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QUestion:

let $f(x),g(x)$ is continuous on $R$,and such $$f(x-y)=f(x)f(y)-g(x)g(y)$$ and $f(0)=1$

show that: for any $x\in R$, have $g(x)=-g(-x)$

my try: let $x=y=0$,then $$f(0)-[f(0)]^2=g(0)g(0)\Longrightarrow g(0)=0$$

and let $x=0$, note $f(0)=1,g(0)=0$,so $$f(-y)=f(y)$$ since $$g(x)g(y)=f(x)f(y)-f(x-y)$$ so $$g(-x)g(y)=f(-x)f(y)-f(-x-y)=f(x)f(y)-f(x+y)$$ since $$f(x+y)=f(x)f(-y)-g(x)g(-y)\Longrightarrow g(x)g(-y)=f(x)f(y)-f(x+y)$$ so $$g(-x)g(y)=g(x)g(-y)$$

But I can't have $g(x)=-g(-x)$

math110
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  • Well you can already set $y:=-x$ to get that their squares are equal. And then, I suppose the continuity hypothesis must be used to prove they have same sign. – xavierm02 Sep 05 '14 at 14:46

4 Answers4

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Since the proof comes directly from Kelenner's comment, I post this answer in community wiki, so as to not earn undeserved points.

We assume $f(0)=1$ and

$$\forall x,y \in \Bbb R, f(x-y)=f(x)f(y)-g(x)g(y)$$

Notice that we use nowhere the assumption that $f$ or $g$ is continuous.

Then, for all $x\in\Bbb R$,

$$1=f(0)=f(x-x)=f(x)^2-g(x)^2$$

And

$$f(2x)=f(x)f(-x)-g(x)g(-x)$$

Also, $\forall x,y\in \Bbb R$,

$$f(y-x)=f(x)f(y)-g(x)g(y)=f(x-y)$$

Hence $\forall x\in\Bbb R$, $f(x)=f(-x)$, that is, $f$ is even.

Therefore, $f(-x-y)=f(x+y)$, or, $\forall x,y\in \Bbb R$,

$$f(-x)f(y)-g(-x)g(y)=f(x)f(-y)-g(x)g(-y)$$

Hence, $\forall x,y\in \Bbb R$, $g(-x)g(y)=g(x)g(-y)$.

Now, let's suppose $g(y_0)\neq0$ for some $y_0 \in \Bbb R$.

We have then, $\forall x\in\Bbb R$, such that $g(x)\neq0$, $$\frac{g(-x)}{g(x)}=\frac{g(-y_0)}{g(y_0)}=a$$

Where $a$ is a constant real number.

And if $g(x)=0$, then from $g(-x)g(y_0)=g(x)g(-y_0)=0$ and $g(y_0)\neq0$, we have that $g(-x)=0$ too.

That is, $\forall x\in\Bbb R$, $g(-x)=ag(x)$.

But then $\forall x\in\Bbb R$, $g(x)=ag(-x)=a^2g(x)$, and since $g(y_0)\neq0$, we can simplify and $a^2=1$, or $a=\pm1$. Thus the function $g$ is either even or odd.

If $g$ is even, then $\forall x\in\Bbb R$,

$$f(2x)=f(x)^2-g(x)^2=1$$

So $f$ is constant ($=1$).

But then $\forall x\in\Bbb R$, $g(x)^2=f(x)^2-f(2x)=0$.

Hence, if $g$ is even, it's the null function.

But that means $g$ is always odd, since either it is odd, either it is null (thus also odd).

Jean-Claude Arbaut
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Hint (using that $f(-x)=f(x)$ and $g(x)g(-y)=g(-x)g(y)$ the results form the question above).

$x$ should be arbitrary in the whole calculation.

With $\forall x,y\in R$ $$g(x)g(-y)=g(-x)g(y)\tag{i}$$ set $y\to -x$ then $$g(x)^2=g(-x)^2\tag{ii}$$

Because of $g$ is real $$g(-x)=\pm g(x)\tag{iii}$$ If there is any $x_+$ so that $g(-x_+)=g(x_+)$, then with (i) $\qquad\forall y\in R: g(-y)=g(y)$.

Or there is a $x_-$ with $g(x_-)=-g(x_-)$, then with (i) $\qquad\forall y\in R:g(-y)=-g(y)$

Thus the sign in $(iii)$ must be the same for all values of $x$.

Now starting with $$f(x-y)=f(x)f(y)-g(x)g(y) \tag{1}$$ Set $y\to x$ $$1=f(0)=f(x)^2-g(x)^2\tag{2}$$

Now look at $$f(x+y)=f(x)f(y)-g(x)g(-y)\tag{3}$$ Assume that $\forall x_+\in R:g(-x_+)=+g(x_+)$

Then set $y\to x_+$ $$f(2x_+)=f(x_+)^2-g(x_+)^2\tag{4}$$ Comparing $(2)$ and $(4)$ you see then $\forall x_+\in R:f(2x_+)=1$.

With $\forall x_+\in R:f(x_+)=1$ and $(2)$ you get $\forall x_+\in R$ $$g(-x_+)=g(x_+)=0=-g(x_+) \tag{5}$$ or the assumption was wrong.

In summary you get $$\forall x\in R:g(-x)=-g(x) \tag{6}$$ Did I make any bad mistakes or is something missing?

Matthias
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  • I see one problem: you assume that either $g(x)=g(-x)$ for all $x$, either $g(x)=-g(-x)$ for all $x$. But it could also "change sign". – Jean-Claude Arbaut Sep 05 '14 at 15:02
  • I assumed this and lead this to a contradiction, if $g(x)$ is not null everywhere. Thus $g(-x)=-g(x)$. Or what do you mean? – Matthias Sep 05 '14 at 15:04
  • But there are functions that are neither odd or even, you can't simply assume it's not possible. – Jean-Claude Arbaut Sep 05 '14 at 15:05
  • I agree with Jean-Claude Arbaut. To resolve the problem, you can suppose that $g$ is not the zero function; then there exists $y_0$ such that $g(y_0)\not=0$. You get $g(-x)=ag(x)$ for all $x$, $a=g(-y_0)/g(y_0) $; replace $x$ by $-x$ and simplify. – Kelenner Sep 05 '14 at 15:05
  • You also can use $g(-x)g(y)=g(x)g(-y) $ and set $x=-y$ as xaviermo2 said. Then you have $g(x)^2=g^2(-x)$ I used that implicit, if g is real, what I assumed by the tag of real-Analysis. – Matthias Sep 05 '14 at 15:07
  • @Matthias Yes, and that does not tell you that $g$ must be even or odd. – Jean-Claude Arbaut Sep 05 '14 at 15:08
  • @Jean-Claude Arbaut: We have $1=f(x)^2-g(x)^2$ for all $x$ ($x=y$ in the functional equation) – Kelenner Sep 05 '14 at 15:09
  • @Jean-ClaudeArbaut But compare the second and the 4th equation. – Matthias Sep 05 '14 at 15:09
  • @Kelenner Ah, ok for this point. But the other remains ;-) – Jean-Claude Arbaut Sep 05 '14 at 15:10
  • @Matthias: The problem is that in general $u(x)v(x)=0$ does not imply $u=0$ or $v=0$. (you try to use $(g(x)-g(-x))(g(x)+g(-x))=0$) – Kelenner Sep 05 '14 at 15:10
  • Actually, it implies that, for all $x$, either $u(x)=0$ or $v(x)=0$, not that $u(x)=0, \forall x$ of $v(x)=0,\forall x$. Problem with ordering of quantifier. It works with polynomials though (using an argument on roots) – Jean-Claude Arbaut Sep 05 '14 at 15:12
  • Sorry I am to stupid to see where I used $u(x)v(x)=0$ implies $v=0$ or $u=0$. Tell where – Matthias Sep 05 '14 at 15:14
  • @Jean-Claude Arbaut: Using $g(x)=ag(-x)$ as in my previous comment gives $a^2=1$, hence $g(-x)=g(x)$ or $g(-x)=-g(x)$. – Kelenner Sep 05 '14 at 15:14
  • @Kelenner Maybe I'm a bit slow today, but I don't see why the same $a$ is guaranteed to work for all $x$ ;-) – Jean-Claude Arbaut Sep 05 '14 at 15:20
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    You get $f(2x) = 1$ only for the $x$ that satisfied $g(x) = g(-x)$. – J. J. Sep 05 '14 at 15:20
  • @Matthias The problem is you write "Assume that $g(−x)=+g(x)$", does it mean "for all $x$"? If yes, how can you infer that, if it is wrong, then "$g(−x)=−g(x)$ for all $x$"? There is a problem with quantifiers. That something is true for one $x$ does not imply it's true for all. – Jean-Claude Arbaut Sep 05 '14 at 15:21
  • @Jean-Claude Arbaut: I have fixed a $y_0$ such that $g(y_0)\not =0$. As for all $x,y$ we have $g(x)g(-y)=g(y)g(-x)$, we get $g(-x)=ag(x)$ for all $x$ with $a=g(-y_0)/g(y_0)$. – Kelenner Sep 05 '14 at 15:22
  • @Kelenner Oh, yes!!! Very nice! With this it's ok then. (but only with this ;-)) – Jean-Claude Arbaut Sep 05 '14 at 15:23
  • $x$ can be arbitrary at every equation I wrote. – Matthias Sep 05 '14 at 15:24
  • @Matthias No, it can't. Write quantifiers explicitly, and you will see what you write is wrong: as J.J. explains, that $g(x)^2=g(-x)^2$ does not imply that $g$ is either even or odd. The problem is you don't write quantifiers, so what works for a priori only one $x$, you assume is works for all $x$. What do you exactly mean by "Assume that $g(−x)=+g(x)$"? Then only thing you can infer from $g(x)^2=g(-x)^2$ is that, for all $x$, either $g(x)=-g(-x)$ of $g(x)=g(-x)$. But for only this $x$. – Jean-Claude Arbaut Sep 05 '14 at 15:26
  • Try this with $g(x)=|\sin(x)|$ for $x<0$ and $g(x)=\sin(x)$ for $x\geq0$. You have then certainly that $g(x)^2=g(-x)^2$ for all $x$. So what? – Jean-Claude Arbaut Sep 05 '14 at 15:29
  • How do you conclude that $g(x_+)=0$? And then how do you conclude that $\forall x\neq x_+$, $g(-x)=-g(x)$? Where did you get that $f(x_+)^2=1$? – Jean-Claude Arbaut Sep 05 '14 at 15:44
  • @Jean-ClaudeArbaut Maybe with the use of contintuity and that $f(0)=1>0$ you can show that $\forall x: g(-x)=-g(x)$ or $\forall x: g(-x)=g(x)$ with the first $\forall x:g(x)=0=-g(-x)$ this is still odd – Matthias Sep 05 '14 at 16:06
  • Kelenner gave a nice solution above, to overcome these problems. – Jean-Claude Arbaut Sep 05 '14 at 16:36
  • @Jean-ClaudeArbaut should I add this part of proof in the answer? – Matthias Sep 05 '14 at 16:40
  • @Jean-ClaudeArbaut is the whole proof now ok? – Matthias Sep 05 '14 at 16:48
  • I have given the answer Kelenner probably had in mind, see below. – Jean-Claude Arbaut Sep 05 '14 at 17:07
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I like these problems, so let me give it a try and write it down neatly.

Since the right hand side is invariant under the swapping $x$ and $y$ we have $f(-t)=f(t)$. Applying the identity with $-x$ and $-y$, using that $f$ is even, gives that $g(-x)g(-y) = g(x)g(y)$, for all $x$ and $y$. (1) Setting $x=y$ implies that $g(-x)\in \{ g(x), -g(x)\}$ for all x.

If $g\not\equiv 0$, then take any $y$ such that $g(y)\neq 0$. Then either $g(y)=g(-y)$ or $g(y)=-g(-y)$, and substituting and dividing by $g(y)$ in equation (1) shows that $g$ is either even or odd. (Or identically zero but that's ok because both even and odd.)

If $g$ is even then applying the first identity with $x$ and $-y$, using that $f$ and $g$ are both even, results in $f(x-y)=f(x+y)$, thus $f$ would be a constant namely $f\equiv 1$. Contradiction because then $g(x) g(y) = 0$ for all $x$ and $y$ , thus $g\equiv 0$ but we already dismissed that case.

Myself
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First note that by using $f(x)=f(-x)$ $$ f(0) = f(x-x) = f(x)^2- g(x)^2 \\ f(0) = f(-x-(-x))=f(x)^2-g(-x)^2 \\ g(x)^2 = g(-x)^2 $$ Thus let one define into two sets (note an element may be in both) $$ \Gamma^+=\{x \in \mathbb{R} : g(x) = g(-x) \} \\ \Gamma^-=\{x \in \mathbb{R} : g(x) = -g(-x) \} \\ $$ Clearly $\mathbb{R}=\Gamma^+ \cup \Gamma^-$, such that $g$ is even for $x\in\Gamma^+$, and odd for $x\in\Gamma^-$.

Consider $x\in\Gamma^+$, one has $$ f(x-y)=f(x)f(-y)-g(x)g(-y) \\ \therefore f(x-y)=f(x+y)\Rightarrow f(x-x)=f(x+x) \Rightarrow f(x)=1 \:\forall x\textrm{ as }f(0)=1 \\ \Rightarrow g(x)=0 \:\forall x \in\Gamma^+ $$ Thus $x\in\Gamma^+\Rightarrow g(x)=0 \Rightarrow g(x)=-g(-x)\Rightarrow x\in\Gamma^-$. Thus $\mathbb{R}=\Gamma^-$.

destroyer
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