$F(x)=\int_{-2}^{2} dy f(x,y)$ is an even function, is $G(x)=\int_{-2}^{2} dy [f(x,y)]^2$ even?

Question

I have a real valued function in two real variables $f(x,y)$ which is essentially a black box. The only thing I really know is that $$ F(x)=\int_{-2}^{2} f(x,y) dy $$ is even and that $$ \int_{-\infty}^{\infty} F(x) dx=1 $$ Can I conclude that $$ G(x)=\int_{-2}^{2} \left[f(x,y)\right]^2 dy $$ is also even? I looked for a counterexample, but couldn't find one. I intuitively feel like this should be true, but am struggling to make that more rigorous. Any ideas? Thanks!

Answer 1

I asked the question but after doing some more work I have found a counterexample.

$$f(x,y)=\Theta (x) \left(x y^2-\frac{4 x}{3}\right)$$

Where $\Theta(x)$ is the step function, then

$$ F(x)=\int_{-2}^2 \Theta (x) \left(x y^2-\frac{4 x}{3}\right) \, dy=0 $$

Is an even function, but

$$ G(x)=\int_{-2}^2 \left(\Theta (x) \left(x y^2-\frac{4 x}{3}\right)\right)^2 \, dy= \frac{256 x^2 \Theta (x)}{45} $$

Is not. However I believe that if $f(x,y)=f(-x,\pm y)$ then $G(x)=G(-x)$. I'd be glad to hear if anyone can give more general conditions on $f(x,y)$.