Young's Inequality | A Geometric Proof of Young's Inequality
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- Опубліковано 24 лют 2020
- #youngsinequality #minkowski #holder
A nice proof of Young's Inequality for Products using Young's Inequality for Increasing Functions.
Typically comes up in lp space work in analysis.
Heavily related to Young's Inequality for inner products, for integral operators, Minkowski's Inequality and Holder's Inequality
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Useful in deriving Hölder’s, nice video Prof!!
I had seen the dry unmotivated algebraic proof of this inequality and it did not really stick for me.
The inverse function integral interpretation way you proved is easier to grasp. Also connection to AM-GM a nice mnemonic. I liked your “hold your head this way” to be aware of inverse function without really graphing it 😀
Do you know much about the mathematician Young?
That graphical approach gives a good sense of what that inequality means, This can also be proved by weighted AM GM Inequality too, Thank you sir
very interesting and nice explained :)
Thanks for sharing this! It helped a lot!
Definitely!
Really nice, this way is very easy
Thank you for the explanation 🤩🤩🤩🤩
Always welcome
Well One question I wanted to ask
Do we study these in higher mathematics as well or are these restricted to olympiads.
Because whenever I searched for Holders or Minkowski or such inequalities for olympiad purposes, I used to see terms like Metric spaces and stuff which I don't know about.
Could u illuminate plz.
Yes these inequalities are very much related to metric spaces, it’s cool stuff!
Desus got fuck smart.
Well this can also be derived directly from weighted am gm with weights 1/p and 1/q which in turn comes from Jensen's.
Nice alternate proof prof.
Thanks Yash
Cool
Thanks Terry!
cool！ a picture proof is 1000 times better than a dry proof using AM GM
Sir why we take here a^p = b^q,
If I'm understanding correctly, equality occurs when b=f(a) and they gives a^p=b^q
hello there, thanks for ur video!
why would you choose the function x^(p-1)? why not x^(p-2) or x^(p-3) or just a random function, containing the exponent p, somehow.
Greetings!
Great question. I think the intuition is that when you integrate x^{p-1} you get a constant times x^p and the inequality in question is related to p-th powers of numbers.
@@ProfOmarMath Thanks a lot Sir, that helped!
i think if you choose p=1.00001 then x^p-2 or x^p-3 will not be increasing function
You want the solution to the integral to be x^p/p, so you take its derivative.
p and q should be > 1 rather than just positive
Yeah Sir it must be.
And Sir your videos are too good! You make maths more attractive to me. And this Young's inequality proof was super cool!
@@charanthiru5761 Thanks! What are your favorite topics in mathematics?
@@ProfOmarMath Never thought that u would reply to me. Blessing.
Calculus, Trig, Vector Algebra Sir.
Hej awesome, I knew something! Thanks for confirming in the comments. I love your stuff btw.
@@Oliver-cn5xx Thanks!
Why a , b 》0
Gets funny with negatives if p and q are odd