All right, I was working on a problem set yesterday, and I came up with the following proof of an absurd conclusion. Even though it’s not directly related to the homework, I was curious as to where the error is. I can’t find it, so I figure I’m being dense and the best thing to do is to put it to y’all.

First, I need some notation. Let f be a real-valued function which is integrable on the closed interval [a, b]. Let f[sub]+/sub be equal to f(x) if f(x) > 0, and 0 otherwise. Let f[sub]–/sub be equal to -f(x) if f(x) < 0, and 0 otherwise. Let J(f) be the integral of f over [a, b]. I would use I(f), but that doesn’t show up well in absolute value expressions, which I will be using extensively.

It follows that f[sub]+/sub > 0, and f[sub]–/sub > 0. Moreover, f(x) = f[sub]+/sub - f[sub]–/sub. These are pretty simple proofs, so I won’t bother including them.

So here’s my fallacious proof:

We consider the quantity |J(f)|. f = f[sub]+[/sub] - f[sub]–[/sub], so |J(f)| = |J(f[sub]+[/sub] - f[sub]–[/sub])|. J(f[sub]+[/sub] - f[sub]–[/sub]) = J(f[sub]+[/sub]) - J(f[sub]–[/sub]), so |J{f}| = |J(f[sub]+[/sub]) - J(f[sub]–[/sub])|. By the triangle inequality, |J(f[sub]+[/sub]) - J(f[sub]–[/sub])| < |J(f[sub]+[/sub])| - |J(f[sub]–[/sub])|. Both J(f[sub]+[/sub]) and J(f[sub]–[/sub]) are non-negative, so |J(f[sub]+[/sub])| - |J(f[sub]–[/sub])| = J(f[sub]+[/sub]) - J(f[sub]–[/sub]). Therefore, we have that |J(f)| < J(f[sub]+[/sub]) - J(f[sub]–[/sub]). But J(f[sub]+[/sub]) - J(f[sub]–[/sub]) = J(f[sub]+[/sub] - f[sub]–[/sub]), which is equal to J(f). Therefore, |J(f)| < J(f).

Clearly, this result is absurd, but I’ll be double-darned if I can spot the error in my logic. Anybody see it?