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first show you only need to consider squares of functions as $f\cdot g = \frac{1}{4} \left[(f+g)^2 - (f-g)^2\right]$. Show then that you only need to consider only positive valued functions becuase $f(x)\cdot g(x)=\left|f(x)\right|^2$. then, if $0 \leq f(x) \leq M$ on $\left[a,b\right]$ show that $f^2(x) - f^2(y) \leq 2M \left(\,f(x)-f(y)\right)$. does anyone know how I would answer this ?? |
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This is not a problem I would assign as homework (at least, not without substantial guidance). Rather, it is one of the fundamental results of the subject -- the subject being advanced calculus / elementary real analysis -- and as such I would expect any instructor / textbook to supply a proof. For instance, Rudin's Principles of Mathematical Analysis covers this. For those readers who do not have access to any basic analysis textbooks, see Corollary 8 of http://math.uga.edu/~pete/243integrals2.pdf As Robin says, the result also follows from Lebesgue's criterion of Riemann integrability: now that's something I would leave as an exercise, since finding this short argument on one's own helps to drive home the power of the Lebesgue criterion. |
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This follows from Lebesgue's characterization of Riemann integrable functions as bounded functions continuous outside a set of Lebesgue measure zero. This characterization is usually the swiftest way of deciding on the Riemannn integrability of a function. |
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