f,g in L2 => fg in L2 ??

[sto]

Let (X,_S_,m) be a measure space and L2(X,m) be the space of real valued
functions square integrable on X.

If f,g in L2, then how does the inequality

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

imply that fg in L2?

Clearly the RHS is integrable and finite, in which case, since integration
is monotonic, | fg | is integrable and therefore in L1. But what does that
say about fg being in L2?

Thanks
-sto

[sto]

sto <s...@invalid.net> wrote:
> If f,g in L2, then how does the inequality

This means f in L2 and g in L2.

In other words, if f and g are functions in L2, then how does this imply
the function fg is also in L2?

[W^3]

In article
<327551462319682117.8...@news.east.earthlink.net>,
sto <s...@invalid.net> wrote:

fg need not be in L^2. Let f(x) = g(x) = x^{-1/4} on (0,1) for example.

[sto]

I don't understand this counterexample.
The square of x^{-1/4} is x^{-1/2}, which has antiderivative 2x^{1/2},
which in turn integrates to 2 on (0,1) so it *is* in L2.

[W^3]

In article
<1748465756319757355....@news.east.earthlink.net>,
sto <s...@invalid.net> wrote:

f(x)g(x) = x^{-1/2}, which is not in L^2(0,1).

[sto]

OK I get it. You were saying f(x) = x^{-1/4} in L2(0,1), g(x) =
x^{-1/4} in L2(0,1), but fg(x) = x^{-1/2} *not* in L2(0,1)

-thanks