Showing a function is integrable

[gbean]

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.

 

[gbean]

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)?

 

[snipez90]

Obviously you can, I mean that's the reason for factoring $(f(x))^2 - (f(y))^2$ in the first place.