but somehow accurate, because it becomes false when the coefficients are non constant, even with analytic coefficients (H. Levy H. Lewy counter-example).

Dick: At first glance, the Fourier transform reduces the question to the resolution of an algebraic equation $P(\xi)\hat u(\xi)=\hat f(\xi)$. The difficulty is whether $\xi\mapsto\hat f(\xi)/P(\xi)$ is the Fourier transform of a distribution. Because $P$ may vanish, and $P^{-1}(0)$ can be quite singular, this is not a piece of cake. Malgrange had to prove his division theorem to solve it.

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A Theorem is 'obvious' when one does not see an immediate obstruction (for instance a counter-example). Of course it may be true or false, depending on how you are lucky or not. An obvious true theorem whose proof is notoriously difficult is the existence of non-trivial solutions to linear PDEs for constant coefficients operators. I don't mean elliptic, hyperbolic, parabolic PDEs, or PDEs of principal type. No, just PDEs. It is not only true but somehow accurate, because it becomes false when the coefficients are non constant, even with analytic coefficients (H. Levy counter-example).