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