INEQUALITY'S BLOG

Update inequality

15 Comments »

  1. 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

  2. yes !!
    we have \displaystyle \frac{1}{p}+\frac{1}{q}=\frac{1}{r}\Leftrightarrow \frac{1}{\frac{p}{r}}+\frac{1}{\frac{q}{r}}=1
    then let \displaystyle k=\frac{p}{r}; k'=\frac{q}{r} then use Holder’s inequality.

    Comment by vankhea — October 3, 2011 @ 3:01 am

  3. 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

  4. first solution: it not difficult you just use Holder’s 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

  5. 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

  6. 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

  7. Never mind i will solution it in next timeπŸ˜€

    Comment by vankhea — October 14, 2011 @ 4:51 am

  8. Thanks I could really solve it

    Comment by aya hussein — October 15, 2011 @ 6:32 pm

  9. Oh really good.

    Comment by vankhea — October 16, 2011 @ 12:47 am

  10. 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

  11. sorry i don’t know too. but u can search that books from ebook.

    Comment by vankhea — October 23, 2011 @ 11:27 am

  12. 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

  13. sorry i can’t.

    Comment by vankhea — December 5, 2011 @ 7:17 pm

  14. 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


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