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Young's Inequality | A Geometric Proof of Young's Inequality
Рет қаралды 11,233
Күн бұрын#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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@MyOneFiftiethOfADollar 2 жыл бұрын
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?
@yashvardhan2093 3 жыл бұрын
Useful in deriving Hölder’s, nice video Prof!!
@shrivardhank.2418 2 жыл бұрын
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
@AxiomTutor 2 жыл бұрын
Desus got fuck smart.
@yashvardhan6521 3 жыл бұрын
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.
@ProfOmarMath 3 жыл бұрын
Thanks Yash
@yeast4529 7 ай бұрын
Really nice, this way is very easy
@Szynkaa 2 жыл бұрын
very interesting and nice explained :)
@ProfOmarMath 4 жыл бұрын
p and q should be > 1 rather than just positive
@charanthiru5761 4 жыл бұрын
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!
@ProfOmarMath 4 жыл бұрын
@@charanthiru5761 Thanks! What are your favorite topics in mathematics?
@charanthiru5761 4 жыл бұрын
@@ProfOmarMath Never thought that u would reply to me. Blessing. Calculus, Trig, Vector Algebra Sir.
@Oliver-cn5xx 4 жыл бұрын
Hej awesome, I knew something! Thanks for confirming in the comments. I love your stuff btw.
@ProfOmarMath 4 жыл бұрын
@@Oliver-cn5xx Thanks!
@annaluiza3592 3 жыл бұрын
Thanks for sharing this! It helped a lot!
@ProfOmarMath 3 жыл бұрын
Definitely!
@yashvardhan6521 3 жыл бұрын
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.
@ProfOmarMath 3 жыл бұрын
Yes these inequalities are very much related to metric spaces, it’s cool stuff!
@ailemac9421 3 жыл бұрын
Thank you for the explanation 🤩🤩🤩🤩
@ProfOmarMath 3 жыл бұрын
Always welcome
@pipilu3055 Жыл бұрын
cool！ a picture proof is 1000 times better than a dry proof using AM GM
@mayurjawane9463 4 жыл бұрын
Sir why we take here a^p = b^q,
@ProfOmarMath 4 жыл бұрын
If I'm understanding correctly, equality occurs when b=f(a) and they gives a^p=b^q
@imenlakrout6982 3 жыл бұрын
Why a , b 》0
@ProfOmarMath 3 жыл бұрын
Gets funny with negatives if p and q are odd
@terryendicott2939 3 жыл бұрын
Cool
@ProfOmarMath 3 жыл бұрын
Thanks Terry!
@davie-uz1dl 4 жыл бұрын
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!
@ProfOmarMath 4 жыл бұрын
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.
@davie-uz1dl 4 жыл бұрын
@@ProfOmarMath Thanks a lot Sir, that helped!
@mathmaticalproblemandsolution 2 жыл бұрын
i think if you choose p=1.00001 then x^p-2 or x^p-3 will not be increasing function
@violetasuklevska9074 Жыл бұрын
You want the solution to the integral to be x^p/p, so you take its derivative.
Mohamed Omar
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