A question on positive distributions.

Question

How do do we show that is $\int_{\Omega}|f(x)|=0$ if and only if $\int_{\Omega}|f^{+}(x)|=0$ and $\int_{\Omega}|f^{-}(x)|=0$.

It is known that $f(x)$ has a unique decomposition where in it equals $f(x)= f^{+}(x)-f^{-}(x)$. I know that if there is a positive distribution T then $\int f^{-}(x) dx $ is 0. Really appreciate the help.

Answer 1

Denote $f^+$ and $f^-$ the positive and negative part of $f$, that is : $f^+=\frac{|f|+f}{2}$ and $f^-=\frac{|f|-f}{2}$

Then $f=f^+-f^-$and $|f|=f^++f^-.$

So $\int_{\Omega}|f(x)|dx=0\Longleftrightarrow \int_{\Omega}f^+(x)dx +\int_{\Omega}f^-(x) dx=0$.

Since both $\int_{\Omega}f^+(x) dx$ and $\int_{\Omega}f^-(x) dx$ are non negative (recall that $f^+$ and $f^-$ are both non negative functions), they both must be equal to zero.

Answer 2

$$X=X^{+} - X^{-}$$

$$|X|=X^{+} + X^{-}$$

$$X^{+} \geq 0 \hspace{1cm} $X^{-} \geq 0 $$

so

$$0=\int_{\Omega} |X| dx=\int_{\Omega} X^{+} dx + \int_{\Omega} X^{-} dx\Rightarrow 0=\int_{\Omega} X^{+} dx = \int_{\Omega} X^{-} dx$$