Let $a,b$ be elements in a $C^\ast$-algebra $A$, such that $$a^\ast ab^\ast b=b^\ast ba^\ast a=0$$ $$a^\ast abb^\ast=bb^\ast a^\ast a=0$$ $$aa^\ast b^\ast b=b^\ast b aa^\ast =0$$ $$aa^\ast bb^\ast=bb^\ast aa^\ast=0$$ Why is $ab=ba=a^\ast b=ab^\ast=0$?
I tried to do different things, for example I started with: $$abb^\ast a^\ast=ab(ab)^\ast\ge 0.$$ It is $bb^\ast a^\ast a=0$, which implies $abb^\ast a^\ast a=0$. But this try is pointless.
Or $aa^\ast b^\ast b=aa^\ast bb^\ast$, which implies $aa^\ast (b^\ast b-bb^\ast)=0, $ but it doesn't help.
Maybe I have to use a little bit theory for example continuous functional calculus, but I really have no idea. Today I'm having a bad day :( Could you help me? Regards
From $a^*ab^*b=0$ we get $$0=a^*ab^*ba^*a=(ba^*a)^*ba^*a, $$ so $ba^*a =0$. But then $ba^*ab^*=0$, which implies $ab^*=ba^*=0$
The others are similar.