# Thread: g,h are bounded

1. ## g,h are bounded

Hello.
I am trying to make the next problem:
If g and h are continuous and real functions and g(x-y)= g(x)g(y) + h(x)h(y) for all x and y real, proove that g and h are bounded.

I have realised that, for example g=cos and h=sin is a possible example.
Also, I have thought to give values to x and y for example: y=0 -->g(x)= g(x)g(0) + h(x)h(0); x=0 --> g(-y)=g(0)g(y) + h(0)h(y); y=x-->g(0)=g^2(x)+h^2(x)=g^2(y)+h^(y) and y=-x --> g(-2y)=g(x)g(-x) + h(x)h(-x) but I don't get any conclusion.
Could you help me?

Thanks for every aportation.

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Ok. I think that as g(0)=g^2(x)+h^2(x) we have that for all x the point (g(x), h(x)) is in the circle of center 0 and radius squared root of g(0) so they are bounded,
is it a good argument?

How could I find examples of g and h differents of "cos" and "sin"??

Thanks again.

2. Originally Posted by Lolyta
Hello.
I am trying to make the next problem:
If g and h are continuous and real functions and g(x-y)= g(x)g(y) + h(x)h(y) for all x and y real, proove that g and h are bounded.

I have realised that, for example g=cos and h=sin is a possible example.
Also, I have thought to give values to x and y for example: y=0 -->g(x)= g(x)g(0) + h(x)h(0); x=0 --> g(-y)=g(0)g(y) + h(0)h(y); y=x-->g(0)=g^2(x)+h^2(x)=g^2(y)+h^(y) and y=-x --> g(-2y)=g(x)g(-x) + h(x)h(-x) but I don't get any conclusion.
Could you help me?

Thanks for every aportation.

---
Ok. I think that as g(0)=g^2(x)+h^2(x) we have that for all x the point (g(x), h(x)) is in the circle of center 0 and radius squared root of g(0) so they are bounded,
is it a good argument?

How could I find examples of g and h differents of "cos" and "sin"??

Thanks again.
As you noted, g^2+h^2 must be a non-negative constant and have value g(0) and g^2(x),h^2(x) <= c for all x. Let d=h(0). You have: |h(x)|=sqrt(c-g^2(x)) where c=g(0). I will assume h is positive for the moment, but the same holds in the opposite case.

Now, g(x-0)=g(x)g(0)+h(x)h(0) which implies g(x)(1-c)=d*h(x) = d*sqrt(c-g^2(x)). Squaring both sides yields g^2(x)(1-c)^2 = d^2(c-g^2(x)), or, g^2(x)(1-2c+c^2+d^2) = d^2c. hence either g is constant, or we must have d^2c=0. But c=0 implies that g(x)=0. Therefore if g is not constant, d=0. So g(x)=cg(x), which means c=1.

Therefore we have g^2(x)+h^2(x)=1. You can now see that g(x) looks like cosine, and h(x) looks like sine. Though they may not be exactly this way, for instance, g(x)=cos(2x) and h(x)=sin(2x) also work.

I'll add that g(x) can be the constant function 1, and h(x) the constant function 0.

3. Originally Posted by daon2
Squaring both sides yields g^2(x)(1-c)^2 = d^2(c-g^2(x)), or, g^2(x)(1-2c+c^2+d^2) = d^2c. hence either g is constant, or we must have d^2c=0. But c=0 implies that g(x)=0. Therefore if g is not constant, d=0. So g(x)=cg(x), which means c=1.
Why do you know that "either g is constant, or we must have d^2c=0" ?

4. Originally Posted by Lolyta
Why do you know that "either g is constant, or we must have d^2c=0"
$g(0)=g^2(x)+h^2(x), \forall x.$ and $g^2(x)+h^2(x)\ge g^2(x)\ge 0$
Thus $\sqrt{g(0)}\ge |g(x)|.$

5. Originally Posted by pka
$g(0)=g^2(x)+h^2(x), \forall x.$ and $g^2(x)+h^2(x)\ge g^2(x)\ge 0$
Thus $\sqrt{g(0)}\ge |g(x)|.$
But... we don't know that g(0) is >0 so... $\sqrt{g(0)}$ could not exist and... if $g(x) \in (0,1)$ then $g^2(x)+h^2(x)\ge g^2(x)\ge 0$ hasn't has to be true because $g^2(x)\le g(x)$

6. Originally Posted by Lolyta
But... we don't know that g(0) is >0 so... $\sqrt{g(0)}$ could not exist and... if $g(x) \in (0,1)$ then $g^2(x)+h^2(x)\ge g^2(x)\ge 0$ hasn't has to be true because $g^2(x)\le g(x)$
You certainly know that $g(0)\ge 0$.
BTW: $\forall a~\&~b,~a^2+b^2\ge a^2.$

7. Originally Posted by pka
You certainly know that $g(0)\ge 0$.
BTW: $\forall a~\&~b,~a^2+b^2\ge a^2.$
Sorry, I feel embarrased. You were absolutely right.

8. .

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