In Tao's textbook An Epsilon of Room I, the exercise 1.4.6 claims that it is possible to reduce to the case when $f,g$ are non-negative in order to prove that $$\|f+g\|_p^p+\|f-g\|_p^p\ge(\|f\|_p+\|g\|_p)^p+|\|f\|_p-\|g\|_p|^p$$ for all $p\in[1,2]$ and for every $f,g\in L^p(X,\mathcal{X},\mu)$ the Lebesgue space of $p$-integrable complex-valued functions on any measure space $(X,\mathcal{X},\mu)$.
To begin with, I simply can't see why it is possible to reduce to the real case. From the real case to the non-negative case, I'm ok.
Can you help me with this?
This is self-study of Hanner inequalities, not homework.
Thanks.
It suffices to show $$ \|f+g\|_p^p+\|f-g\|_p^p=\int|f+g|^p+\int|f-g|^p\ge\int||f|+|g||^p+\int||f|-|g||^p $$ which can be shown if for $A,B\in \mathbb C$, $$ |A+B|^p+|A-B|^p\ge||A|+|B||^p+||A|-|B||^p. $$ Since the both side are invarient under rotation, one may assume $A=a$ is real and $B=be^{\text{i}\theta}$, $\theta\in(-\pi/2,\pi/2]$. So $$ |A+B|^p+|A-B|^p=(a^2+b^2+2ab\cos\theta)^{p/2}+(a^2+b^2-2ab\cos\theta)^{p/2}:=H(\theta). $$ Some calculation is needed here to show that $H(\theta)$ attain its maximum at $\theta=0$, this gives $$ |A+B|^p+|A-B|^p\ge H(0)=|a+b|^p+|a-b|^p=||A|+|B||^p+||A|-|B||^p. $$ Complete.