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Prove that $$ \lambda _{1}\lambda _{2}^{*}\varphi _{1}\varphi _{2}^{*}+\lambda _{1}^{*}\lambda _{2}\varphi _{1}^{*}\varphi _{2} \leq \left | \lambda _{1} \right |\left | \lambda _{2} \right |\left \{ \left | \varphi _{1} \right |^{2}+\left | \varphi _{2} \right |^{2} \right \} $$ where all symbols are complex numbers.

I encountered this while trying to prove that the set of all square integrable functions form a vector space.

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This relation does not make sense for arbitrary complex numbers on the lhs since the field of complex numbers is not ordered. –  André Nov 14 '12 at 14:19
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You're wrong, André. Both LHS and RHS are real. –  Luboš Motl Nov 14 '12 at 14:30
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@André: The left hand side is always real. –  Arnold Neumaier Nov 14 '12 at 14:31
    
Ups... Sorry, you are of course right ;) –  André Nov 14 '12 at 14:51
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2 Answers

OP's inequality (v3) is obvious if either $\lambda_1=0$ or $\lambda_2=0$, so we may assume that $\lambda_1\neq 0$ and $\lambda_2\neq 0$.

Define $$\phi_1:=\sqrt{\frac{\lambda_1\lambda_2^*}{|\lambda_1\lambda_2|}}\varphi_1,$$

and

$$\phi_2:=\sqrt{\frac{\lambda_1^*\lambda_2}{|\lambda_1\lambda_2|}}\varphi_2.$$

Then OP's inequality becomes

$$2{\rm Re}(\phi_1\phi_2^*) \leq |\phi_1|^2 + |\phi_2|^2, $$

or equivalently

$$ |\phi_1-\phi_2 |^2 \geq 0, $$

which is true.

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Thanks , Can it also be proven using the Schwarz inequality? –  ahmed Nov 14 '12 at 16:19
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The vector space property yes, but using it for what you made of it would not be natural. –  Arnold Neumaier Nov 14 '12 at 16:57
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Define $\vec{\phi}:=(\phi_1,\phi_2)\in\mathbb{C}^2$, and $\vec{\psi}:=(\phi_2,\phi_1)\in\mathbb{C}^2$. Then the ineq. $|2{\rm Re}(\phi_1\phi_2^*)|\leq |\phi_1|^2 + |\phi_2|^2$ is the Cauchy-Schwarz ineq. $|\langle\vec{\phi},\vec{\psi}\rangle| \leq ||\vec{\phi}||_2 ~||\vec{\psi}||_2$. –  Qmechanic Nov 14 '12 at 17:11
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The formula is positive homogeneous in the $\lambda_j~$. Thus it is enough to prove the result for $|\lambda_j|=1~$ for $j=1,2~$. For this case, the result follows from $|\lambda_1\phi_1-\lambda_2\phi_2|^2\ge 0~$ by using the definition $|x|^2=x^*x~$ and expanding the brackets.

Thus the result holds without any restriction.

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