Functions whose second derivative is of the same sign $$\frac{\mathrm{d^{2}} }{\mathrm{d} x^{2}}\psi=\frac{2m}{\hbar^{2}}\left [ V\left ( x \right ) -E\right ]\psi$$
I must show that $E>V\left(x\right)$ for all $x$ for every normalisable solution to the time-independent Schrödinger equation.
Let's suppose $\psi$ is a function of $x$, and if $\psi(x)>0$ then $\psi''(x)>0$. Similarly, if $\psi(x)<0$ then $\psi''(x)<0$.
A positive derivative indicates an increasing rate of change (convex up and local minima exists) and a negative derivative indicates a decreasing rate of change(concave down and local maxima exists) but why isn't the function able to start and end at $x=0$?
 A: It is not required that $E> V(x)$ for all $x$; only that $E > V(x)$ for some $x$. 
Proposition: If $\psi$ is a normalizable (i.e., square-integrable) solution for the time-independent Schrödinger equation, then $E > V_{\text{min}}$.
Proof: If not, then $V(x) - E$ is positive for all $x$, hence the time-independent Schrödinger equation written in the problem implies $\psi$ and $\psi''$ have the same sign everywhere. Let $g = \psi^2$. Then $$g''(x) = \frac{d^2}{dx^2} \psi(x)^2 = 2(\psi'(x)^2 + \psi(x) \psi''(x)) > 0$$ for all $x$. This implies $g'(x)$ is increasing everywhere, so $$\int_{b}^{N} g(x) \, dx = \int_{b}^{N} \int_{a}^{x} g'(y) \, dy + g(a) \, dx = \int_{b}^{N} \int_{a}^{x} g'(y) \, dy \, dx + (N-b) g(a) $$ $$ \ge \int_{b}^{N} \int_{a}^{x} g'(a) \, dy \, dx + (N-b) g(a) = g'(a) \int_{b}^{N} (x-a) \, dx + (N-b) g(a) = g'(a) \left( \frac{N^2 - b^2}{2} - a(N-b) \right) + (N-b) g(a)$$ for all $a\le b$. If we choose some $a$ such that $g'(a) > 0$, then taking $b>a$ and $N \to \infty$ shows $\psi$ is not normalizable. Such an $a$ must exist since otherwise $g$ is everywhere monotonic (and no monotonic function is $L^1$-integrable). $\blacksquare$
