-1
$\begingroup$

first show you only need to consider squares of functions as f.g = 1/4 [(f+g)sqr - (f-g)sqr]. show then that you only need to consider only positive valued functions becuase f(x).g(x)=|f(x)|sqr. then , if 0 <=f(x) <= M on [a,b] show that f sqr(x) - f sqr(y) <= 2M (f(x)-f(y)).

does anyone know how i would answer this ??

$\endgroup$
2
  • $\begingroup$ If this is a homework problem, then -- as stated in the FAQ mathoverflow.net/faq#whatnot -- your question would be better suited to one of the sites mentioned there. $\endgroup$
    – Yemon Choi
    Sep 29, 2010 at 7:58
  • 1
    $\begingroup$ This question has been reasked on MSE, so could be closed here without loss. $\endgroup$ Sep 29, 2010 at 8:07

2 Answers 2

2
$\begingroup$

It follows from Lebesgue's characterization of Riemann integrable functions as bounded functions continuous outside a set of Lebesgue measure zero.

$\endgroup$
2
  • $\begingroup$ thanks for your advice, but is there a simpler approach because i am only a second year student and we have not covered Lebesgue's characterization of Riemann integrable functions at all. $\endgroup$
    – sam
    Sep 29, 2010 at 7:53
  • 1
    $\begingroup$ So this was a homework question? $\endgroup$ Sep 29, 2010 at 18:23
1
$\begingroup$

If $f$ and $g$ are Riemann integrable over the interval $[a,b]$ then there is an $M$ such that $|f|$ and $|g|$ are both $\le M$ on $[a,b]$. The Riemann integrability of $f g$ then immediately follows from the inequality $$|f(x)g(x)-f(x')g(x')|\le |f(x)-f(x')||g(x)|+|f(x')||g(x)-g(x')|$$ $$\le M(|f(x)-f(x')| +|g(x)-g(x')|) $$ for all $x, x'\in [a,b]$.

$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.