let p>1 ,q>1 such that 1/p+1/q=1/r
and let {ai},{bi} be real or coplex numbers then
|summation ai ^r bi^r |^1/r <= (summation |ai|^p )^1/p (summation |bi|^q)^1/q
i want to prove this
what can i do?
Comment by aya hussein — October 2, 2011 @ 10:50 pm
yes !!
we have
then let then use Holder’s inequality.
thanxs
if i want to prove |summation ai^s bi^s ci ^s|^1/s <= (summation |ai|^p)^1/p*(summation |bi|^q)^1/q*(summation |ci|^r)^1/r
where 1/p+1/q+1/r=1/s can you help me
Comment by aya hussein — October 8, 2011 @ 7:05 pm
first solution: it not difficult you just use Holderβs inequality with 1/a1+1/a2+1/a3=1 ; with a1=p/s ; a2=q/s ; a3=r/s
remember that |summation ai^s bi^s ci ^s|^1/s<=(summation |ai^sbi^sci^s|)^1/s.
thank you for your efforts but
i still have a problem
i just reach for this step
summation |ai ^s bi^s|<=(SUMMATION|ai|^p)^s/p * (SUMMATION|bi|^(qr/q+r))^(s(q+r)/qr)
summation|bi^s ci^s|<=(summation|bi|^(pq/p+q)^(s(p+q)/pq) * (summation|ci|^r)^s/r
how can i complete
Sorry to harass you
Comment by aya hussein — October 13, 2011 @ 8:35 pm
forgive me , can i ask you for a good website or book explaining banach space and banach algebra, normed space and inner product .in a good way
thank you very much
Comment by aya hussein — October 22, 2011 @ 11:54 am
sorry i don’t know too. but u can search that books from ebook.
if i want to show that l^infinitly is not closed in the space s of all sequences i search to find sequence of l ^infinity convergent to seq ins but not in l^infinitly , can you help me? thanks
Comment by aya hussein — December 3, 2011 @ 9:40 pm
i found sequence of sequences in l^infinity (1,0,0,…..)
(1,2,0,0,…………..)
and so on(1,1/2,…,1/n,0,0,0,0,0) it converge to (1,2,3,4,…………..)not in l^infinity
but found in s
if you want to know the answer
Comment by aya hussein — December 6, 2011 @ 2:58 pm
let p>1 ,q>1 such that 1/p+1/q=1/r
and let {ai},{bi} be real or coplex numbers then
|summation ai ^r bi^r |^1/r <= (summation |ai|^p )^1/p (summation |bi|^q)^1/q
i want to prove this
what can i do?
Comment by aya hussein — October 2, 2011 @ 10:50 pm
yes !!
we have
then let then use Holder’s inequality.
Comment by vankhea — October 3, 2011 @ 3:01 am
thanxs
if i want to prove |summation ai^s bi^s ci ^s|^1/s <= (summation |ai|^p)^1/p*(summation |bi|^q)^1/q*(summation |ci|^r)^1/r
where 1/p+1/q+1/r=1/s can you help me
Comment by aya hussein — October 8, 2011 @ 7:05 pm
first solution: it not difficult you just use Holderβs inequality with 1/a1+1/a2+1/a3=1 ; with a1=p/s ; a2=q/s ; a3=r/s
remember that |summation ai^s bi^s ci ^s|^1/s<=(summation |ai^sbi^sci^s|)^1/s.
Comment by vankhea — October 9, 2011 @ 1:54 am
second solution: use inequality 04 van khea click link below
https://vankheakh.wordpress.com/category/van-khea-04/
Comment by vankhea — October 9, 2011 @ 1:56 am
u should learn more about inequality in some link below
http://rattanakrith.blogspot.com/2011/06/van-kheas-inequality-and-applications.html
http://mathbookee.blogspot.com/2011/03/van-khea-inequality-and-applications.html
https://vankheakh.wordpress.com/2011/04/02/van-kheas-inequality-and-application/
Comment by vankhea — October 9, 2011 @ 2:01 am
thank you for your efforts but
i still have a problem
i just reach for this step
summation |ai ^s bi^s|<=(SUMMATION|ai|^p)^s/p * (SUMMATION|bi|^(qr/q+r))^(s(q+r)/qr)
summation|bi^s ci^s|<=(summation|bi|^(pq/p+q)^(s(p+q)/pq) * (summation|ci|^r)^s/r
how can i complete
Sorry to harass you
Comment by aya hussein — October 13, 2011 @ 8:35 pm
Never mind i will solution it in next time π
Comment by vankhea — October 14, 2011 @ 4:51 am
Thanks I could really solve it
Comment by aya hussein — October 15, 2011 @ 6:32 pm
Oh really good.
Comment by vankhea — October 16, 2011 @ 12:47 am
forgive me , can i ask you for a good website or book explaining banach space and banach algebra, normed space and inner product .in a good way
thank you very much
Comment by aya hussein — October 22, 2011 @ 11:54 am
sorry i don’t know too. but u can search that books from ebook.
Comment by vankhea — October 23, 2011 @ 11:27 am
if i want to show that l^infinitly is not closed in the space s of all sequences i search to find sequence of l ^infinity convergent to seq ins but not in l^infinitly , can you help me? thanks
Comment by aya hussein — December 3, 2011 @ 9:40 pm
sorry i can’t.
Comment by vankhea — December 5, 2011 @ 7:17 pm
i found sequence of sequences in l^infinity (1,0,0,…..)
(1,2,0,0,…………..)
and so on(1,1/2,…,1/n,0,0,0,0,0) it converge to (1,2,3,4,…………..)not in l^infinity
but found in s
if you want to know the answer
Comment by aya hussein — December 6, 2011 @ 2:58 pm