How one can show that $\int _0 ^1 ((h')^2-h^2)dx \ge0 $ for all $h\in C^1[0,1]$ and $h(0) = 0$? How one can show that $\int _0 ^1 ((h')^2-h^2)dx \ge0 $ for all $h\in C^1[0,1]$ and $h(0) = 0$?                                                   

By the fundamental theorem of Calculus:
$$
\left|h(x)\right| = \left|\int_0^x h'(t) \,dt\right| \le \int_0^x \left|h'(t)\right| \,dt \le \int_0^1 \left|h'(x)\right| \,dx
$$
By Cauchy-Schwarz (or Jensen's) inequality:
$$
\left|h(x)\right|^2 \le \left(\int_0^1 \left|h'(x)\right| \,dx\right)^2 \le \int_0^1 \left|h'(x)\right|^2 \,dx
$$
Integrate both sides with respect to $x$ from $0$ to $1$ to get:
$$
\int_0^1 \left|h(x)\right|^2\,dx \le \int_0^1 \left|h'(x)\right|^2 \,dx
$$
