Verify: For every Darboux Integrable function

Question

How to Verify:

For every Darboux Integrable function $f: [0,2] \to R$,

$$(\int_{0}^{2}f(x) dx )^2 \leq 2\int_{0}^{2}(f(x) dx)^2 $$

Answer 1

You could look up any standard proof of the Cauchy-Schwarz inequality. Most begin with noting that $$ 0\le \|f-c·g\|^2=\|f\|^2-2c·\langle f,g\rangle+c^2·\|g\|^2. $$

In this case $$ 0\le \int_0^2(f(x)-c)^2\,dx=\int_0^2f(x)^2dx-2c·\int_0^2 f(x)dx+2c^2. $$

Now compute the minimal point of the parabola in $c$. Multiply by 2 and complete the square in $2c$, $$ 0\le 2\int_0^2f(x)^2dx-\left(\int_0^2 f(x)dx\right)^2+\left(2c-\int_0^2 f(x)dx\right)^2 $$ There is a value of $c$ that turns the last term to zero, and the remainder is the requested inequality.