Given a potential $V(x)$ that is constant in time, to find the wavefunctions that solve the Shrodinger's equation, first we look for stationary states, by assuming that $$\Psi(x,t) = \psi(x) \phi(t)$$ for some functions $\psi$ (a function of space alone) and $\phi$ (a function of time alone). By substituting this into Shrodinger's equation we get: $$i \hbar \frac{1}{\phi} \frac{d \phi}{dt} = -\frac{\hbar^2}{2m} \frac{1}{\psi}\frac{d^2\psi}{dx^2} + V$$ Note that since the left hand side is a function in time alone, and the right hand side a function in space alone, both sides must be equal to some constant in space and time $E$. So we have for the left hand side: $$i \hbar \frac{1}{\phi} \frac{d \phi}{dt} = E$$ $$\frac{d\phi}{dt} = -\frac{iE}{\hbar}\phi$$ $$\phi(t) = e^{-iEt/\hbar}$$ (The constant doesn't matter since we care about $\Psi = \psi\phi$ and we can renormalise later). Now the right hand side gives us the time-independent Shrodinger equation: $$-\frac{\hbar^2}{2m}\frac{d^2\psi}{dx^2} + V\psi = E\psi$$ By solving this equation and finding $\Psi = \psi\phi$, we find our stationary states corresponding to our potential function. These are called stationary states because every expectation value is constant in time. Moreover, the energy of each stationary state is precisely given by $E$ (the standard deviation is zero in this measurement). For the piece de resistance; it turns out that every solution to the Shrodinger equation is a linear combination of stationary states.

Problem 2.1

Prove the following three theorems:

For normalizable solutions, the separation constant $E$ is real.
The time-independent wave function $\psi(x)$ can always be taken to be real. This doesn't mean that every solution to the time-independent Schrodinger equation is real; what it says is that if you've got one that is not, it can always be expressed as a liner combination of solutions (with the same energy) that are.
If $V(x)$ is an even function, then $\psi(x)$ can always be taken to be either even or odd (in an analogous manner to the second part of this question).

Solution:

Suppose that the separation constant is not real, i.e. $E = E_0 + i\Gamma$ where $E_0,\Gamma \in \mathbb{R}$. Then for the time-dependent portion of the Shrodinger equation: $$i\hbar \frac{d\phi}{dt} = (E_0 + i\Gamma)\phi$$ $$\phi(t) = e^{\Gamma t/\hbar}e^{-iE_0t/\hbar}$$ So we've picked up an additional $e^{\Gamma t/\hbar}$ term that grows with time. This cannot be true for normalizable functions (as they are supposed to stay normalizable, but this implies the function grows and grows with time)
Consider the time-independent Shrodinger equation: $$-\frac{\hbar^2}{2m}\frac{d^2\psi}{dx^2} + V\psi = E\psi$$ Notice that if $\psi$ is a solution to this equation (for some $E$), then $\psi^*$ is also a solution for the same $E$. This implies $(\psi+\psi^*)$ and $i(\psi-\psi^*)$ are both also solutions to the equation for the same $E$. Notice however that both of these solutions are purely real (!!) This implies: $$\psi = \frac{1}{2}(\psi + \psi^*) + \frac{1}{2}i^3 (i(\psi - \psi^*))$$ is a linear combination of real solutions to the time-independent Shrodinger equation, hence arbitrary solution $\psi$ can always been expressed as a linear combination of real solutions.
Consider the time-independent Shrodinger equation $$-\frac{\hbar^2}{2m}\frac{d^2\psi}{dx^2} + V\psi = E\psi$$ Suppose $V$ is an even function, and that $psi(x)$ is one solution to the equation. Notice then that: $$-\frac{\hbar^2}{2m}\frac{d^2\psi(-x)}{dx^2} + V(x)\psi(-x) = -\frac{\hbar^2}{2m}\frac{d^2\psi(-x)}{dx^2} + V(-x)\psi(-x)=E\psi(-x)$$ Therefore $\psi(-x)$ is also a solution for the same $E$. Notice however that $(\psi(x)+\psi(-x))$ is an even function, and that $(\psi(x)-\psi(-x))$ is an odd function. Hence $$\psi(x) = \frac{1}{2}(\psi(x)+\psi(-x)) + \frac{1}{2}(\psi(x)-\psi(-x))$$ for arbitrary $\psi(x)$ expresses it as a linear combination of even and odd functions.

Solution:
Consider the time-independent Shrodinger equation: $$-\frac{\hbar^2}{2m}\frac{d^2\psi}{dx^2} + V\psi = E\psi$$ Rearranging, we have: $$\frac{d^2\psi}{dx^2} = \frac{2m}{\hbar^2}[V(x) - E]\psi$$ If $E < V_{\text{min}}$, then $\psi$ and its second derivative always have the same sign. This function cannot possibly be normalized, since if $\psi$ is ever positive, then it will have positive curvature and curve away from the axis, and become increasing positive. Similarly, if $\psi$ is ever negative, it will have negative curvature and curve away from the axis, and become increasingly negative.