Verified answer

if f_n -> f uniformly on D, Definition set of the f_n's and f,

for any e > 0 there is n0(e) integer (function of e ONLY)

so that

for any n € N

if n >= n0 then for any x € D |f_n(x) - f(x0) | < e

now, if you observe the proposition here above (a bit complex)

as it's true for any x € D

then (same beginning hypothesis)

for any e > 0 there is n0(e) integer (function of e ONLY)

so that

for any n >= n0

we have : ---> Sup |f_n(x) - f(x0) | < e

(as it is true for any x € D)

therefore

M_n ---> 0

the reciprocal goes the same way,

as the core and key argument is concerned :

if M_n ---> 0 then

for any e > 0 there exits n0(e) so that

for any n >= n0

Sup_(for any x € D) | f_n(x) - f(x) | < e

therefore

for any x € D, | f_n(x) - f(x) | < e

and finally

f_n converges uniformly towards f on D

PS: the "Sup_norm" is therefore called

norme de la convergence uniforme

(in French... yes, i am French... nobody's perfect...)

so i will try the translation:

"uniform convergence norm".

et voilà !!

hope it'll help, Felix !!

(French is a bit based on Latin ! Ciao !)

"sup" is ------??

(waited)....time to abandon the question.