Prove Schwarz inequality in $R^2$ Can someone please show me how you would prove the following in $R^2$
$\int f(x)* g(x) dx  \leqslant \int f(x)^2 dx * \int g(x)^2 dx  $
starting from 
$\int [\lambda*f(x) - g(x)]^2 dx  \geqslant  0$, where $\lambda$ is real
Also, can someone confirm that $\int f(x)^2 dx = \int f(x) dx \int f(x) dx$
 A: We have that $\int_{a}^{b}\left(\lambda f(x)-g(x)\right)^2dx\ge0$ for all $\lambda$, so 
$\displaystyle\lambda^{2}\left(\int_{a}^{b}(f(x))^2dx\right)-2\lambda\left(\int_{a}^{b}f(x)g(x)dx\right)+\int_{a}^{b}(g(x))^2dx\ge0$ for all $\lambda$.
Therefore $b^2-4ac=\displaystyle4\left(\int_{a}^{b}f(x)g(x)dx\right)^2-4\left(\int_{a}^{b}(f(x))^2dx\right)\left(\int_{a}^{b}(g(x))^2dx\right)\le0$, 
so $\;\;\;\displaystyle\left(\int_{a}^{b}f(x)g(x)dx\right)^2\le\left(\int_{a}^{b}(f(x))^2dx\right)\left(\int_{a}^{b}(g(x))^2dx\right)$.
A: I guess you miss a square in the first inequality. (Tell me if i am wrong)(Since if $f= 1/2, g= 1/2 $ , on the interval [0,1], we would have $1/4 \leqslant 1/4 * 1/4$)
$\int [\lambda*f(x) - g(x)]^2 dx  \geqslant  0$, where $\lambda$ is real.
Then $\int \lambda^2f(x)^2 dx + \int g(x)^2dx \geqslant 2\int \lambda f(x) g(x)$, which is true for all real $\lambda$.
If the right hand side of the inequality is negative, take a negative $\lambda$ so that we can have $\int \lambda^2f(x)^2 dx + \int g(x)^2dx \geqslant 2\int \lambda f(x) g(x) \geqslant 0$.
Then we can square both side
$(\int \lambda^2 f(x)^2 dx)^2 + (\int g(x)^2 dx)^2 + 2 \int g(x)^2 dx * \int \lambda^2 f(x)^2 dx \geqslant 4 \lambda^2 (\int  f(x) g(x))^2  \geqslant 2 \lambda^2 (\int  f(x) g(x))^2 \geqslant 0$.
$2 \int g(x)^2dx * \int \lambda^2 f(x)^2 dx \geqslant 2 \lambda^2 (\int  f(x) g(x))^2$
Thus $ \int g(x)^2dx * \int f(x)^2 dx \geqslant  (\int  f(x) g(x))^2$
