🚀 go-pugleaf

let C = 1/8,
let f be L1 and L2 on [0,\infty),
and let \int_0^\infty f(x) x dx < C \int_0^\infty f(x) dx.

Then
       \int_0^C f(x) (C-x) dx >= \int_C^\infty f(x) (x-C) dx
implies with a slight amount of work
       \int_0^{2 C} f(x) dx >= \int_{2 C}^\infty f(x) dx,
meaning that the median is not more than twice the mean.

Apply Cauchy-Schwartz to f and the characteristic function on [0,2C].
Then,
       \int_0^\infty f^2 >= ( \int_0^{2 C} f )^2 / (2 C)
                         >= ( 1/2 \int_0^\infty f )^2 / (2 C)
                          = ( \int_0^\infty f )^2.

Vincent Broman,  code 572 bayside,                                  =   o     
Naval Command Control and Ocean Surveillance Center, RDT&E Div.   =  _ /- _   
San Diego, CA  92152-5000,  USA                                  =  (_)> (_)  
Internet Email: broman@nosc.mil    Fax: +1 619 553 1635   Phone: +1 619 553 1641