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

[sto]

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

[Arturo Magidin]

(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

[Arturo Magidin]

On Feb 16, 9:18 pm, Arturo Magidin <magi...@member.ams.org> wrote:

Sigh. Quick correction:

> (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)).

Should be

-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)).

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

[Tim Norfolk]

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

[Arturo Magidin]

On Feb 16, 9:57 pm, Tim Norfolk <timsn...@aol.com> wrote:
> On Feb 16, 10:20 pm, Arturo Magidin <magi...@member.ams.org> wrote:
>
>
>
> > On Feb 16, 9:18 pm, Arturo Magidin <magi...@member.ams.org> wrote:
>
> > Sigh. Quick correction:
>
> > > (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)).
>
> > Should be
>
> > -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)).
>
> > Should be
>
> > f(x)g(x) <= (1/2)(f(x)f(x) + g(x)g(x)).
>
> > The cumulative effect is the same, though.

>

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

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

[Tim Norfolk]

> Arturo Magidin- Hide quoted text -
>
> - Show quoted text -

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

[sto]

Got it, thanks.