This question caught my eye from "How to integrate it" by Sean M. Stewart. I have attempted it for fun and am now stuck.
Let $f$ and $g$ be continuous bounded functions on some interval $[a,b]$ such that $\int_a^b|f(x)-g(x)|dx=0$. Show that $\int_a^b|f(x)-g(x)|^2dx=0$.
Note: $0\le |\int_a^bf(x)-g(x)dx |\le\int_a^b|f(x)-g(x)|dx=0$
We conclude $|\int_a^bf(x)-g(x)dx |=0$
Multiply both sides by $|\int_a^bf(x)-g(x)dx |$ giving $|\int_a^bf(x)-g(x)dx |^2=0$
This is where I'm stuck. Specifically how do I get both the square and the modulus back past the pesky $\int dx$
Possibly $|\left(\int_a^bf(x)-g(x)dx\right)^2|=0$ leading to $0=|\left(\int_a^bf(x)-g(x)dx\right)^2|\le?$
I feel I'm missing something with obvious with modulus but it is escaping me. Of course I could be on the wrong track altogether.
Side question: the afore mention book only has solutions to selected problems so if anyone knew of a full solution set somewhere that'd greatly help.
Thanks in advance for your help.
I now realise I had discarded $f(x)=g(x)$ because I thinking was thinking (incorrectly) of $\int_a^b f(x)-g(x)dx=0$ whereby I imagined a situation such as $f(x)=-x$ and $g(x)=x$ over an interval such as $[-1,1]$ I have accepted the answer that includes the case for discontinuous functions but appreciate all other efforts.