Analysis question regarding properties of the Riemann integral. Consider $f:[0,1]\mapsto\mathbb{R}$ such that $|f|^2 \in R([0,1])\cap C^2([0,1])$ and $f(0) = 0$, $f(1) = 0$. Then prove that
$$
\left( \int_{0}^{1}|f|^2 dt \right)^\frac{1}{2} \leq 2 \left( \int_{0}^{1}\left| \frac{\partial f}{\partial t} \right|^2 dt \right)^\frac{1}{2}
$$
Professor gave a hint and told us to use the fundamental theorem of calculus, but I'm not entirely sure how to start. $R([0,1])$ is the set of Riemann integrable functions over the closed interval $[0,1]$, and $C^2([0,1])$ is the set of all functions which are continuous over the closed interval $[0,1]$ when squared.

We have using FTC and Cauchy-Schwarz
$$|f(x)|^2 = \left||f(0)|^2 + \int_0^x 2f(t)f'(t) \, dt\right| \\= 2\left|\int_0^x f(t)f'(t) \, dt \right| \\ \leqslant 2\left(\int_0^x |f(t)|^2 \, dt \right)^{1/2}\left(\int_0^x |f'(t)|^2 \, dt \right)^{1/2}\\ \leqslant 2\left(\int_0^1 |f(t)|^2 \, dt \right)^{1/2}\left(\int_0^1 |f'(t)|^2 \, dt \right)^{1/2}$$
Now integrate again over $[0,1]$:
$$\int_0^1|f(t)|^2 \, dt \leqslant 2\left(\int_0^1 |f(t)|^2 \, dt \right)^{1/2}\left(\int_0^1 |f'(t)|^2 \, dt \right)^{1/2}\int_0^1\,dt \\ = 
2\left(\int_0^1 |f(t)|^2 \, dt \right)^{1/2}\left(\int_0^1 |f'(t)|^2 \, dt \right)^{1/2}.$$
If the integral of $|f|^2$ is zero then the inequality is trivially true. Otherwise, divide to obtain
$$\left(\int_0^1|f(t)|^2 \, dt\right)^{1/2} \leqslant 2\left(\int_0^1 |f'(t)|^2 \, dt \right)^{1/2}$$
