$F(x)=int_{-2}^{2} dy f(x,y)$ is an even function, is $G(x)=int_{-2}^{2} dy [f(x,y)]^2$ even?
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!
calculus real-analysis multivariable-calculus even-and-odd-functions
asked Nov 20 at 0:11
bRost03
34319
add a comment |
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!
calculus real-analysis multivariable-calculus even-and-odd-functions
asked Nov 20 at 0:11
bRost03
34319
add a comment |
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!
calculus real-analysis multivariable-calculus even-and-odd-functions
asked Nov 20 at 0:11
bRost03
34319
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!
calculus real-analysis multivariable-calculus even-and-odd-functions
calculus real-analysis multivariable-calculus even-and-odd-functions
asked Nov 20 at 0:11
bRost03
34319
asked Nov 20 at 0:11
bRost03
34319
edited Nov 20 at 2:18
asked Nov 20 at 0:11
bRost03
34319
asked Nov 20 at 0:11
bRost03
34319
asked Nov 20 at 0:11
bRost03
34319
34319
add a comment |
add a comment |
1 Answer
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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 $f(x,y)$.
answered Nov 20 at 3:10
bRost03
34319
add a comment |
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1 Answer
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active
oldest
votes
1 Answer
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oldest
votes
active
oldest
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active
oldest
votes
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 $f(x,y)$.
answered Nov 20 at 3:10
bRost03
34319
add a comment |
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 $f(x,y)$.
answered Nov 20 at 3:10
bRost03
34319
add a comment |
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 $f(x,y)$.
answered Nov 20 at 3:10
bRost03
34319
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 $f(x,y)$.
answered Nov 20 at 3:10
bRost03
34319
answered Nov 20 at 3:10
bRost03
34319
answered Nov 20 at 3:10
bRost03
34319
answered Nov 20 at 3:10
bRost03
34319
34319
add a comment |
add a comment |
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