# Product rule in normed space

1. Nov 17, 2012

### cummings12332

1. The problem statement, all variables and given/known data
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)

2. Relevant equations
counterexample , you may wish to examine the case f=g=0 and choose fn=gn for
some piecewise linear functions.

3. The attempt at a solution
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||=max|fn|*max|gn|=max|fn||gn=max|fngn|=||fngn|| and ||f||*||g||=||fg|| ( by definition of norm) so ||fn*gn||->||fg||
i dont know it is right or wrong

and by the (C[0,1],||.||1) i have no idea to get the counterexample

Last edited: Nov 17, 2012
2. Nov 17, 2012

### gopher_p

How would you prove the "product rule" for convergent sequences of real numbers? i.e how do you prove, given $a_n\rightarrow a\in\mathbb{R}$ and $b_n\rightarrow b\in\mathbb{R}$, that $a_nb_n\rightarrow ab$? Notice that $|\cdot|$ is a norm on the real vector space $\mathbb{R}$.

What part of that proof (if any) goes wrong if you try to apply it to the normed vector spaces that you're working with?

3. Nov 18, 2012

### cummings12332

i proved the product rule by an->a bn->b then (an-a)(bn-b)->0 and a(bn-b)->0 b(an-a)->0 i.e. (an-a)(bn-b)+a(bn-b)+b(an-a)=anbn-ab->0 . should i prove that ||fn||*||gn||->||f||*||g|| instead of ||fngn||->||fg||??? but if it is , i dont know what is the differences for the case for index infinite and index 1?????

4. Nov 18, 2012

### HallsofIvy

Staff Emeritus
I don't see anywhere that you are using the difference between those norms and the usual norm on functions.

What is the precise definition of those norms?

5. Nov 18, 2012

### cummings12332

for ||fn||1 that is the sum of |fn| , for ||fn||infinte that is the max of |fn|