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Homework Help: Young's Inequality alternative proof

  1. Oct 16, 2011 #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.
     
  2. jcsd
  3. Oct 18, 2011 #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
     
  4. Oct 18, 2011 #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
     
  5. Oct 18, 2011 #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.
     
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