Eigenvalues of $\frac{d^2}{dx^2}$ in $C^2(\mathbb{R})$

Question

Consider the eigenvalue problem \begin{equation} \left\{ \begin{array}{l} \Phi \in C^{2}(\mathbb{R}) \ \text{and bounded }\\ -\Phi^{''}(x)=\lambda\Phi(x), \ x\in \mathbb{R}. \end{array} \right. \end{equation}

and assume that the solution is given by $\Phi(x)=Ae^{i\sqrt{\lambda}x}+Be^{-i\sqrt{\lambda}x}$.

How can one prove that $\lambda$ is real and not negative?

Answer 1

If it is not true that $\lambda \geq 0$, then $\operatorname{Im} \sqrt{\lambda} \neq 0$, i.e. $\sqrt{\lambda} = a + ib$ with $b \neq 0$. Thus $$C\exp(i \sqrt{\lambda} x) = C \exp(iax)\exp(-bx)$$ is unbounded as $x \to \infty$ or as $x \to -\infty$ depending on the sign of $b$. Thus if a solution of the form you propose is bounded, then either $\lambda \geq 0$ or $A = B = 0$.

(Note that it is not possible for one term to sufficiently cancel out the other unless $b = -b$ i.e. $\lambda \geq 0$).

Answer 2

If $\lambda<0,$ then $\sqrt{\lambda}=i \Im{\sqrt{\lambda}} + \Re{\sqrt{\lambda}},$ then $\Phi(x)=Ae^{i\sqrt{\lambda}x}+Be^{-i\sqrt{\lambda}x} $ can be written by $\Phi(x)=Ae^{-\Im{\sqrt{\lambda}}x + i\Re{\sqrt{\lambda}}x }+B e^{-\Im{\sqrt{\lambda}} x+ i\Re{\sqrt{\lambda}}x }.$ But $e^{-\Im{\sqrt{\lambda}}x}$ is not bounded. Thus, $\lambda$ is not negative.