[math analysis] sup f< sup g=>∫f^n<∫g^n

  • Thread starter Thread starter mathdunce
  • Start date Start date
  • Tags Tags
    Analysis
mathdunce
Messages
16
Reaction score
0

Homework Statement


1.JPG



Homework Equations


2.JPG



The Attempt at a Solution


3.JPG
 

Attachments

  • 1.JPG
    1.JPG
    7.8 KB · Views: 554
  • 2.JPG
    2.JPG
    13.5 KB · Views: 471
  • 3.JPG
    3.JPG
    25.7 KB · Views: 592
Last edited:
Physics news on Phys.org
We know that

\int_0^1{f^n}\leq (\sup f)^n.

So we need to show that

(\sup f)^n\leq \int_0^1{g^n}

Since sup(f)<sup(g), there exists a neighbourhood ]a,b[ such that

\forall x\in ]a,b[:~\sup(f)&lt;g(x)

Now we can use

\int_0^1{g_n}\geq \int_a^b{g_n}\geq (b-a)\inf_{x\in ]a,b[}{g^n(x)}.

so you must prove now that there exists an n such that

(\sup f)^n&lt;(b-a)\inf_{x\in ]a,b[}{g^n(x)}
 
micromass said:
We know that

\int_0^1{f^n}\leq (\sup f)^n.

So we need to show that

(\sup f)^n\leq \int_0^1{g^n}

Since sup(f)<sup(g), there exists a neighbourhood ]a,b[ such that

\forall x\in ]a,b[:~\sup(f)&lt;g(x)

Now we can use

\int_0^1{g_n}\geq \int_a^b{g_n}\geq (b-a)\inf_{x\in ]a,b[}{g^n(x)}.

so you must prove now that there exists an n such that

(\sup f)^n&lt;(b-a)\inf_{x\in ]a,b[}{g^n(x)}
Hi Micromass. Thank you for your help with https://www.physicsforums.com/showthread.php?t=451462
I now know how to solve the first question, but I still do not know know to link them with e^{mn+c}. I tried the mean value theorem of integral without success. Could you please give me another hint? Thank you!
 
mathdunce said:
Hi Micromass. Thank you for your help with https://www.physicsforums.com/showthread.php?t=451462
I now know how to solve the first question, but I still do not know know to link them with e^{mn+c}. I tried the mean value theorem of integral without success. Could you please give me another hint? Thank you!

Oh, I think I know how to do the second one, too. Thanks. I have not written it down formally yet.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
Back
Top