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

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

1. ### stoGuest

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

2. ### Arturo MagidinGuest

(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

3. ### Arturo MagidinGuest

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
4. ### Tim NorfolkGuest

Just to nit-pick, why not just look at (|f|-|g|)^2 >=0?

Tim Norfolk, Feb 17, 2011
5. ### Arturo MagidinGuest

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
6. ### Tim NorfolkGuest

I just hate prrof by cases, because I all too often forget one.

Tim Norfolk, Feb 17, 2011
7. ### stoGuest

Got it, thanks.

sto, Feb 17, 2011