Let f:[0, ∞) → ℝ be a decreasing function. Assume that |[itex]\int^{\infty}_{0}[/itex]f(x)dx| < ∞ Show that [itex]lim_{x→∞}[/itex]xf(x) = 0. Attempt: By hypothesis f is integrable for every b>0, as is the function x, so that for every such b, xf(x) is integrable on [0,b]. From here I tried passing to sums by considering a partition P such that U(P,xf(x)) - L(P,xf(x)) < ε, but I'm probably barking up the wrong tree there. I also can't see a way to use integration by parts to generate an xf(x) term since I don't have differentiability or continuity of f. I'm pretty much just lost and need a slight nudge in the right direction.
No, integration by parts can't help, for just the reason you mention. lim f(x) as x->infinity has to be zero, right? That should be easy to see. So you can assume lim f(x)>0. Prove it by contradiction. So suppose lim x->infinity x*f(x) is NOT zero. That means there is an e>0 such that for any N>0 there is c such that c*f(c)>e. That should be enough of a slight nudge to get you started.
I thought I was good, but I'm having trouble seeing where the contradiction lies. I expected that maybe I could choose N large so that f(x)<ε, yet x*f(x)>ε, but I can't see how will work. I know that ∫x*f(x) need not converge so there's nothing there. I also looked at the same N large under the same conditions to see if choosing a partition P such that U(P,x*f(x))-L(P,x*f(x))<ε might imply x*f(x)<ε, but that didn't work either. I must be making dunce move and missing something super obvious here. :(
Ok, pick a c such that c*f(c)>e. That means the integral of f(x) from 0 to c is greater than e, right? Now pick a c'>2c such that c'*f(c')>e. Can you get a lower bound for the integral from c to c' of f(x)?
Ah! Got it! I was completely missing out on the fact that f decreasing implies x*f(x) < ∫f dx. Kept focusing too hard on other irrelevant things. Thanks!