analisis funcional

sup[ 0 [< x=y [< 1 ]( ∬ d[x]d[y] ) = 1 & x=1 & y=1
sup[ 0 [< x=y [< 1 ]( ∬ [xy] d[x]d[y] ) = (1/4) & x=1 & y=1


med[ 0 [< x=y [< 1 ]( ∬ d[x]d[y] ) = (1/2) & x=(1/2^{(1/2)}) & y=(1/2^{(1/2)})
med[ 0 [< x=y [< 1 ]( ∬ [xy] d[x]d[y] ) = (1/8) & x=(1/2^{(1/4)}) & y=(1/2^{(1/4)})


sup[ 0 [< x=y [< 1 ]( ∫ [ (1+(-x)) ] d[y] ) = (1/4) & x=(1/2) & y=(1/2)
med[ 0 [< x=y [< 1 ]( ∫ [ (1+(-x)) ] d[y] ) = (1/8) & x=(1/2)( 1+(1/2)^{(1/2)} ) & y=(1/2)( 1+(1/2)^{(1/2)} )
x^{2}+(-x)+(1/8)=0