I am interested in the following inequality: $$\left (\displaystyle\int_0^1 u_1(x)v_1(x)+u_2(x)v_2(x)\ dx\right )^2+\left (\displaystyle\int_0^1 u_2(x)v_1(x)-u_1(x)v_2(x)\ dx\right )^2\leq \left ( \displaystyle\int_0^1 u_1^2(x)+u_2^2(x)\ dx\right )\cdot\left ( \displaystyle\int_0^1 v_1^2(x)+v_2^2(x)\ dx\right )$$
$\bullet$ I proved that $LHS\leq 2\cdot RHS$ using Cauchy inequality, but $2$ is not the best posible constant.
$\bullet$ Also, if one of the four function is $0$ or two of them are equal then the inequality follows with ease via Cauchy Inequality.
I wonder if this inequality holds anytime.
Best Answer
With $u = u_1 + iu_2$ and $v = v_1 + iv_2$ we have $$ u \overline v = \bigl( u_1 v_1 + u_2 v_2 \bigr) + i \bigl( u_2 v_1 - u_1 v_2 \bigr) \, , \\ |u|^2 = u_1^2 + u_2^2 \, ,\\ |v|^2 = v_1^2 + v_2^2 \, , $$ so that your inequality is equivalent to $$ \left | \int_0^1 u(x) \overline{v(x)}\, dx\right|^2 \le \int_0^1 |u(x)|^2 \, dx \cdot \int_0^1 |v(x)|^2 \, dx \, . $$ That is the Cauchy-Schwarz inequality for complex-valued functions.