Showing a function is integrable

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SUMMARY

The discussion centers on proving the integrability of the function \( f^2 \) given that \( f \) is a positive, increasing function on the interval [0, 1] with \( f(x) \leq 1 \). It is established that since \( f \) is monotone, \( f^2 \) is also integrable on [0, 1]. Furthermore, the conclusion is drawn that \( \int f^2(x) \, dx \leq \int f(x) \, dx \) due to the properties of increasing functions and the bounded nature of \( f \).

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  • Familiarity with the properties of definite integrals
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Students and educators in calculus, particularly those focusing on real analysis and the properties of integrable functions. This discussion is beneficial for anyone looking to deepen their understanding of function behavior in integration.

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Homework Statement


A function f is positive and increasing on [0, 1]. f(x) \leq 1 \forall x in [0, 1]. Show that f^2 is integrable, and that \int f^2(x) dx \leq \int f(x) dx.


Homework Equations





The Attempt at a Solution


Since f is increasing and positive, it is also monotone. If f(x) is monotone on [a, b], then f(x) is integrable on [a, b]. Also, when x<y, then f(x) < f(y).

f^2(x) - f^2(y) = (f(x) - f(y))(f(x) + (f(y))
Since f is increasing, (f(x) - f(y)) < 0 and (f(x) + f(y)) > 0.

Apparently, the last 2 lines are supposed to show that f^2(x) is increasing, but I don't understand the reasoning.

And if I show that f^2(x) is increasing, then it is also integrable, which solves one part of the question.

Then since f^2(x) \leq f(x) for x>0, and f(x)\leq 1 by the given => \int f^2(x) dx \leq \int f(x) dx, by inequality of functions indicating also the inequality of the integrals. This solves the second part of the question.
 
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Does it have anything to with the fact that since f(x) - f(y) < 0, and f(x) + f(y) > 0, then the quantity becomes (f(x) < f(y))(f(x) < f(y))? Can I multiply through, so that f^2(x) < f^2(y)?
 
Obviously you can, I mean that's the reason for factoring (f(x))^2 - (f(y))^2 in the first place.
 

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