| fg | <= 1/2 ( f*f + g*g ) ???

Discussion in 'Undergraduate Math' started by sto, Feb 17, 2011.

  1. sto

    sto Guest

    Let (X,m) me a measure space and f,g be in L2(X,m)

    Presumably it is obvious that for every x in X,

    |f(x)g(x)| <= 1/2 ( f(x)*f(x) + g(x)*g(x) )

    How do you prove this rigorously? I've tried looking at the RHS as the
    area of a rectangle, I've tried the formula for the long side of an
    obtuse triangle, I've tried looking at the diagonals of a parallelogram
    and nothing works.
    Thanks,
    -sto
     
    sto, Feb 17, 2011
    #1
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  2. (f(x) + g(x))^2 >=0

    so

    f(x)f(x) + 2f(x)g(x) + g(x)g(x) >=0

    so

    -2f(x)g(x) <= f(x)f(x) + g(x)g(x).

    so

    f(x)g(x) <= (1/2)(f(x)f(x) + g(x)g(x)).

    Doing the same thing with (f(x)-g(x))^2 gives

    -f(x)g(x) <= (1/2)(f(x)f(x) + g(x)g(x)).

    Since |f(x)g(x)| is either f(x)g(x) or -f(x)g(x), in either case it is
    less than or equal to (1/2)(f(x)f(x)+g(x)g(x)).

    Nothing to do with them being L_2 function, all you need is for the
    values to be real.
     
    Arturo Magidin, Feb 17, 2011
    #2
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  3. Should be

    -f(x)g(x) <= (1/2)(f(x)f(x) + g(x)g(x))
    Should be

    f(x)g(x) <= (1/2)(f(x)f(x) + g(x)g(x)).

    The cumulative effect is the same, though.
     
    Arturo Magidin, Feb 17, 2011
    #3
  4. sto

    Tim Norfolk Guest

    Just to nit-pick, why not just look at (|f|-|g|)^2 >=0?
     
    Tim Norfolk, Feb 17, 2011
    #4
  5. Because I wanted to open the possibility of screwing up the signs? (-;

    Didn't really think about finessing it; if I had, I probably would
    have.
     
    Arturo Magidin, Feb 17, 2011
    #5
  6. sto

    Tim Norfolk Guest

    I just hate prrof by cases, because I all too often forget one.
     
    Tim Norfolk, Feb 17, 2011
    #6
  7. sto

    sto Guest

    Got it, thanks.
     
    sto, Feb 17, 2011
    #7
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