Young's Inequality alternative proof

  • #1
Young's Inequality can be restated as:

s^(x)t^(1-x)<=xs + (1-x)t where s,t>=0 and 0<x<1.

Basically i've been asked to prove this. I've been fiddling about with it for a couple of hours
to no avail.

I've tried to substitute t=e^u and s=e^v and then use partial differentiation w.r.t to v on st, but i'm not getting the required form. (I can't assume that exp is a convex function - otherwise it follows trivially)

Thanks in advance.
 

Answers and Replies

  • #2
I don't know how to prove it yet, but my 2 cents would be:
[tex](\frac{s}{t})^x t \leq sx + (1-x)t[/tex]
[tex](\frac{s}{t})^x \leq \frac{s}{t} x + 1-x [/tex]
[tex] a^x \leq a x + 1 - x [/tex]
for [itex]a > 0 [/itex]
This way, you have one variable less to worry about
 
  • #3
Actually, think about it, consider the function
[tex] f(x)=ax + 1 - x - a^x [/tex]
[tex] f(0)=1-1=0 [/tex]
[tex] f(1) = a + 1 - 1 - a = 0 [/tex]
[tex] f'(x_0)= a - 1 - \ln a a^{x_0} = 0 [/tex]
[tex] a^{x_0} = \frac{a - 1}{\ln a} [/tex]
Now assume [itex]a>1[/itex]. Then if you can show that the rhs is smaller than a, you have shown that [itex] 0 < x_0 < 1 [/itex]. so
[tex] a-1 < \ln a a [/tex]
[tex] 1 - \frac{1}{a} < \ln a [/tex]
I think you can find this property somewhere.
[tex] f''(x)=- (\ln a) ^2 a^{x}[/tex] which is smaller than 0 everywhere, particularly at [itex]x_0[/itex].
So for a > 1 you have already proven everything. For a < 1, you have to show that the rhs is larger than a. Let [itex] a = 1/b [/itex] with [itex] b > 1 [/itex].
[tex] \frac{\frac{1}{b}-1}{\ln \frac{1}{b}} > \frac{1}{b} [/tex]
[tex] \frac{\frac{1}{b}-1}{-\ln b} > \frac{1}{b} [/tex]
[tex] 1 - b < \ln b [/tex]
since b > 1, this should hold trivially.
Anyhow, it all comes down to [itex] 1 - \frac{1}{a} < \ln a [/itex]. You should look it up (and make sure I haven't made any mistakes) lol
 
  • #4
[tex] f(a)=\ln a a +1 - a \geq 0[/tex]
For [itex] a = 1 [/itex] you have f(1)=0, so it's correct.
[tex] f'(a)= \frac{a}{a} + \ln a - 1 = \ln a [/tex]
which is always greater than 0, so you have a monotonically growing function.
 

Related Threads on Young's Inequality alternative proof

  • Last Post
Replies
1
Views
1K
  • Last Post
Replies
8
Views
1K
  • Last Post
Replies
10
Views
2K
  • Last Post
Replies
8
Views
808
  • Last Post
Replies
6
Views
1K
  • Last Post
Replies
4
Views
2K
  • Last Post
Replies
1
Views
1K
  • Last Post
Replies
2
Views
846
  • Last Post
Replies
3
Views
975
  • Last Post
Replies
7
Views
5K
Top