the product rule fn->f , gn->g implies fngn->fg true in the normed

vector space (C[0,1],||.||) depends on the the norm||.||. Give a proof or a

counterexample for (C[0,1],||.||infinite),(C[0,1].||.||1)

HINTS:For

counterexample , you may wish to examine the case f=g=0 and choose fn=gn for

some piecewise linear functions.

can someone help me to solve out this

question????? many thanks

what i did for ||.|| infinite is that ||fn||->||f||, ||gn||->||g|| ,then (||fn||-||f||）*（||gn||-||g||)->0 ,||g||(||fn||-||f||)->0,||f||(||gn||-||g||)->0

then get (||fn||-||f||）*（||gn||-||g||)+||g||(||fn||-||f||)+||f||(||gn||-||g||)=||fn||*||gn||-||f||*||g||->0

therefore ||fn||*||gn||->||f||*||g||

for it is infinite so we get ||fn|||*||gn||=||fngn|| and ||f||*||g||=||fg|| ( by definition of norm) so ||fn*gn||->||fg||

i dont know it is right or wrong

and by the ||.||1 i have no idea to get the counterexample