Schwarz inequality for multiple integrals: $left[int_A f(x)g(x) dxright]^2 le int_A f(x)^2 dx cdot int_A g(x)^2 dx$

If $f,g:Ato mathbb{R}$ are integrable, prove the Schwarz inequality

$$left[int_A f(x)g(x) dxright]^2 le int_A f(x)^2 dx cdot
int_A g(x)^2 dx$$

This is that type of question that isn’t easy to think about. I’ve found a solution for one dimension integrals here

But now I’m working on multiple integrals, so I’m not integrating over the real line, but in blocks. $int_A f(x) dx$ is defined as integrable when $underline{int_A} f(x) dx = overline{int_A} f(x) dx$, and these integrals are the sup and inf of those $M_B$ and $m_b$ things.

Does the proof I found apply for this case? It still makes sense for me when I switch $int_a^b$ by $int_A$, because the center of the proof is the equality $(|f|+lambda|g|)^2 = |f|^2 + 2lambda |f||g| + lambda^2 |g|^2$


The integral of a non-negative, integrable function is non-negative. It follows that for any $lambdainmathbb{R}$ we have $int_{A}left(f(x)-lambda g(x)right)^2,dx geq 0$, so the discriminant of the quadratic polynomial

$$ p(lambda) = left(int_A g(x)^2,dxright)lambda^2 -2left(int_A f(x),g(x),dxright)^2 lambda + left(int_A f(x)^2,dxright)$$
is non-positive. Just write down the discriminant in terms of the coefficients and you are done.

Anyway, since in general $L^1(A)notsubseteq L^2(A)$, I believe that the correct assumptions should be about the square-integrability of $f$ and $g$. Or about $f,gin L^1(A)$, plus $mu(A)<+infty$.

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