I have an engineering background, but not deep mathematics background. So, I am having difficulty proving an equation regarding density function as shown in the following problem, and I hope to get some help.
- Problem statement: Given densities $f_1$ and $f_2$, set $g=f_1-f_2$. Then, $c=||g^+||=||g^-||=\frac{1}{2}||g||$ where $||\cdot||$ is L$1$-norm and $c > 0$.
For your reference, density function $f$ is $f \geq 0$, $||f||=1$, and $f \in L^1$ where $L^1$ is L$1$ space corresponding to measure space $(X,\mathcal{A},\mu)$, i.e., all real-valued measurable functions $f:X\rightarrow R$ s.t. $||f|| = \int_X |f(x)| \mu(dx) < \infty$.
- My tries: I tried the followings, but I am not sure if it is correct.
Since $f_1 \in L^1$ and $f_2 \in L^1$, $g=f_1-f_2 \in L^1$, and, if we assume $||g^+||=||g^-||=c$ with $c > 0$, \begin{align} 2c=||g^+|| + ||g^-|| &= ||(f_1-f_2)^+|| + ||(f_1-f_2)^-||\\ &= ||(f_1-f_2)^+ + (f_1-f_2)^-||\\ &= ||g|| \end{align} where $f^+ := \max(0,f)$ and $f^- := \max(0,-f)$, and as a result, $f=f^+-f^-$. Also, one more useful property regarding L$1$ norm is that, if $f_1>0$, $f_2>0$, and $f_1,f_2 \in L^1$, then $||f_1 + f_2|| = ||f_1|| + ||f_2||$.
Thank you for your help in advance.
To prove $\|g^+\|=\|g^-\|,$ we first use the fact that $\|f_1\|=\|f_2\|=1$ and that $f_1,f_2\ge0$ to get \begin{align} 0=\|f_1\|-\|f_2\|&=\int_X |f_1| \ \mu(dx) - \int_X |f_2| \ \mu(dx) \\ &=\int_X \left(|f_1|-|f_2|\right) \ \mu(dx) \\ &= \int_X \left(f_1-f_2\right) \ \mu(dx). \end{align} But $$\int_X \left(f_1-f_2\right) \ \mu(dx)=\int_X (f_1-f_2)^+ \ \mu(dx)-\int_X (f_1-f_2)^- \ \mu(dx)=0,$$ which implies that $$ \int_X (f_1-f_2)^+ \ \mu(dx)=\int_X (f_1-f_2)^- \ \mu(dx).$$ Since $(f_1-f_2)^+$ and $(f_1-f_2)^-$ are non-negative functions, we have that $\|g^+\|=\|g^-\|$.
The rest of the proof looks correct.