Hello, Really good videos. I have a question: in 8:57 you do ai = (xi + yi)^(p - 1). I do not understand why is this possible, because I think is necessary to show that (xi + yi)^(p - 1) belongs to a lq space in order to say that the Holder's inequality is valid. Sorry if this is a silly question, I'm not a matematician. Thanks in advice!
I thought the same. Until I realized that if |ai|^p is a member of p, |ai|^(p-1) is a member of Lq. You can turn any Lp member to an Lq member: if a is a member of Lp, then we know the sum of |ai|^p is finite. So let's define b to be the series of |ai|^(p-1), then the sum of ||ai|^(p-1)|^(p/(p-1) i.e. to the power of q is finite, as it is equal to the sum in the case of a. Now, this means b is a member of Lq.
Very thorough! One thing I find is that you voice every single tiny bit of notation as you write it. You're writing it! We can see it! So rather than saying exactly what you're writing, maybe say something more illuminating, a rephrase of what you're writing perhaps.
It was wonderful.Like you said: Anyone from anywhere can learn anything... Thank you from Turkey...
Great proof. Straight-forward explanation. I highly appreciate the whole list.
Thanks! Very helpful
Excellent
well done! Thanks!
Thanks 👍
excellent
Well explained.
Thank you very much. Only small misspelled at 21:53 -- you have lost in the denominator "p" in elements of the summ.
Hello,
Really good videos. I have a question: in 8:57 you do ai = (xi + yi)^(p - 1). I do not understand why is this possible, because I think is necessary to show that (xi + yi)^(p - 1) belongs to a lq space in order to say that the Holder's inequality is valid.
Sorry if this is a silly question, I'm not a matematician.
Thanks in advice!
I thought the same. Until I realized that if |ai|^p is a member of p, |ai|^(p-1) is a member of Lq. You can turn any Lp member to an Lq member: if a is a member of Lp, then we know the sum of |ai|^p is finite. So let's define b to be the series of |ai|^(p-1), then the sum of ||ai|^(p-1)|^(p/(p-1) i.e. to the power of q is finite, as it is equal to the sum in the case of a. Now, this means b is a member of Lq.
nice! thank you!
Why repeat same stuff 5 times?
you missed out a ^p on the fraction you wrote at 19:15 . Still an excellent video as per.
Very thorough! One thing I find is that you voice every single tiny bit of notation as you write it. You're writing it! We can see it! So rather than saying exactly what you're writing, maybe say something more illuminating, a rephrase of what you're writing perhaps.
Everything great. But the writing of Left parentheses as a straight line is really annoying.