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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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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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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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