Recent content by kilimanjaro

Yes, f(x) = 0 is in L^1. The issue is whether f_n converges to f in the L^1 norm.

If a function f is in L^2([a,b]), then it is in L^1([a,b]) by the Cauchy-Schwarz inequality (just consider the L^1 norm of f as being the inner product of |f| and 1). They key point here is that [a,b] has finite measure; the argument wouldn't work on something like the real line, and specific...

Well, let me start with a bit of background info. My transcript is an absolute mess, I have dropped out twice (one withdrawal, and one semester of all F's). So basically I am finishing a 4 year degree in 5. Since my whole dropout phase I have been at 3 schools total, a community college, a local...